Motion In A Line (One Dimension)
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Exercises For Motion In A Straight Line
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Vertical Shots And Free Fall
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Projectile Motion
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[{"Name":"Motion In A Line (One Dimension)","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Instantaneous Velocity and Average Velocity","Duration":"4m 59s","ChapterTopicVideoID":8934,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:04.290","Text":"Hello. In this lecture we\u0027ll be talking about the relationship"},{"Start":"00:04.290 ","End":"00:07.965","Text":"between speed, position and acceleration."},{"Start":"00:07.965 ","End":"00:11.220","Text":"First, we\u0027re going to talk about 1-dimensional movement."},{"Start":"00:11.220 ","End":"00:14.985","Text":"That means we\u0027ll be talking about the x-axis and ignoring the y-axis for now,"},{"Start":"00:14.985 ","End":"00:17.640","Text":"you can move positively or negatively back-and-forth in"},{"Start":"00:17.640 ","End":"00:21.910","Text":"the x-axis and even go past an initial point and return to another point."},{"Start":"00:22.220 ","End":"00:25.995","Text":"Let\u0027s assume that I have x as a function of t,"},{"Start":"00:25.995 ","End":"00:28.740","Text":"meaning my position as a function of time."},{"Start":"00:28.740 ","End":"00:31.740","Text":"In this example, we\u0027ll say that x is a function of t,"},{"Start":"00:31.740 ","End":"00:36.550","Text":"is equal to t^3 power plus 5."},{"Start":"00:37.460 ","End":"00:41.970","Text":"Now I want to find the velocity which is defined as the derivative"},{"Start":"00:41.970 ","End":"00:47.285","Text":"of the location over the time dx over dt."},{"Start":"00:47.285 ","End":"00:50.795","Text":"In physics, this is denoted by the symbol x dot."},{"Start":"00:50.795 ","End":"00:55.280","Text":"In our example, it will be set as equal to 3t^2,"},{"Start":"00:55.280 ","End":"00:57.155","Text":"our 5 from above drops off."},{"Start":"00:57.155 ","End":"01:00.270","Text":"We ended up with 3t^2 plus 0."},{"Start":"01:01.750 ","End":"01:04.190","Text":"This equation, this function,"},{"Start":"01:04.190 ","End":"01:07.610","Text":"which we\u0027ll call the function of t, as you can see there,"},{"Start":"01:07.610 ","End":"01:09.635","Text":"gives us the instantaneous velocity,"},{"Start":"01:09.635 ","End":"01:11.990","Text":"which is called the instantaneous velocity because it"},{"Start":"01:11.990 ","End":"01:15.460","Text":"gives us the speed or velocity at any given moment."},{"Start":"01:15.460 ","End":"01:20.660","Text":"If I set t is equal to 1 or 2 or 3 or any other number,"},{"Start":"01:20.660 ","End":"01:23.005","Text":"I can actually find the velocity."},{"Start":"01:23.005 ","End":"01:25.789","Text":"We can also speak about the average velocity."},{"Start":"01:25.789 ","End":"01:29.900","Text":"This value is symbolized with a line over the top of your value."},{"Start":"01:29.900 ","End":"01:31.280","Text":"If it had an arrow in the end,"},{"Start":"01:31.280 ","End":"01:32.690","Text":"it would be a vector like this."},{"Start":"01:32.690 ","End":"01:34.685","Text":"It\u0027s just your average velocity."},{"Start":"01:34.685 ","End":"01:36.965","Text":"If I want to calculate average velocity,"},{"Start":"01:36.965 ","End":"01:41.510","Text":"the equation I use is Delta x over Delta t."},{"Start":"01:41.510 ","End":"01:44.690","Text":"The difference between instantaneous velocity and"},{"Start":"01:44.690 ","End":"01:48.634","Text":"average velocity is with instantaneous velocity, you\u0027re dividing derivatives."},{"Start":"01:48.634 ","End":"01:51.395","Text":"Whereas with average velocity you\u0027re dividing the difference,"},{"Start":"01:51.395 ","End":"01:52.925","Text":"the change over time."},{"Start":"01:52.925 ","End":"01:55.220","Text":"For the average velocity,"},{"Start":"01:55.220 ","End":"01:59.600","Text":"you\u0027ll take your terminal velocity minus"},{"Start":"01:59.600 ","End":"02:05.020","Text":"your initial velocity divided by your terminal time and your initial time."},{"Start":"02:05.020 ","End":"02:09.440","Text":"For example, if I want to find my average velocity over"},{"Start":"02:09.440 ","End":"02:13.585","Text":"the course of time when t=1 and t=3,"},{"Start":"02:13.585 ","End":"02:17.930","Text":"meaning my average velocity over that 2 second interval."},{"Start":"02:17.930 ","End":"02:21.575","Text":"I need to find the value of x,"},{"Start":"02:21.575 ","End":"02:24.935","Text":"meaning my velocity when t=3,"},{"Start":"02:24.935 ","End":"02:28.350","Text":"that\u0027s my terminal velocity."},{"Start":"02:28.350 ","End":"02:33.380","Text":"I need to subtract it from that my initial velocity when t=1."},{"Start":"02:33.380 ","End":"02:36.230","Text":"We then divide that by the difference in time,"},{"Start":"02:36.230 ","End":"02:38.105","Text":"which in this case is 3 minus 1."},{"Start":"02:38.105 ","End":"02:39.410","Text":"If we want to solve this example,"},{"Start":"02:39.410 ","End":"02:41.435","Text":"we look for the x value when t=3,"},{"Start":"02:41.435 ","End":"02:47.860","Text":"3^3 is 27 plus 5 gives us a terminal x value of 32."},{"Start":"02:47.860 ","End":"02:51.800","Text":"We subtract from that the value of x when t=1,"},{"Start":"02:51.800 ","End":"02:55.040","Text":"which is 1 of a third or 1 plus 5, which equals 6,"},{"Start":"02:55.040 ","End":"02:57.275","Text":"we put that over 2 3 minus 1,"},{"Start":"02:57.275 ","End":"02:59.855","Text":"and we end up with the value of 13."},{"Start":"02:59.855 ","End":"03:04.115","Text":"If in this example we\u0027re measuring our distance in terms of meters,"},{"Start":"03:04.115 ","End":"03:06.665","Text":"then we\u0027d write this as 13 meters over seconds,"},{"Start":"03:06.665 ","End":"03:09.485","Text":"or 13 meters per second as our average velocity."},{"Start":"03:09.485 ","End":"03:12.425","Text":"Of course, if we\u0027re measuring a different interval of time, say,"},{"Start":"03:12.425 ","End":"03:16.490","Text":"between t1 and t4 or t2 and t5,"},{"Start":"03:16.490 ","End":"03:20.990","Text":"we\u0027d end up with a different set of values in a different average velocity."},{"Start":"03:20.990 ","End":"03:24.800","Text":"Thus far we\u0027ve covered instantaneous velocity and average velocity."},{"Start":"03:24.800 ","End":"03:28.190","Text":"Notice the instantaneous velocity equation deals in derivatives,"},{"Start":"03:28.190 ","End":"03:31.010","Text":"and the average velocity equation is"},{"Start":"03:31.010 ","End":"03:34.130","Text":"a near copy that deals in the difference between the 2,"},{"Start":"03:34.130 ","End":"03:36.680","Text":"your terminal and your initial velocity."},{"Start":"03:36.680 ","End":"03:39.230","Text":"Again, note for your average velocity that it"},{"Start":"03:39.230 ","End":"03:41.420","Text":"doesn\u0027t really matter how you get to the end point,"},{"Start":"03:41.420 ","End":"03:42.665","Text":"it just matters that you get there."},{"Start":"03:42.665 ","End":"03:47.600","Text":"You can go slowly along 1 line or you can go past your end point and return."},{"Start":"03:47.600 ","End":"03:50.495","Text":"It doesn\u0027t matter, it\u0027s still logging your average velocity."},{"Start":"03:50.495 ","End":"03:52.550","Text":"What matters here is your initial point and"},{"Start":"03:52.550 ","End":"03:55.655","Text":"your endpoint because that\u0027s what the average velocity is measuring."},{"Start":"03:55.655 ","End":"03:58.100","Text":"Now there is 1 case when"},{"Start":"03:58.100 ","End":"04:02.210","Text":"your instantaneous velocity will be the same as your average velocity."},{"Start":"04:02.210 ","End":"04:06.720","Text":"That\u0027s in a case where your velocity is in fact constant."},{"Start":"04:07.880 ","End":"04:10.895","Text":"We can in fact give an example of this."},{"Start":"04:10.895 ","End":"04:18.190","Text":"Let\u0027s assume for a moment that x is equal to 3 t plus 2."},{"Start":"04:18.190 ","End":"04:21.530","Text":"This is linear, meaning it doesn\u0027t have any exponents involved."},{"Start":"04:21.530 ","End":"04:24.380","Text":"We know that no matter what\u0027s going on,"},{"Start":"04:24.380 ","End":"04:26.465","Text":"v as a function of t,"},{"Start":"04:26.465 ","End":"04:30.440","Text":"meaning our velocity as a function of time is equal to 3."},{"Start":"04:30.440 ","End":"04:33.535","Text":"In this case, if you look for the average velocity,"},{"Start":"04:33.535 ","End":"04:34.760","Text":"doesn\u0027t matter what the times are."},{"Start":"04:34.760 ","End":"04:36.215","Text":"Let\u0027s say it\u0027s 2 and 6."},{"Start":"04:36.215 ","End":"04:38.300","Text":"You\u0027ll come up with the same average velocity,"},{"Start":"04:38.300 ","End":"04:41.560","Text":"which is 3, the same as your constant velocity."},{"Start":"04:41.560 ","End":"04:45.395","Text":"So far we\u0027ve talked about velocity, instantaneous velocity,"},{"Start":"04:45.395 ","End":"04:50.105","Text":"which is found from derivatives and average velocity found from the difference in times."},{"Start":"04:50.105 ","End":"04:53.420","Text":"Of course, they are the same when your velocity is constant."},{"Start":"04:53.420 ","End":"04:56.190","Text":"Now let\u0027s go and talk about acceleration."}],"ID":9213},{"Watched":false,"Name":"Instantaneous Acceleration and Average Acceleration","Duration":"3m 19s","ChapterTopicVideoID":8935,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:05.760","Text":"The relationship between velocity and acceleration"},{"Start":"00:05.760 ","End":"00:12.310","Text":"is the same as the relationship between position and velocity."},{"Start":"00:12.440 ","End":"00:15.915","Text":"We\u0027ll start with instantaneous acceleration."},{"Start":"00:15.915 ","End":"00:23.880","Text":"The definition of instantaneous acceleration is dv over dt."},{"Start":"00:23.880 ","End":"00:29.700","Text":"This is the derivative of velocity over the derivative of time, symbolized by v-dot."},{"Start":"00:29.700 ","End":"00:32.835","Text":"If we look at the example on the left of 3t^2,"},{"Start":"00:32.835 ","End":"00:36.775","Text":"then our instantaneous acceleration equals 6t."},{"Start":"00:36.775 ","End":"00:40.790","Text":"Oftentimes we also write down the connection between the acceleration and"},{"Start":"00:40.790 ","End":"00:44.075","Text":"the position because it\u0027s the derivative of velocity."},{"Start":"00:44.075 ","End":"00:49.204","Text":"Acceleration is then the 2nd derivative of position x-double-dot."},{"Start":"00:49.204 ","End":"00:52.025","Text":"The 2 dots being for 2nd derivative,"},{"Start":"00:52.025 ","End":"00:54.590","Text":"whereas 1 dot is 1st derivative."},{"Start":"00:54.590 ","End":"01:01.630","Text":"We can also write that as d^2x over dt^2."},{"Start":"01:01.630 ","End":"01:06.815","Text":"This is the instantaneous acceleration, 6t."},{"Start":"01:06.815 ","End":"01:11.285","Text":"We can also talk about the average acceleration, a-line."},{"Start":"01:11.285 ","End":"01:13.430","Text":"Again, the line being for average."},{"Start":"01:13.430 ","End":"01:16.035","Text":"The equation is Delta v,"},{"Start":"01:16.035 ","End":"01:19.175","Text":"change in velocity, over Delta t, the change in time."},{"Start":"01:19.175 ","End":"01:22.235","Text":"In other words, this is the terminal velocity"},{"Start":"01:22.235 ","End":"01:25.145","Text":"minus initial velocity over the change in time."},{"Start":"01:25.145 ","End":"01:30.025","Text":"By this we mean your terminal time minus your initial time."},{"Start":"01:30.025 ","End":"01:33.649","Text":"In the same way that we calculated our instantaneous acceleration,"},{"Start":"01:33.649 ","End":"01:36.335","Text":"we can now calculate our average acceleration from the left."},{"Start":"01:36.335 ","End":"01:41.615","Text":"Our average acceleration over the time 1 second to 3 seconds"},{"Start":"01:41.615 ","End":"01:49.530","Text":"equals the change in velocity between those 2 times v t,"},{"Start":"01:49.530 ","End":"01:56.690","Text":"3 minus v t equals 1 over the change in times 3 minus 1."},{"Start":"01:56.690 ","End":"02:00.080","Text":"If we calculate this, we get 3t."},{"Start":"02:00.080 ","End":"02:07.000","Text":"The second is 27 minus 3 over 2."},{"Start":"02:07.000 ","End":"02:11.495","Text":"Our units here are meters over seconds squared."},{"Start":"02:11.495 ","End":"02:16.925","Text":"With the understanding that we\u0027re measuring in meters for distance and seconds for time."},{"Start":"02:16.925 ","End":"02:19.610","Text":"We\u0027ve calculated our average acceleration."},{"Start":"02:19.610 ","End":"02:21.440","Text":"If you want to calculate it out all the way,"},{"Start":"02:21.440 ","End":"02:23.360","Text":"you get 12 meters per second squared."},{"Start":"02:23.360 ","End":"02:26.960","Text":"You also have your instantaneous acceleration,"},{"Start":"02:26.960 ","End":"02:28.430","Text":"which is dependent on time."},{"Start":"02:28.430 ","End":"02:32.325","Text":"Just like with velocity, if your acceleration,"},{"Start":"02:32.325 ","End":"02:35.120","Text":"your average acceleration is measured over a different time,"},{"Start":"02:35.120 ","End":"02:36.635","Text":"your results will change."},{"Start":"02:36.635 ","End":"02:39.400","Text":"Just like with the velocity,"},{"Start":"02:39.400 ","End":"02:41.725","Text":"if your acceleration is constant,"},{"Start":"02:41.725 ","End":"02:47.420","Text":"then your average acceleration will be the same as your instantaneous acceleration."},{"Start":"02:47.970 ","End":"02:53.695","Text":"You\u0027ll have a constant velocity in scenarios where x is equal to t^2."},{"Start":"02:53.695 ","End":"02:58.960","Text":"For example, x equal to 4t^2 plus 3t plus 5."},{"Start":"02:58.960 ","End":"03:01.105","Text":"In this scenario, if you calculate it out,"},{"Start":"03:01.105 ","End":"03:04.510","Text":"you find that your instantaneous acceleration as a function of"},{"Start":"03:04.510 ","End":"03:10.080","Text":"t is equal to 8 meters per second squared."},{"Start":"03:10.080 ","End":"03:12.295","Text":"If you calculate out the hallway,"},{"Start":"03:12.295 ","End":"03:16.330","Text":"you\u0027ll find that your average acceleration is the same,"},{"Start":"03:16.330 ","End":"03:18.740","Text":"regardless of the time you\u0027re measuring."}],"ID":9214},{"Watched":false,"Name":"Midway Summary","Duration":"58s","ChapterTopicVideoID":8936,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:04.770","Text":"To summarize, we\u0027ve talked about the instantaneous velocity,"},{"Start":"00:04.770 ","End":"00:08.010","Text":"which is the derivative of position over time."},{"Start":"00:08.010 ","End":"00:10.740","Text":"We\u0027ve also talked about the average velocity being"},{"Start":"00:10.740 ","End":"00:13.875","Text":"the change in position over the change in time."},{"Start":"00:13.875 ","End":"00:19.530","Text":"We should write those both as equations below as you can see."},{"Start":"00:19.530 ","End":"00:21.575","Text":"We also talked about acceleration."},{"Start":"00:21.575 ","End":"00:26.420","Text":"The instantaneous acceleration is the derivative of velocity over time and"},{"Start":"00:26.420 ","End":"00:32.210","Text":"the average acceleration as change in velocity over change in time."},{"Start":"00:32.210 ","End":"00:35.705","Text":"We want to write the equations for these."},{"Start":"00:35.705 ","End":"00:39.425","Text":"We\u0027ve talked about velocity as a function of position,"},{"Start":"00:39.425 ","End":"00:43.550","Text":"and acceleration as a function of both velocity and position."},{"Start":"00:43.550 ","End":"00:46.640","Text":"What we\u0027re able to do now is find"},{"Start":"00:46.640 ","End":"00:51.245","Text":"the velocity or the acceleration using the position as the given data point."},{"Start":"00:51.245 ","End":"00:53.930","Text":"Next, we\u0027ll find out how we can use our acceleration to"},{"Start":"00:53.930 ","End":"00:57.660","Text":"find our velocity or use our velocity to find our position."}],"ID":9215},{"Watched":false,"Name":"How to Find Velocity Using Acceleration","Duration":"3m 59s","ChapterTopicVideoID":8937,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.995","Text":"Hello. In this video,"},{"Start":"00:01.995 ","End":"00:05.820","Text":"I\u0027ll explain how we use our acceleration as a function"},{"Start":"00:05.820 ","End":"00:10.260","Text":"of time to find our velocity as a function of time."},{"Start":"00:10.260 ","End":"00:13.470","Text":"How afterwards from there we can use our velocity as a function of"},{"Start":"00:13.470 ","End":"00:17.760","Text":"time to find our position as a function of time."},{"Start":"00:17.760 ","End":"00:19.635","Text":"In the previous video,"},{"Start":"00:19.635 ","End":"00:23.160","Text":"we went from position to velocity using a derivative."},{"Start":"00:23.160 ","End":"00:26.564","Text":"We also used derivatives to go from velocity to acceleration."},{"Start":"00:26.564 ","End":"00:28.140","Text":"Now going in the opposite direction,"},{"Start":"00:28.140 ","End":"00:30.810","Text":"we\u0027ll need to do the opposite called an integral."},{"Start":"00:30.810 ","End":"00:32.655","Text":"Let\u0027s do an example."},{"Start":"00:32.655 ","End":"00:35.310","Text":"In this case, we\u0027ll say our acceleration as a function of"},{"Start":"00:35.310 ","End":"00:38.685","Text":"time is equal to just choose something,"},{"Start":"00:38.685 ","End":"00:43.690","Text":"anything, 3t plus 2."},{"Start":"00:44.270 ","End":"00:46.470","Text":"To find our velocity,"},{"Start":"00:46.470 ","End":"00:50.960","Text":"what we\u0027re going to do is take an integral of a,"},{"Start":"00:50.960 ","End":"00:58.900","Text":"acceleration as a function of t. We\u0027re going to multiply that by dt."},{"Start":"01:04.540 ","End":"01:08.375","Text":"Now what we need to do to find our velocity, V,"},{"Start":"01:08.375 ","End":"01:11.675","Text":"is plug in t to this integral equation and add on C,"},{"Start":"01:11.675 ","End":"01:14.570","Text":"which is the integral constant."},{"Start":"01:14.570 ","End":"01:18.080","Text":"A lot of students will make a mistake"},{"Start":"01:18.080 ","End":"01:21.995","Text":"and automatically set C as equal to the initial velocity."},{"Start":"01:21.995 ","End":"01:23.600","Text":"That\u0027s not necessarily the case."},{"Start":"01:23.600 ","End":"01:25.805","Text":"It can be equal to the initial velocity,"},{"Start":"01:25.805 ","End":"01:28.765","Text":"but does not have to be the same."},{"Start":"01:28.765 ","End":"01:30.830","Text":"Going back to our example,"},{"Start":"01:30.830 ","End":"01:37.295","Text":"what we do here is to find V as a function of t. We would plug in our t to the integral,"},{"Start":"01:37.295 ","End":"01:39.020","Text":"and it would end up looking like this."},{"Start":"01:39.020 ","End":"01:43.670","Text":"An integral of 3t plus 2 dt."},{"Start":"01:43.670 ","End":"01:46.865","Text":"If we multiply that out and do the function,"},{"Start":"01:46.865 ","End":"01:54.655","Text":"we end up with 3t squared over 2 plus 2t."},{"Start":"01:54.655 ","End":"01:58.145","Text":"Now what we need to do is find that C, that constant."},{"Start":"01:58.145 ","End":"02:02.180","Text":"The way we find the constant is we do need to find our initial velocity."},{"Start":"02:02.180 ","End":"02:05.765","Text":"Not that will necessarily be the same as C, but it helps us."},{"Start":"02:05.765 ","End":"02:08.210","Text":"We find our initial velocity,"},{"Start":"02:08.210 ","End":"02:12.530","Text":"which is the same as Vt equals 0."},{"Start":"02:12.530 ","End":"02:14.495","Text":"For the sake of our example,"},{"Start":"02:14.495 ","End":"02:20.330","Text":"let\u0027s assume that V at t equals 0 equals 2 meters per second."},{"Start":"02:20.360 ","End":"02:23.390","Text":"The question is, what do we do to find C?"},{"Start":"02:23.390 ","End":"02:28.025","Text":"Well, we set the equation at t equals 0 and see what we come up with."},{"Start":"02:28.025 ","End":"02:31.025","Text":"V at t equals 0,"},{"Start":"02:31.025 ","End":"02:34.820","Text":"equals 0 plus 0 plus"},{"Start":"02:34.820 ","End":"02:39.770","Text":"C. But we know that our initial velocity has to be 2 meters per second."},{"Start":"02:39.770 ","End":"02:43.165","Text":"We know in this case that C equals 2."},{"Start":"02:43.165 ","End":"02:48.545","Text":"We can now do is take our C as 2 and plug it back into the full equation."},{"Start":"02:48.545 ","End":"02:59.430","Text":"In this case, what we\u0027ll get is V of t equals 3t squared over 2 plus 2t plus 2."},{"Start":"02:59.430 ","End":"03:04.160","Text":"In our example, our C in fact did equal the initial velocity."},{"Start":"03:04.160 ","End":"03:06.170","Text":"But there are times when that is not the case."},{"Start":"03:06.170 ","End":"03:09.740","Text":"For example, if V as a function of t, sorry,"},{"Start":"03:09.740 ","End":"03:16.060","Text":"if your acceleration as a function of t equals e^2t,"},{"Start":"03:16.060 ","End":"03:19.415","Text":"you\u0027ll find that it\u0027s not the case."},{"Start":"03:19.415 ","End":"03:21.890","Text":"Just to summarize a little,"},{"Start":"03:21.890 ","End":"03:26.255","Text":"what we found here is that if you want to find velocity from acceleration,"},{"Start":"03:26.255 ","End":"03:30.020","Text":"you do an integral of a dt plus the constant."},{"Start":"03:30.020 ","End":"03:31.370","Text":"To find that constant,"},{"Start":"03:31.370 ","End":"03:35.495","Text":"the easiest way to do it is set t at 0 and"},{"Start":"03:35.495 ","End":"03:40.400","Text":"go through to find your value for the constant and then plug it back in."},{"Start":"03:40.400 ","End":"03:44.150","Text":"You can find your constant using a different time."},{"Start":"03:44.150 ","End":"03:46.805","Text":"Could be t equals really any amount."},{"Start":"03:46.805 ","End":"03:50.210","Text":"But oftentimes you\u0027ll find it at t equals 0."},{"Start":"03:50.210 ","End":"03:51.890","Text":"The same procedure applies though."},{"Start":"03:51.890 ","End":"03:54.865","Text":"You just plug t in and go from there."},{"Start":"03:54.865 ","End":"03:59.820","Text":"Now let\u0027s see how we find our position using our velocity."}],"ID":9216},{"Watched":false,"Name":"How to Find Position Using Velocity","Duration":"2m 57s","ChapterTopicVideoID":8938,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"If we want to find"},{"Start":"00:02.025 ","End":"00:07.720","Text":"our position using our velocity we\u0027ll do the exact same thing we did before."},{"Start":"00:10.520 ","End":"00:16.605","Text":"We\u0027ll set x as a function of t equal to an integral of"},{"Start":"00:16.605 ","End":"00:22.785","Text":"v as a function of t times d t. If we calculate that out,"},{"Start":"00:22.785 ","End":"00:25.950","Text":"if we look at the last equation,"},{"Start":"00:25.950 ","End":"00:30.930","Text":"give us t^3 over 2"},{"Start":"00:30.930 ","End":"00:40.640","Text":"plus t^2 plus 2t."},{"Start":"00:40.640 ","End":"00:42.830","Text":"Now we just need to add on the constant C and to find"},{"Start":"00:42.830 ","End":"00:45.560","Text":"the constant C. We\u0027re going to do the same operation as before."},{"Start":"00:45.560 ","End":"00:49.530","Text":"We\u0027re going to set our t=0."},{"Start":"00:49.530 ","End":"00:52.230","Text":"We can do x as a function of t=0,"},{"Start":"00:52.230 ","End":"00:57.375","Text":"which can also be written as x sub 0, your initial position."},{"Start":"00:57.375 ","End":"01:01.455","Text":"We\u0027re going to find it by plugging in t=0 in our equation. What do we get?"},{"Start":"01:01.455 ","End":"01:06.300","Text":"We get 0 plus 0 plus 0 plus"},{"Start":"01:06.300 ","End":"01:12.470","Text":"C. Now we would get some data given to us before."},{"Start":"01:12.470 ","End":"01:17.585","Text":"For example, let\u0027s say that we were told that our initial position is 3 meters."},{"Start":"01:17.585 ","End":"01:21.130","Text":"Then C in this case equals 3."},{"Start":"01:21.130 ","End":"01:23.800","Text":"We can plug that into the equation."},{"Start":"01:23.800 ","End":"01:26.885","Text":"To give you a little quick summary, to find our x,"},{"Start":"01:26.885 ","End":"01:31.910","Text":"we\u0027re going to do an integral of v t d t plus the constant C,"},{"Start":"01:31.910 ","End":"01:32.975","Text":"and to find that constant,"},{"Start":"01:32.975 ","End":"01:34.850","Text":"we\u0027re going to set t=0."},{"Start":"01:34.850 ","End":"01:37.040","Text":"It can be another value, but usually,"},{"Start":"01:37.040 ","End":"01:38.915","Text":"0, plug that in and what we get,"},{"Start":"01:38.915 ","End":"01:44.840","Text":"we\u0027ll go back in as the constant for C. If we\u0027re going to take this a step further,"},{"Start":"01:44.840 ","End":"01:47.195","Text":"if we want to go from our acceleration"},{"Start":"01:47.195 ","End":"01:50.150","Text":"as a function of t to our position as a function of t,"},{"Start":"01:50.150 ","End":"01:52.760","Text":"a t to x t. We actually have to do it in 2 steps."},{"Start":"01:52.760 ","End":"01:55.940","Text":"We\u0027re going to first do an integral to find our velocity,"},{"Start":"01:55.940 ","End":"01:59.600","Text":"and then do an integral of the velocity to find our position."},{"Start":"01:59.600 ","End":"02:03.140","Text":"The one thing that we need is we need some pre-given point,"},{"Start":"02:03.140 ","End":"02:07.490","Text":"usually, it would be v t=0 and x at t=0."},{"Start":"02:07.490 ","End":"02:08.930","Text":"Again, we can use a different value,"},{"Start":"02:08.930 ","End":"02:10.870","Text":"although those are the most common."},{"Start":"02:10.870 ","End":"02:14.525","Text":"Now you really do need 2 given data points if you want to find"},{"Start":"02:14.525 ","End":"02:20.660","Text":"an exact value for your velocity or your position,"},{"Start":"02:20.660 ","End":"02:23.945","Text":"using the integrals, you don\u0027t need to be given"},{"Start":"02:23.945 ","End":"02:28.580","Text":"your initial position or initial velocity,"},{"Start":"02:28.580 ","End":"02:29.900","Text":"you can be given other values."},{"Start":"02:29.900 ","End":"02:33.540","Text":"For example, you could be given x at time=2."},{"Start":"02:34.040 ","End":"02:36.210","Text":"If that was 5 meters,"},{"Start":"02:36.210 ","End":"02:40.500","Text":"let\u0027s say we could use that and find that for our last equation,"},{"Start":"02:40.500 ","End":"02:43.380","Text":"or instead of being given a V and an x, you could be given 2 Xs,"},{"Start":"02:43.380 ","End":"02:44.810","Text":"and actually, even with that,"},{"Start":"02:44.810 ","End":"02:47.590","Text":"you can still find an exact value."},{"Start":"02:47.590 ","End":"02:50.990","Text":"Really what you just need is 2 data points and from there you can go."},{"Start":"02:50.990 ","End":"02:53.090","Text":"Now the next thing we\u0027re going to do is find out"},{"Start":"02:53.090 ","End":"02:56.580","Text":"what we can do when our acceleration is constant."}],"ID":9217},{"Watched":false,"Name":"Specific Case of Constant Acceleration","Duration":"1m 3s","ChapterTopicVideoID":8939,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.535","Text":"Now there is 1 special case."},{"Start":"00:02.535 ","End":"00:05.190","Text":"If our acceleration is constant,"},{"Start":"00:05.190 ","End":"00:08.460","Text":"we\u0027re going to symbolize that with a sub 0."},{"Start":"00:08.460 ","End":"00:12.525","Text":"Then we will find some really familiar equations that you might have seen in"},{"Start":"00:12.525 ","End":"00:16.650","Text":"the other lectures that we can use in this special situation."},{"Start":"00:16.650 ","End":"00:19.215","Text":"To set up our velocity equation,"},{"Start":"00:19.215 ","End":"00:24.090","Text":"we do V as a function of time equals v0 plus"},{"Start":"00:24.090 ","End":"00:31.020","Text":"a(t) which is our equation for velocity and with constant acceleration,"},{"Start":"00:31.020 ","End":"00:36.150","Text":"and x as a function of t equals x_0 plus v_0,"},{"Start":"00:36.150 ","End":"00:39.885","Text":"t plus 1.5 a_0 t^2."},{"Start":"00:39.885 ","End":"00:46.610","Text":"Again, this is the positional equation for constant acceleration."},{"Start":"00:46.610 ","End":"00:52.235","Text":"These are equations that can only be used if you have a constant acceleration,"},{"Start":"00:52.235 ","End":"00:54.650","Text":"otherwise, you have to do your 2 integrals."},{"Start":"00:54.650 ","End":"00:59.885","Text":"But they\u0027re very useful for saving time if you find that your acceleration is constant."},{"Start":"00:59.885 ","End":"01:03.660","Text":"That\u0027s the end of this short video series."}],"ID":9218},{"Watched":false,"Name":"Position and Time Exercise","Duration":"2m 49s","ChapterTopicVideoID":8940,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.630 ","End":"00:03.430","Text":"Hello. In this question,"},{"Start":"00:03.430 ","End":"00:05.860","Text":"we\u0027re being told that the position of a body traveling in"},{"Start":"00:05.860 ","End":"00:09.550","Text":"a straight line is given by the equation x(t) is"},{"Start":"00:09.550 ","End":"00:15.700","Text":"equal to 32te^negative t. Now in the first question,"},{"Start":"00:15.700 ","End":"00:20.095","Text":"we\u0027re being told to find the time at which the body stops moving."},{"Start":"00:20.095 ","End":"00:21.835","Text":"How do we do this?"},{"Start":"00:21.835 ","End":"00:29.730","Text":"The body will stop moving when our velocity is equal to 0."},{"Start":"00:29.730 ","End":"00:38.580","Text":"Now what we have to do is we have to derive this function and then set it equal to 0."},{"Start":"00:38.580 ","End":"00:44.615","Text":"Our velocity with respect to time is equal to x dot,"},{"Start":"00:44.615 ","End":"00:49.220","Text":"which is equal to, so first,"},{"Start":"00:49.220 ","End":"00:51.425","Text":"we\u0027ll derive it according to this t"},{"Start":"00:51.425 ","End":"00:59.765","Text":"so 32e^negative t plus and now with respect to this,"},{"Start":"00:59.765 ","End":"01:05.280","Text":"so we\u0027re going to have in fact negative 32te^negative"},{"Start":"01:05.280 ","End":"01:14.315","Text":"t. Now of course we know that this is equal to 0."},{"Start":"01:14.315 ","End":"01:20.735","Text":"Then we can say that 32e^negative t is equal to,"},{"Start":"01:20.735 ","End":"01:22.540","Text":"let\u0027s write this down,"},{"Start":"01:22.540 ","End":"01:29.885","Text":"is equal to 32te^negative t. I just moved this to the other side of the equal sign."},{"Start":"01:29.885 ","End":"01:31.880","Text":"Then the 32 can cross out,"},{"Start":"01:31.880 ","End":"01:34.115","Text":"and then this can cross out,"},{"Start":"01:34.115 ","End":"01:36.570","Text":"leaving just a 1 here."},{"Start":"01:36.570 ","End":"01:39.735","Text":"Then we\u0027ll get that 1 is equal to t,"},{"Start":"01:39.735 ","End":"01:43.030","Text":"so t equals 1 second,"},{"Start":"01:43.030 ","End":"01:46.055","Text":"the body will stop moving."},{"Start":"01:46.055 ","End":"01:48.020","Text":"Now in the second question,"},{"Start":"01:48.020 ","End":"01:53.105","Text":"we\u0027re asked to find the distance of the body at this time from the origin."},{"Start":"01:53.105 ","End":"01:54.605","Text":"Now for this question,"},{"Start":"01:54.605 ","End":"01:59.530","Text":"we\u0027re going to assume that our t starts at t=0 seconds."},{"Start":"01:59.530 ","End":"02:05.150","Text":"If we substitute into our x when t=0,"},{"Start":"02:05.150 ","End":"02:06.550","Text":"at the start,"},{"Start":"02:06.550 ","End":"02:12.395","Text":"we will get that it equals 0 because 32 times 0 will equal 0."},{"Start":"02:12.395 ","End":"02:18.565","Text":"Now we have to substitute in our t=1."},{"Start":"02:18.565 ","End":"02:25.740","Text":"For that distance, and then we\u0027ll just get 32 divided by e,"},{"Start":"02:25.740 ","End":"02:30.000","Text":"because it\u0027s 32 times 1 times e^negative 1,"},{"Start":"02:30.000 ","End":"02:36.740","Text":"which means 1 over e. Then we just would have to do 32 divided by e minus 0,"},{"Start":"02:36.740 ","End":"02:41.390","Text":"which will just equal 32 divided by e. This is"},{"Start":"02:41.390 ","End":"02:46.780","Text":"the distance of the body at t=1 second from the origin."},{"Start":"02:46.780 ","End":"02:49.370","Text":"That\u0027s the end of our lesson."}],"ID":9219},{"Watched":false,"Name":"Velocity As A Function Of Position","Duration":"11m 16s","ChapterTopicVideoID":8941,"CourseChapterTopicPlaylistID":5374,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.950 ","End":"00:03.855","Text":"Hello. In this question,"},{"Start":"00:03.855 ","End":"00:07.650","Text":"we\u0027re told that a particle is moving in the positive direction of"},{"Start":"00:07.650 ","End":"00:12.765","Text":"the x-axis from x=0 until infinity"},{"Start":"00:12.765 ","End":"00:16.080","Text":"and its velocity is given by velocity in"},{"Start":"00:16.080 ","End":"00:22.485","Text":"the x direction equals c multiplied by the square root of x,"},{"Start":"00:22.485 ","End":"00:25.845","Text":"c being some constant which is bigger than 0."},{"Start":"00:25.845 ","End":"00:28.620","Text":"At the time t=0,"},{"Start":"00:28.620 ","End":"00:32.475","Text":"the particle is at a position x=0,"},{"Start":"00:32.475 ","End":"00:42.325","Text":"so let\u0027s just write this down at xt=0, this equals 0."},{"Start":"00:42.325 ","End":"00:45.250","Text":"Now we\u0027re being asked in the first question,"},{"Start":"00:45.250 ","End":"00:47.560","Text":"what are the units of c?"},{"Start":"00:47.560 ","End":"00:55.705","Text":"Now, we\u0027re given this equation that v_x is equal to c multiplied by root x."},{"Start":"00:55.705 ","End":"01:01.250","Text":"Now, we can rewrite that as c multiplied by x^1/2."},{"Start":"01:01.490 ","End":"01:07.555","Text":"We know that our units for velocity is meters per second,"},{"Start":"01:07.555 ","End":"01:11.965","Text":"I\u0027m going to put it in square brackets because that\u0027s how it\u0027s expected to be written."},{"Start":"01:11.965 ","End":"01:15.090","Text":"Then c, we\u0027re trying to find this,"},{"Start":"01:15.090 ","End":"01:20.810","Text":"so I\u0027m going to leave this empty multiplied by"},{"Start":"01:20.880 ","End":"01:29.490","Text":"x^1/2 which is meters to the half."},{"Start":"01:29.490 ","End":"01:32.255","Text":"If I now want to find this,"},{"Start":"01:32.255 ","End":"01:36.205","Text":"I have to divide both sides by this."},{"Start":"01:36.205 ","End":"01:37.870","Text":"This is c,"},{"Start":"01:37.870 ","End":"01:41.365","Text":"I need to divide both sides by m^1/2."},{"Start":"01:41.365 ","End":"01:48.661","Text":"Then I\u0027ll get meters per second multiplied by m to"},{"Start":"01:48.661 ","End":"01:54.800","Text":"the negative 1/2 equals"},{"Start":"01:55.460 ","End":"01:59.370","Text":"c. Therefore,"},{"Start":"01:59.370 ","End":"02:02.230","Text":"I can say I\u0027ll have"},{"Start":"02:02.720 ","End":"02:10.275","Text":"m^1/2 because it\u0027s meters to the 1 multiplied by meters to the negative 1/2,"},{"Start":"02:10.275 ","End":"02:17.565","Text":"so it\u0027s meters to the 1/2 multiplied by seconds because"},{"Start":"02:17.565 ","End":"02:21.510","Text":"we\u0027re dividing here by seconds so it\u0027s to the negative 1 which are"},{"Start":"02:21.510 ","End":"02:25.670","Text":"the units of c. Now in the second question,"},{"Start":"02:25.670 ","End":"02:30.830","Text":"we\u0027re being told to find the velocity and acceleration as a function of time."},{"Start":"02:30.830 ","End":"02:37.054","Text":"Now what we\u0027re given here in this equation is the velocity as a function of the position."},{"Start":"02:37.054 ","End":"02:41.235","Text":"Now the position will be as a function of time,"},{"Start":"02:41.235 ","End":"02:43.435","Text":"so we\u0027re trying to find what that is."},{"Start":"02:43.435 ","End":"02:52.450","Text":"How we\u0027ve seen that x(t) is equal to t^2 plus 2, for example."},{"Start":"02:52.450 ","End":"02:55.485","Text":"This is x as a function of time,"},{"Start":"02:55.485 ","End":"02:59.015","Text":"so then if we were to substitute that in,"},{"Start":"02:59.015 ","End":"03:06.535","Text":"we would get that v and the x direction would be c root t^2 plus 2."},{"Start":"03:06.535 ","End":"03:09.905","Text":"We would substitute in our position as a function of time,"},{"Start":"03:09.905 ","End":"03:12.455","Text":"and then we would get a velocity as a function of time."},{"Start":"03:12.455 ","End":"03:13.850","Text":"Now, this was just an example,"},{"Start":"03:13.850 ","End":"03:15.985","Text":"so I\u0027m going to rub this out."},{"Start":"03:15.985 ","End":"03:18.270","Text":"That is what we\u0027re trying to do,"},{"Start":"03:18.270 ","End":"03:25.855","Text":"we\u0027re trying to find this equation as a function of time with variable t_0x."},{"Start":"03:25.855 ","End":"03:27.675","Text":"How do we solve this?"},{"Start":"03:27.675 ","End":"03:30.910","Text":"Get ready, we\u0027re going to do differential equations."},{"Start":"03:30.910 ","End":"03:35.490","Text":"I know it sounds really difficult and scary, but we\u0027ll do it."},{"Start":"03:35.690 ","End":"03:38.900","Text":"That\u0027s the only way we can solve this question."},{"Start":"03:38.900 ","End":"03:45.050","Text":"I know that my velocity is my dx by dt."},{"Start":"03:45.050 ","End":"03:49.825","Text":"Remember that\u0027s our definition for velocity that we learned in the first lesson."},{"Start":"03:49.825 ","End":"03:55.400","Text":"This equals to our c multiplied by x^1/2."},{"Start":"03:55.400 ","End":"03:59.060","Text":"Notice whenever you are doing integration or differentiation,"},{"Start":"03:59.060 ","End":"04:02.240","Text":"it\u0027s a lot easier instead of working in this format of"},{"Start":"04:02.240 ","End":"04:07.285","Text":"this square root sine x to actually write it as x^1/2."},{"Start":"04:07.285 ","End":"04:11.780","Text":"Now what I\u0027m going to do is I\u0027m going to multiply both sides by dt."},{"Start":"04:11.780 ","End":"04:13.850","Text":"I\u0027m going to get this dt over here,"},{"Start":"04:13.850 ","End":"04:15.830","Text":"so multiply both sides by dt,"},{"Start":"04:15.830 ","End":"04:20.135","Text":"and then I want to get this x^1/2 to this side so I\u0027m going to"},{"Start":"04:20.135 ","End":"04:25.795","Text":"multiply by x to the negative 1/2 or divide by x^1/2."},{"Start":"04:25.795 ","End":"04:30.425","Text":"What I\u0027m going to do here is I\u0027m going to multiply by"},{"Start":"04:30.425 ","End":"04:36.930","Text":"dt and I\u0027m going to also multiply by x to the negative 1/2."},{"Start":"04:37.700 ","End":"04:41.721","Text":"Then what I\u0027m going to get is I\u0027ll get that x"},{"Start":"04:41.721 ","End":"04:50.510","Text":"to the negative 1/2dx=cdt."},{"Start":"04:50.510 ","End":"04:56.855","Text":"Now, all I have to do is do an integration on both of the sides."},{"Start":"04:56.855 ","End":"05:00.665","Text":"Now what I\u0027ve done here is I\u0027ve separated out my variables,"},{"Start":"05:00.665 ","End":"05:02.825","Text":"so all my t variables,"},{"Start":"05:02.825 ","End":"05:06.245","Text":"I mean here I don\u0027t have t variables I have this, but it doesn\u0027t matter."},{"Start":"05:06.245 ","End":"05:12.020","Text":"All my t variables go in 1 side are all my variables that aren\u0027t my x variables,"},{"Start":"05:12.020 ","End":"05:14.825","Text":"and then all my x variables go in the other side,"},{"Start":"05:14.825 ","End":"05:18.120","Text":"which match with the dx, then I integrate."},{"Start":"05:18.120 ","End":"05:21.195","Text":"What is the integration of x to the negative 1/2?"},{"Start":"05:21.195 ","End":"05:30.000","Text":"It\u0027s going to be"},{"Start":"05:30.000 ","End":"05:34.240","Text":"2x^1/2=ct plus some constant."},{"Start":"05:34.940 ","End":"05:40.790","Text":"Now of course we know that I have to find what this constant is."},{"Start":"05:40.790 ","End":"05:45.950","Text":"The only way I can do it is with my initial conditions."},{"Start":"05:45.950 ","End":"05:47.855","Text":"Now what was my initial condition?"},{"Start":"05:47.855 ","End":"05:51.665","Text":"It\u0027s this, when t=0, x=0."},{"Start":"05:51.665 ","End":"05:52.880","Text":"Let\u0027s rearrange this,"},{"Start":"05:52.880 ","End":"05:57.840","Text":"so my x=0 so this is 2 times"},{"Start":"05:58.060 ","End":"06:06.569","Text":"0=c multiplied by t=0 so c times 0 plus a constant."},{"Start":"06:06.569 ","End":"06:11.925","Text":"That means that I have 0 equals constant."},{"Start":"06:11.925 ","End":"06:14.555","Text":"My constant is equal 0 and therefore,"},{"Start":"06:14.555 ","End":"06:21.200","Text":"I know that my equation is going to look something like 2x to the 1/2"},{"Start":"06:21.200 ","End":"06:30.165","Text":"equals ct which we can then rearrange to be x equals,"},{"Start":"06:30.165 ","End":"06:35.240","Text":"so we\u0027ll divide both sides by 2 and then square both sides in order to just get"},{"Start":"06:35.240 ","End":"06:36.950","Text":"x so then we\u0027ll"},{"Start":"06:36.950 ","End":"06:45.700","Text":"get c over 2t^2."},{"Start":"06:45.700 ","End":"06:51.540","Text":"This is our x equation as a function of time."},{"Start":"06:51.860 ","End":"06:54.990","Text":"Now let\u0027s answer the question,"},{"Start":"06:54.990 ","End":"06:57.770","Text":"find the velocity as a function of time."},{"Start":"06:57.770 ","End":"07:00.000","Text":"Then I can say that,"},{"Start":"07:00.000 ","End":"07:01.480","Text":"I\u0027m just going to scroll down a bit,"},{"Start":"07:01.480 ","End":"07:09.010","Text":"that my velocity in the x direction as a function of time is equal to c,"},{"Start":"07:09.010 ","End":"07:12.345","Text":"multiplied by the square root of x."},{"Start":"07:12.345 ","End":"07:14.685","Text":"That will be the square root of this,"},{"Start":"07:14.685 ","End":"07:19.510","Text":"so it will be c over 2t^2,"},{"Start":"07:19.510 ","End":"07:21.290","Text":"but then it\u0027s the square root, so it\u0027s that,"},{"Start":"07:21.290 ","End":"07:26.165","Text":"so I can say that it c^2 over 2t."},{"Start":"07:26.165 ","End":"07:29.255","Text":"This is my velocity as a function of time."},{"Start":"07:29.255 ","End":"07:33.780","Text":"Now we\u0027re also being asked to find the acceleration as a function of time."},{"Start":"07:33.800 ","End":"07:40.025","Text":"As we know, the acceleration is simply just the derivative of the velocity."},{"Start":"07:40.025 ","End":"07:43.700","Text":"I can say that my acceleration in the x direction as"},{"Start":"07:43.700 ","End":"07:47.615","Text":"a function of time is equal to the derivative of this,"},{"Start":"07:47.615 ","End":"07:52.880","Text":"which will just be c^2 over 2, that\u0027s it."},{"Start":"07:52.880 ","End":"07:56.959","Text":"Now, let\u0027s answer the third question."},{"Start":"07:56.959 ","End":"07:59.390","Text":"In the third question we\u0027re being asked to find"},{"Start":"07:59.390 ","End":"08:03.770","Text":"the average velocity and the time taken for the particle to travel a distance"},{"Start":"08:03.770 ","End":"08:12.035","Text":"s. Let\u0027s say that our particle starts at here at position x=0,"},{"Start":"08:12.035 ","End":"08:15.240","Text":"and at time t=0,"},{"Start":"08:15.240 ","End":"08:16.800","Text":"this is right at the beginning."},{"Start":"08:16.800 ","End":"08:22.095","Text":"Then it moves some distance to here where x=s,"},{"Start":"08:22.095 ","End":"08:23.835","Text":"this is the new position,"},{"Start":"08:23.835 ","End":"08:32.020","Text":"and t=t, say t tag."},{"Start":"08:32.090 ","End":"08:38.195","Text":"What we\u0027re being asked is to find the average velocity from here to here."},{"Start":"08:38.195 ","End":"08:39.920","Text":"Now remember from 1 of"},{"Start":"08:39.920 ","End":"08:44.435","Text":"our first lessons that when we\u0027re working out the average velocity,"},{"Start":"08:44.435 ","End":"08:47.465","Text":"it doesn\u0027t matter how we got there,"},{"Start":"08:47.465 ","End":"08:52.820","Text":"we could have gotten, or like this gotten a squiggly line and ended up here."},{"Start":"08:52.820 ","End":"08:55.700","Text":"We could have gone backwards here,"},{"Start":"08:55.700 ","End":"08:57.290","Text":"round, and back to here."},{"Start":"08:57.290 ","End":"09:00.250","Text":"It doesn\u0027t matter, we just want to know from a to b,"},{"Start":"09:00.250 ","End":"09:06.635","Text":"from 0 to s. Our average velocity we can write is equal to,"},{"Start":"09:06.635 ","End":"09:11.260","Text":"remember Delta x by Delta t,"},{"Start":"09:11.260 ","End":"09:15.860","Text":"which is the difference in their positions divided by the difference in times."},{"Start":"09:15.860 ","End":"09:18.515","Text":"I have s minus 0,"},{"Start":"09:18.515 ","End":"09:22.815","Text":"and t tag minus 0."},{"Start":"09:22.815 ","End":"09:26.375","Text":"I\u0027m just going to get s over t tag."},{"Start":"09:26.375 ","End":"09:28.760","Text":"Now, we don\u0027t want to leave our answer like this"},{"Start":"09:28.760 ","End":"09:32.370","Text":"because we don\u0027t know what this time t is,"},{"Start":"09:32.370 ","End":"09:36.470","Text":"there\u0027s too many variables in this. What can we do?"},{"Start":"09:36.470 ","End":"09:39.575","Text":"We can look back to our original equation,"},{"Start":"09:39.575 ","End":"09:47.155","Text":"which was that x=c over 2t^2."},{"Start":"09:47.155 ","End":"09:50.035","Text":"Let\u0027s get this, let\u0027s isolate out this t,"},{"Start":"09:50.035 ","End":"10:00.860","Text":"so we\u0027ll get that t=x^1/2 multiplied by c divided by 2,"},{"Start":"10:01.700 ","End":"10:06.409","Text":"divided by c. All I\u0027ve done is rearranged"},{"Start":"10:06.409 ","End":"10:13.295","Text":"this equation in order to isolate out my t. Once I have that,"},{"Start":"10:13.295 ","End":"10:19.700","Text":"then I can just substitute in this in place of this t tag."},{"Start":"10:19.700 ","End":"10:25.150","Text":"Now notice that x represents our position and we know that our position is s,"},{"Start":"10:25.150 ","End":"10:29.315","Text":"so we can say that x=s."},{"Start":"10:29.315 ","End":"10:31.940","Text":"Then I just have to rewrite this and say that"},{"Start":"10:31.940 ","End":"10:36.815","Text":"my average velocity is equal to s divided by t,"},{"Start":"10:36.815 ","End":"10:42.410","Text":"so it will be x^1/2 multiplied by 2 over c but our x is our s,"},{"Start":"10:42.410 ","End":"10:50.565","Text":"so s^1/2 multiplied by 2 over c. Then I can rearrange this,"},{"Start":"10:50.565 ","End":"10:57.890","Text":"because this is s^1 multiplied by s to the negative 1/2."},{"Start":"10:57.890 ","End":"11:04.335","Text":"I can just get the s^1/2 multiplied by c over 2."},{"Start":"11:04.335 ","End":"11:08.195","Text":"This is our average velocity, and then we go,"},{"Start":"11:08.195 ","End":"11:11.975","Text":"now we\u0027ve found the average velocity as a function of"},{"Start":"11:11.975 ","End":"11:17.040","Text":"the distance traveled s. That\u0027s the end of the lesson."}],"ID":9220}],"Thumbnail":null,"ID":5374},{"Name":"Exercises For Motion In A Straight Line","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Position in a Constant Velocity Motion","Duration":"7m 52s","ChapterTopicVideoID":8956,"CourseChapterTopicPlaylistID":5375,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8956.jpeg","UploadDate":"2017-03-21T08:44:07.3400000","DurationForVideoObject":"PT7M52S","Description":null,"MetaTitle":"Position in a Constant Velocity Motion: Video + Workbook | Proprep","MetaDescription":"Kinematics - Exercises for Motion in a Straight Line. Watch the video made by an expert in the field. 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In this lesson,"},{"Start":"00:01.910 ","End":"00:05.370","Text":"we\u0027re going to be speaking about free fall and throwing vertically."},{"Start":"00:05.370 ","End":"00:10.244","Text":"Now, free fall is due to acceleration due to gravity."},{"Start":"00:10.244 ","End":"00:13.350","Text":"A body which is released from rest from a height will"},{"Start":"00:13.350 ","End":"00:16.425","Text":"fall to the ground at a constant acceleration."},{"Start":"00:16.425 ","End":"00:22.094","Text":"This constant acceleration is called gravitational acceleration due to gravity."},{"Start":"00:22.094 ","End":"00:26.835","Text":"The acceleration in gravity is always constant."},{"Start":"00:26.835 ","End":"00:28.485","Text":"It\u0027s important to note that."},{"Start":"00:28.485 ","End":"00:31.635","Text":"We know it in day-to-day, our language,"},{"Start":"00:31.635 ","End":"00:35.835","Text":"as falling from the acceleration of gravity,"},{"Start":"00:35.835 ","End":"00:38.655","Text":"but the full name is gravitational acceleration."},{"Start":"00:38.655 ","End":"00:42.140","Text":"Now, Galileo Galilei conducted experiments where he would"},{"Start":"00:42.140 ","End":"00:45.875","Text":"drop objects from the Leaning Tower of Pisa."},{"Start":"00:45.875 ","End":"00:50.045","Text":"He noticed that each object followed the exact same acceleration,"},{"Start":"00:50.045 ","End":"00:54.200","Text":"and this is what brought about this whole discovery."},{"Start":"00:54.200 ","End":"00:56.375","Text":"Now, Galileo was correct."},{"Start":"00:56.375 ","End":"00:59.615","Text":"Gravitational acceleration is the same for any body."},{"Start":"00:59.615 ","End":"01:02.675","Text":"If you throw a tiny rock,"},{"Start":"01:02.675 ","End":"01:04.325","Text":"and you throw an elephant,"},{"Start":"01:04.325 ","End":"01:08.560","Text":"the acceleration when falling down will be the exact same,"},{"Start":"01:08.560 ","End":"01:13.805","Text":"and the acceleration will be 9.8 meters per second square."},{"Start":"01:13.805 ","End":"01:18.925","Text":"Now, note, this value for gravity is for Earth only,"},{"Start":"01:18.925 ","End":"01:23.150","Text":"so the value for the acceleration due to gravity on"},{"Start":"01:23.150 ","End":"01:27.701","Text":"the Moon or on Mars will be different to that of on Earth,"},{"Start":"01:27.701 ","End":"01:32.750","Text":"and it does not take into account resistive forces such as air resistance."},{"Start":"01:32.750 ","End":"01:34.550","Text":"So that\u0027s why if you throw,"},{"Start":"01:34.550 ","End":"01:38.060","Text":"for instance, a rock and you throw a feather,"},{"Start":"01:38.060 ","End":"01:43.340","Text":"the feather will take longer to fall down because it experiences greater air resistance."},{"Start":"01:43.340 ","End":"01:50.177","Text":"So this number doesn\u0027t take into account air resistance and is correct only for Earth."},{"Start":"01:50.177 ","End":"01:53.060","Text":"If you would want to, go on YouTube and look up"},{"Start":"01:53.060 ","End":"01:56.648","Text":"throwing a hammer and a feather from the Moon."},{"Start":"01:56.648 ","End":"01:58.800","Text":"Because on the Moon there\u0027s no air,"},{"Start":"01:58.800 ","End":"02:02.000","Text":"so there\u0027s no air resistance forces, and you can see that,"},{"Start":"02:02.000 ","End":"02:06.890","Text":"although the value for this acceleration is different on the Moon,"},{"Start":"02:06.890 ","End":"02:09.080","Text":"the 2 objects will fall at"},{"Start":"02:09.080 ","End":"02:14.870","Text":"the exact same acceleration because of the lacking of resistance forces."},{"Start":"02:14.870 ","End":"02:19.830","Text":"Now, the direction of g is always towards the center of the Earth."},{"Start":"02:21.110 ","End":"02:23.655","Text":"If this is Earth,"},{"Start":"02:23.655 ","End":"02:27.225","Text":"and someone is standing over here,"},{"Start":"02:27.225 ","End":"02:29.570","Text":"the force of gravity will point to here,"},{"Start":"02:29.570 ","End":"02:30.913","Text":"which is the center of the Earth,"},{"Start":"02:30.913 ","End":"02:33.470","Text":"and if someone is standing over here on the other side,"},{"Start":"02:33.470 ","End":"02:35.610","Text":"say, in Australia,"},{"Start":"02:36.040 ","End":"02:38.990","Text":"their gravitational acceleration is also"},{"Start":"02:38.990 ","End":"02:42.410","Text":"pointing in this direction to the center of the earth,"},{"Start":"02:42.410 ","End":"02:44.525","Text":"the radial direction inwards."},{"Start":"02:44.525 ","End":"02:47.940","Text":"The value of g was derived from experiments,"},{"Start":"02:47.940 ","End":"02:51.487","Text":"so another way to say that is that it\u0027s an empirical value."},{"Start":"02:51.487 ","End":"02:55.404","Text":"It also differs slightly at different places on Earth,"},{"Start":"02:55.404 ","End":"02:57.020","Text":"so the value, for instance,"},{"Start":"02:57.020 ","End":"03:01.730","Text":"in America, for g in certain areas will also be different to the value,"},{"Start":"03:01.730 ","End":"03:04.180","Text":"say, in Australia or in South Africa."},{"Start":"03:04.180 ","End":"03:05.960","Text":"Another thing to know is that,"},{"Start":"03:05.960 ","End":"03:10.715","Text":"although we say that the value for g is 9.8 meters per second,"},{"Start":"03:10.715 ","End":"03:16.550","Text":"a lot of people round it up to 10 meters per second for ease of calculation,"},{"Start":"03:16.550 ","End":"03:19.850","Text":"but the actual number is 9.81,"},{"Start":"03:19.850 ","End":"03:21.230","Text":"and so on, and so forth."},{"Start":"03:21.230 ","End":"03:24.650","Text":"So 9.8 is only to 1 decimal place."},{"Start":"03:24.650 ","End":"03:25.865","Text":"Now, in free fall,"},{"Start":"03:25.865 ","End":"03:30.125","Text":"the motion generated when an object is released from rest at a height,"},{"Start":"03:30.125 ","End":"03:32.323","Text":"so that\u0027s what free fall is,"},{"Start":"03:32.323 ","End":"03:36.150","Text":"so if we have some height over here,"},{"Start":"03:36.150 ","End":"03:37.575","Text":"we\u0027ll call it h,"},{"Start":"03:37.575 ","End":"03:40.100","Text":"and we have some ledge over here,"},{"Start":"03:40.100 ","End":"03:41.629","Text":"and we drop an object,"},{"Start":"03:41.629 ","End":"03:47.630","Text":"it will fall down at the constant acceleration of g. We will choose the axis of"},{"Start":"03:47.630 ","End":"03:50.540","Text":"motion to be the y-axis going in"},{"Start":"03:50.540 ","End":"03:53.870","Text":"the downwards direction because whenever we drop something,"},{"Start":"03:53.870 ","End":"03:57.350","Text":"it drops downwards in the direction of the center of Earth,"},{"Start":"03:57.350 ","End":"03:59.090","Text":"but it drops downwards."},{"Start":"03:59.090 ","End":"04:04.820","Text":"So we can say that this axis is the y-axis,"},{"Start":"04:04.820 ","End":"04:10.449","Text":"and our direction of travel is in the downwards direction,"},{"Start":"04:10.449 ","End":"04:14.479","Text":"so our motion will be going down the y-axis until"},{"Start":"04:14.479 ","End":"04:19.487","Text":"the y value is equal to 0 because that means we\u0027ve hit the ground."},{"Start":"04:19.487 ","End":"04:22.280","Text":"The height is 0. Now, we can label"},{"Start":"04:22.280 ","End":"04:26.904","Text":"this positive direction just to make it a little clearer."},{"Start":"04:26.904 ","End":"04:30.720","Text":"Instead of labeling here 0 being"},{"Start":"04:30.720 ","End":"04:34.770","Text":"the ground and our starting position being at a certain height,"},{"Start":"04:34.770 ","End":"04:37.110","Text":"an easier way could be to,"},{"Start":"04:37.110 ","End":"04:38.330","Text":"instead of label height,"},{"Start":"04:38.330 ","End":"04:45.000","Text":"we would label this over here as our starting position of y_0."},{"Start":"04:45.000 ","End":"04:47.610","Text":"We can say that y_0 equals 0."},{"Start":"04:47.610 ","End":"04:51.395","Text":"So then if here is 0, then when it falls,"},{"Start":"04:51.395 ","End":"04:56.174","Text":"it will fall down the y-axis into the negative numbers,"},{"Start":"04:56.174 ","End":"05:02.330","Text":"and then we\u0027ll also get a representation of the motion just with negative numbers,"},{"Start":"05:02.330 ","End":"05:04.550","Text":"but the motion will be the exact same motion."},{"Start":"05:04.550 ","End":"05:05.750","Text":"Once you do the calculations,"},{"Start":"05:05.750 ","End":"05:07.365","Text":"you\u0027ll get the exact same answer."},{"Start":"05:07.365 ","End":"05:10.860","Text":"If we say that it begins from rest,"},{"Start":"05:10.860 ","End":"05:13.165","Text":"it\u0027s just released from rest from a ledge,"},{"Start":"05:13.165 ","End":"05:15.245","Text":"then we will say that our starting velocity,"},{"Start":"05:15.245 ","End":"05:17.840","Text":"or v_0, is equal to 0."},{"Start":"05:17.840 ","End":"05:21.229","Text":"Now here we can see various equations of motion."},{"Start":"05:21.229 ","End":"05:24.755","Text":"This equation and this equation,"},{"Start":"05:24.755 ","End":"05:29.270","Text":"you\u0027ve seen repeated throughout all of the questions and examples above."},{"Start":"05:29.270 ","End":"05:32.360","Text":"We\u0027ve used these 2 equations quite often."},{"Start":"05:32.360 ","End":"05:33.950","Text":"This equation we\u0027ve also used,"},{"Start":"05:33.950 ","End":"05:36.790","Text":"but these 2 have been our main ones."},{"Start":"05:36.790 ","End":"05:40.250","Text":"Now, what we have on the other side of"},{"Start":"05:40.250 ","End":"05:46.415","Text":"these arrows are these equations just relating to free fall."},{"Start":"05:46.415 ","End":"05:49.325","Text":"For instance, if we look at this first example,"},{"Start":"05:49.325 ","End":"05:53.600","Text":"our velocity is equal to our starting velocity plus 80."},{"Start":"05:53.600 ","End":"05:57.575","Text":"Now, we know that in free fall, usually,"},{"Start":"05:57.575 ","End":"05:59.225","Text":"we\u0027re going to be starting from rest,"},{"Start":"05:59.225 ","End":"06:02.440","Text":"meaning that our starting velocity is going to be equal to 0,"},{"Start":"06:02.440 ","End":"06:04.760","Text":"so that\u0027s why it doesn\u0027t appear here."},{"Start":"06:04.760 ","End":"06:06.950","Text":"Then 80, our acceleration,"},{"Start":"06:06.950 ","End":"06:09.635","Text":"is g, so it\u0027s just gt."},{"Start":"06:09.635 ","End":"06:11.510","Text":"It\u0027s the exact same equation."},{"Start":"06:11.510 ","End":"06:17.510","Text":"Now notice that if we are saying that our y-axis is this,"},{"Start":"06:17.510 ","End":"06:19.250","Text":"and this is the positive direction,"},{"Start":"06:19.250 ","End":"06:21.125","Text":"if we release an object from here,"},{"Start":"06:21.125 ","End":"06:23.150","Text":"and it\u0027s traveling in this downwards direction,"},{"Start":"06:23.150 ","End":"06:27.013","Text":"then it\u0027s in the negative direction,"},{"Start":"06:27.013 ","End":"06:30.641","Text":"so then we would add a minus over here,"},{"Start":"06:30.641 ","End":"06:31.940","Text":"and then of course,"},{"Start":"06:31.940 ","End":"06:33.980","Text":"there\u0027s a minus here because,"},{"Start":"06:33.980 ","End":"06:37.880","Text":"if you remember, if we have our y-axis like this,"},{"Start":"06:37.880 ","End":"06:40.085","Text":"and this is the positive direction,"},{"Start":"06:40.085 ","End":"06:43.640","Text":"and we have an option that is falling in this downwards direction,"},{"Start":"06:43.640 ","End":"06:45.455","Text":"which is the negative direction,"},{"Start":"06:45.455 ","End":"06:49.036","Text":"so in order to show that we have a negative over here."},{"Start":"06:49.036 ","End":"06:52.730","Text":"Then our next equation is for our position."},{"Start":"06:52.730 ","End":"06:56.310","Text":"Our position, our displacement is equal to"},{"Start":"06:56.310 ","End":"07:04.790","Text":"our starting position plus our starting velocity multiplied by t plus 1/2at^2."},{"Start":"07:04.790 ","End":"07:07.625","Text":"Now again, when we\u0027re in free fall,"},{"Start":"07:07.625 ","End":"07:10.790","Text":"we\u0027re starting from position 0,"},{"Start":"07:10.790 ","End":"07:13.985","Text":"remember y_0 equals 0,"},{"Start":"07:13.985 ","End":"07:18.065","Text":"so our starting position is going to be 0, like our origin."},{"Start":"07:18.065 ","End":"07:20.720","Text":"Then, because it\u0027s being released from rest,"},{"Start":"07:20.720 ","End":"07:23.330","Text":"our starting velocity will again be equal 0,"},{"Start":"07:23.330 ","End":"07:24.635","Text":"just like over here."},{"Start":"07:24.635 ","End":"07:29.525","Text":"Then again, our a is going to equal g,"},{"Start":"07:29.525 ","End":"07:32.585","Text":"so we\u0027ll have 1/2gt^2."},{"Start":"07:32.585 ","End":"07:35.410","Text":"Here, of course, we can add the negative,"},{"Start":"07:35.410 ","End":"07:41.570","Text":"just like in this question because our negative direction is downwards."},{"Start":"07:41.570 ","End":"07:45.755","Text":"Then our third equation, again, similarly,"},{"Start":"07:45.755 ","End":"07:48.680","Text":"because our v_0 will be 0,"},{"Start":"07:48.680 ","End":"07:50.890","Text":"because we\u0027re releasing from rest,"},{"Start":"07:50.890 ","End":"07:53.270","Text":"so it disappears, and then we\u0027re left."},{"Start":"07:53.270 ","End":"07:55.475","Text":"Then our a, of course, becomes g,"},{"Start":"07:55.475 ","End":"07:57.410","Text":"because our acceleration is g,"},{"Start":"07:57.410 ","End":"08:01.505","Text":"and then this whole equation becomes this equation."},{"Start":"08:01.505 ","End":"08:07.805","Text":"Now, these 2 equations are the most important equations,"},{"Start":"08:07.805 ","End":"08:11.045","Text":"and you should really remember them off by heart."},{"Start":"08:11.045 ","End":"08:14.165","Text":"However, if you remember these equations off by heart,"},{"Start":"08:14.165 ","End":"08:23.700","Text":"then you simply substitute in v_0 equals 0 when you\u0027re talking about free fall,"},{"Start":"08:23.800 ","End":"08:27.230","Text":"x_0 will equal your y_0,"},{"Start":"08:27.230 ","End":"08:28.811","Text":"which is equal to 0,"},{"Start":"08:28.811 ","End":"08:31.130","Text":"so you\u0027re starting from the origin,"},{"Start":"08:31.130 ","End":"08:34.466","Text":"which will be your y value will equal 0."},{"Start":"08:34.466 ","End":"08:42.680","Text":"Then your acceleration will equal g. So if you remember these 2 equations and this,"},{"Start":"08:42.680 ","End":"08:46.207","Text":"then you can just derive these equations very easily."},{"Start":"08:46.207 ","End":"08:50.620","Text":"Of course, not forgetting to put a negative in front of the g,"},{"Start":"08:50.620 ","End":"08:52.580","Text":"just to denote that you\u0027re going in"},{"Start":"08:52.580 ","End":"08:57.020","Text":"the downwards direction because the downwards direction is the negative direction."},{"Start":"08:57.020 ","End":"09:00.630","Text":"That\u0027s the end of this lesson."}],"ID":9229},{"Watched":false,"Name":"Example-Two Rocks Fall From A Building","Duration":"13m 49s","ChapterTopicVideoID":8975,"CourseChapterTopicPlaylistID":5376,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:04.845","Text":"Hello. The following questions will be practicing what we know"},{"Start":"00:04.845 ","End":"00:08.520","Text":"right now about free-fall using certain data."},{"Start":"00:08.520 ","End":"00:15.610","Text":"In this question, we\u0027re being told that Julie releases a rock from a height of 30 meters."},{"Start":"00:15.650 ","End":"00:18.420","Text":"Let\u0027s draw this."},{"Start":"00:18.420 ","End":"00:25.920","Text":"Here is 30 meters and Edna is standing at a height of 15 meters,"},{"Start":"00:25.920 ","End":"00:29.265","Text":"so here\u0027s Edna at 15 meters."},{"Start":"00:29.265 ","End":"00:32.705","Text":"She releases a rock at the same time as Julie."},{"Start":"00:32.705 ","End":"00:34.515","Text":"There\u0027s another rock here."},{"Start":"00:34.515 ","End":"00:36.735","Text":"Then our first question is,"},{"Start":"00:36.735 ","End":"00:40.290","Text":"at what velocities will the 2 rocks hit the ground?"},{"Start":"00:40.290 ","End":"00:44.205","Text":"Sorry, this is meant to be rocks, hit the ground."},{"Start":"00:44.205 ","End":"00:48.575","Text":"Now first things first we know, I\u0027ll write them over here."},{"Start":"00:48.575 ","End":"00:56.130","Text":"Then we have our equations of v equals v_0 plus at and we also know"},{"Start":"00:56.130 ","End":"01:05.280","Text":"that our x equals x_0 plus v_0t plus 1/2at."},{"Start":"01:05.280 ","End":"01:08.645","Text":"Now we\u0027re going to use these equations in order to solve the question."},{"Start":"01:08.645 ","End":"01:11.150","Text":"The first thing that we\u0027re going to do is we\u0027re going to find"},{"Start":"01:11.150 ","End":"01:16.030","Text":"the velocity with which Julie\u0027s rock will hit the ground."},{"Start":"01:16.030 ","End":"01:23.090","Text":"Our velocity of Julie\u0027s rock is going to equal starting velocity,"},{"Start":"01:23.090 ","End":"01:26.015","Text":"which is 0, because it\u0027s released,"},{"Start":"01:26.015 ","End":"01:28.400","Text":"which usually means released from rest,"},{"Start":"01:28.400 ","End":"01:33.800","Text":"and then our acceleration because it\u0027s just free fall downwards,"},{"Start":"01:33.800 ","End":"01:37.070","Text":"will be g so in our questions,"},{"Start":"01:37.070 ","End":"01:43.715","Text":"we\u0027re going to refer to g as 10 meters per second squared just for ease of calculation."},{"Start":"01:43.715 ","End":"01:49.190","Text":"It\u0027s all around it. Because a positive direction"},{"Start":"01:49.190 ","End":"01:53.690","Text":"is this direction and because this rock will be falling downwards,"},{"Start":"01:53.690 ","End":"01:55.340","Text":"so it\u0027s in the negative direction."},{"Start":"01:55.340 ","End":"01:57.200","Text":"So we have negative acceleration,"},{"Start":"01:57.200 ","End":"02:02.120","Text":"so we have negative 10 multiplied by t. This is the velocity,"},{"Start":"02:02.120 ","End":"02:06.560","Text":"but now we want to find what a t is, a t for Julie."},{"Start":"02:06.560 ","End":"02:10.835","Text":"We\u0027re going to plug this into this equation."},{"Start":"02:10.835 ","End":"02:17.820","Text":"Now we\u0027re going to say that starting our x_0 is going to be 0."},{"Start":"02:17.820 ","End":"02:22.070","Text":"Why is our x_0 0 even though we know that we\u0027re at a height of 30?"},{"Start":"02:22.070 ","End":"02:23.510","Text":"Because for ease of calculation,"},{"Start":"02:23.510 ","End":"02:27.050","Text":"it\u0027s easier to say that our starting position is at the origin,"},{"Start":"02:27.050 ","End":"02:31.820","Text":"meaning that it\u0027s at 0 and then it travels towards its final position,"},{"Start":"02:31.820 ","End":"02:35.065","Text":"which will be then negative 30."},{"Start":"02:35.065 ","End":"02:42.450","Text":"So we say that this is the y-axis and here is the 0,"},{"Start":"02:42.450 ","End":"02:44.190","Text":"so when it gets to the ground,"},{"Start":"02:44.190 ","End":"02:47.700","Text":"it will be a height of negative 30."},{"Start":"02:47.700 ","End":"02:49.580","Text":"Do you see that?"},{"Start":"02:49.580 ","End":"02:55.020","Text":"Because it is for ease of calculation. Let\u0027s see."},{"Start":"02:55.540 ","End":"03:01.130","Text":"Our X will be negative 30 because that\u0027s our final position that we want to be at,"},{"Start":"03:01.130 ","End":"03:04.935","Text":"which will equal our starting position, which is 0,"},{"Start":"03:04.935 ","End":"03:09.500","Text":"because we start at the origin plus v_0 again,"},{"Start":"03:09.500 ","End":"03:11.240","Text":"because it\u0027s being released from rest,"},{"Start":"03:11.240 ","End":"03:17.915","Text":"will be 0 times t plus 1/2 multiplied by acceleration,"},{"Start":"03:17.915 ","End":"03:20.720","Text":"which again will be negative 10,"},{"Start":"03:20.720 ","End":"03:26.305","Text":"because we\u0027re going in the downwards direction, multiplied by t^2."},{"Start":"03:26.305 ","End":"03:28.530","Text":"Now we can simplify this."},{"Start":"03:28.530 ","End":"03:32.060","Text":"This and this cross off because they\u0027re 0 and negative 10"},{"Start":"03:32.060 ","End":"03:36.930","Text":"divided by 2 will equal negative 5t^2."},{"Start":"03:37.420 ","End":"03:42.275","Text":"Then we can divide both sides by negative 5,"},{"Start":"03:42.275 ","End":"03:47.730","Text":"so then we\u0027ll get that the negatives cross off,"},{"Start":"03:47.730 ","End":"03:54.990","Text":"6=t^2, meaning that t is equal to root 6."},{"Start":"03:54.990 ","End":"03:59.085","Text":"If this is our t for Julie,"},{"Start":"03:59.085 ","End":"04:02.595","Text":"so we substitute this in to here."},{"Start":"04:02.595 ","End":"04:05.370","Text":"Now from this equation,"},{"Start":"04:05.370 ","End":"04:14.560","Text":"we can get that our velocity for Julie is equal to negative 10 multiplied by root 6,"},{"Start":"04:14.560 ","End":"04:20.895","Text":"which just equals negative 10 root 6."},{"Start":"04:20.895 ","End":"04:28.485","Text":"This is the velocity that the rock that Julie drops will hit the ground at."},{"Start":"04:28.485 ","End":"04:32.680","Text":"Now we\u0027re going to do the exact same thing for Edna\u0027s rock,"},{"Start":"04:32.680 ","End":"04:39.010","Text":"so we\u0027ll have our velocity of Edna\u0027s rock is equal to our starting velocity,"},{"Start":"04:39.010 ","End":"04:43.315","Text":"which again is 0 because Edna releases her rock from rest as well,"},{"Start":"04:43.315 ","End":"04:46.885","Text":"plus my acceleration multiplied by time."},{"Start":"04:46.885 ","End":"04:50.470","Text":"Now my acceleration is the same for both the rocks,"},{"Start":"04:50.470 ","End":"04:56.390","Text":"so I\u0027m going to have negative 10 multiplied by time Edna."},{"Start":"04:56.390 ","End":"04:57.920","Text":"Again, we don\u0027t know the time,"},{"Start":"04:57.920 ","End":"05:02.630","Text":"so we\u0027re going to use this equation to find out our time."},{"Start":"05:02.630 ","End":"05:10.294","Text":"Once again, if this is the y-axis and this is 0,"},{"Start":"05:10.294 ","End":"05:13.875","Text":"then here will be negative 15."},{"Start":"05:13.875 ","End":"05:16.775","Text":"For the exact same reason we did here,"},{"Start":"05:16.775 ","End":"05:18.530","Text":"because it\u0027s for ease of calculation,"},{"Start":"05:18.530 ","End":"05:26.395","Text":"we can start at 0 at the origin and it drops 15 meters, negative 15."},{"Start":"05:26.395 ","End":"05:29.660","Text":"Let\u0027s see. End destination,"},{"Start":"05:29.660 ","End":"05:31.820","Text":"we want it to be negative 15,"},{"Start":"05:31.820 ","End":"05:34.144","Text":"which equals our starting position,"},{"Start":"05:34.144 ","End":"05:37.580","Text":"which is going to be 0 because we\u0027re starting at the origin"},{"Start":"05:37.580 ","End":"05:42.020","Text":"plus our starting velocity multiplied by time."},{"Start":"05:42.020 ","End":"05:44.870","Text":"Our starting velocity is 0 because it\u0027s released from"},{"Start":"05:44.870 ","End":"05:49.550","Text":"rest plus 1/2 multiplied by acceleration,"},{"Start":"05:49.550 ","End":"05:54.800","Text":"which again is negative 10 multiplied by t of Edna squared."},{"Start":"05:54.800 ","End":"05:56.090","Text":"Let\u0027s see what this is."},{"Start":"05:56.090 ","End":"06:03.515","Text":"This and this cross off and then we have that negative 15 is equal to once again,"},{"Start":"06:03.515 ","End":"06:11.080","Text":"negative 10 divided by 2, negative 5tE^2."},{"Start":"06:11.080 ","End":"06:13.880","Text":"Then if we divide both sides by negative 5,"},{"Start":"06:13.880 ","End":"06:16.010","Text":"we get rid of the negative sign,"},{"Start":"06:16.010 ","End":"06:21.390","Text":"and then we\u0027ll have that 3=t^2,"},{"Start":"06:21.390 ","End":"06:24.680","Text":"and therefore we\u0027ll get that our time for Edna\u0027s rock to"},{"Start":"06:24.680 ","End":"06:28.380","Text":"hit the ground will equal root 3."},{"Start":"06:28.380 ","End":"06:35.055","Text":"Then again, we substitute this value into this equation and then we\u0027ll"},{"Start":"06:35.055 ","End":"06:38.450","Text":"get that our velocity for Edna\u0027s rock when it hits"},{"Start":"06:38.450 ","End":"06:42.770","Text":"the ground will equal negative 10 multiplied by our time,"},{"Start":"06:42.770 ","End":"06:45.645","Text":"which is root 3."},{"Start":"06:45.645 ","End":"06:52.040","Text":"Julie\u0027s rock will hit the ground at a velocity of negative 10 root 6 meters per"},{"Start":"06:52.040 ","End":"06:59.170","Text":"second and Edna\u0027s rock will hit the ground at negative 10 root 3 meters per second."},{"Start":"06:59.170 ","End":"07:02.240","Text":"Now question number 2 is asking what will"},{"Start":"07:02.240 ","End":"07:06.080","Text":"be the time difference between the 3 rocks hitting the ground?"},{"Start":"07:06.080 ","End":"07:08.615","Text":"We just worked out in question number 1,"},{"Start":"07:08.615 ","End":"07:10.900","Text":"that rock number 1,"},{"Start":"07:10.900 ","End":"07:15.380","Text":"that Julie\u0027s rock will hit the ground at root 6 seconds"},{"Start":"07:15.380 ","End":"07:20.450","Text":"and we found out that Edna\u0027s rock will hit the ground at root 3 seconds,"},{"Start":"07:20.450 ","End":"07:24.605","Text":"so the time difference between the 2 rocks hitting the ground"},{"Start":"07:24.605 ","End":"07:29.535","Text":"will be root 6 minus root 3."},{"Start":"07:29.535 ","End":"07:35.005","Text":"That is the answer to question number 2."},{"Start":"07:35.005 ","End":"07:37.370","Text":"Now let\u0027s take a look at question 3."},{"Start":"07:37.370 ","End":"07:44.430","Text":"Now we\u0027re being told that Edna releases her rock only once Julie\u0027s rock passes her."},{"Start":"07:45.470 ","End":"07:50.495","Text":"What will be the time difference between the 2 hits now?"},{"Start":"07:50.495 ","End":"07:54.915","Text":"Julie drops her rock and the rock falls,"},{"Start":"07:54.915 ","End":"08:01.970","Text":"and only once Julie\u0027s rock is mid fall and reaches this point where Edna\u0027s standing,"},{"Start":"08:01.970 ","End":"08:05.560","Text":"Edna will release her rock as well."},{"Start":"08:05.560 ","End":"08:10.115","Text":"I\u0027m going to rub out all of this to make some more space."},{"Start":"08:10.115 ","End":"08:13.550","Text":"Now, we already found out in question number 1 that"},{"Start":"08:13.550 ","End":"08:17.600","Text":"Julie\u0027s rock takes root 6 seconds to hit the ground,"},{"Start":"08:17.600 ","End":"08:23.190","Text":"so t of Julie equals root 6 seconds."},{"Start":"08:23.190 ","End":"08:25.010","Text":"This is unchanging."},{"Start":"08:25.010 ","End":"08:28.985","Text":"The only difference now we have to find is how long"},{"Start":"08:28.985 ","End":"08:34.020","Text":"Edna\u0027s rock will be in flight and when will Edna release her rock."},{"Start":"08:34.020 ","End":"08:35.945","Text":"Then after we find that out,"},{"Start":"08:35.945 ","End":"08:40.890","Text":"we have to find the time difference between the 2 hits."},{"Start":"08:40.890 ","End":"08:42.795","Text":"This is a slightly harder question,"},{"Start":"08:42.795 ","End":"08:45.255","Text":"let\u0027s see how we do this."},{"Start":"08:45.255 ","End":"08:47.850","Text":"Let\u0027s draw this data."},{"Start":"08:47.850 ","End":"08:51.780","Text":"We know that for Julie\u0027s rock to fall 30 meters,"},{"Start":"08:51.780 ","End":"08:57.630","Text":"it took root 6 seconds to fall all of this."},{"Start":"08:57.630 ","End":"09:03.720","Text":"Then we know that Edna\u0027s rock fell 15 meters in root 3 seconds."},{"Start":"09:03.720 ","End":"09:08.900","Text":"Remember, t of Edna was equal to root 3 seconds."},{"Start":"09:08.900 ","End":"09:11.840","Text":"We also worked that out in question number 1."},{"Start":"09:11.840 ","End":"09:22.605","Text":"That means that this distance is root 3 and this distance is root 6."},{"Start":"09:22.605 ","End":"09:28.790","Text":"Now, if Edna\u0027s rock took root 3 seconds to fall 15 meters,"},{"Start":"09:28.790 ","End":"09:34.670","Text":"the difference between Julie\u0027s position and Edna\u0027s position is also 15 meters,"},{"Start":"09:34.670 ","End":"09:36.500","Text":"which is the same 15 meters."},{"Start":"09:36.500 ","End":"09:38.615","Text":"Let me just draw this in."},{"Start":"09:38.615 ","End":"09:43.430","Text":"The distance between here and here is also 15 meters,"},{"Start":"09:43.430 ","End":"09:47.450","Text":"meaning that the distance that Julie\u0027s rock will"},{"Start":"09:47.450 ","End":"09:51.935","Text":"have to fall until Edna releases her rock as which means that the time,"},{"Start":"09:51.935 ","End":"09:54.050","Text":"because it\u0027s the exact same 15 meters,"},{"Start":"09:54.050 ","End":"09:58.255","Text":"will also be root 3 seconds."},{"Start":"09:58.255 ","End":"10:06.295","Text":"Then in order to find the time difference between these 2 hits now that Edna has waited,"},{"Start":"10:06.295 ","End":"10:08.780","Text":"so this is what we\u0027re going to have to do."},{"Start":"10:08.780 ","End":"10:11.665","Text":"The time difference will equal,"},{"Start":"10:11.665 ","End":"10:16.245","Text":"it takes root 6 seconds for Julie\u0027s rock to fall."},{"Start":"10:16.245 ","End":"10:21.120","Text":"So we\u0027re first going to write root 6 and then we"},{"Start":"10:21.120 ","End":"10:28.250","Text":"minus the time it takes for Edna to wait for Julie\u0027s rock to fall,"},{"Start":"10:28.250 ","End":"10:32.025","Text":"which is root 3 seconds because it takes"},{"Start":"10:32.025 ","End":"10:36.245","Text":"root 3 seconds for Julie\u0027s rock to reach Edna\u0027s position,"},{"Start":"10:36.245 ","End":"10:40.467","Text":"so that will be root 3 seconds,"},{"Start":"10:40.467 ","End":"10:44.535","Text":"plus the time it takes for then"},{"Start":"10:44.535 ","End":"10:48.815","Text":"Edna\u0027s to drop her rock and for her rock to hit the ground,"},{"Start":"10:48.815 ","End":"10:53.180","Text":"which is another root 3 seconds."},{"Start":"10:53.180 ","End":"10:57.785","Text":"Then we\u0027re going to get that our time difference is root 6,"},{"Start":"10:57.785 ","End":"11:02.970","Text":"negative 2, root 3."},{"Start":"11:02.970 ","End":"11:05.905","Text":"Let me explain this one more time."},{"Start":"11:05.905 ","End":"11:08.740","Text":"Julie\u0027s rock takes a total of root"},{"Start":"11:08.740 ","End":"11:13.645","Text":"6 seconds to hit the ground from being released at a height of 30."},{"Start":"11:13.645 ","End":"11:16.540","Text":"Edna\u0027s rock takes a time of root"},{"Start":"11:16.540 ","End":"11:21.609","Text":"3 seconds from release until it hits the ground at a height of 15."},{"Start":"11:21.609 ","End":"11:26.365","Text":"The height difference between Julie and Edna is also 15 meters,"},{"Start":"11:26.365 ","End":"11:30.700","Text":"meaning that the same time that it takes for Edna\u0027s rock to fall and hit"},{"Start":"11:30.700 ","End":"11:36.340","Text":"the ground is the same time it will take for Julie\u0027s rock to fall and reach Edna."},{"Start":"11:36.340 ","End":"11:42.090","Text":"We have here, this time taken is root 3,"},{"Start":"11:42.090 ","End":"11:44.330","Text":"and this time taken is also root 3."},{"Start":"11:44.330 ","End":"11:46.640","Text":"Then in order to find the time difference,"},{"Start":"11:46.640 ","End":"11:52.770","Text":"we\u0027re going to say that Julie drops her rock and it takes root 6 seconds,"},{"Start":"11:52.770 ","End":"11:59.735","Text":"negative the time that it takes for Edna to wait until she releases the rock,"},{"Start":"11:59.735 ","End":"12:03.530","Text":"which is until Julie\u0027s rock reaches her position,"},{"Start":"12:03.530 ","End":"12:06.920","Text":"which is root 3 seconds plus"},{"Start":"12:06.920 ","End":"12:12.320","Text":"the time that it takes for Edna\u0027s rock upon being released to hit the ground,"},{"Start":"12:12.320 ","End":"12:17.095","Text":"so it will be root 6 negative 2 times root 3."},{"Start":"12:17.095 ","End":"12:19.530","Text":"That\u0027s the end of question number 3."},{"Start":"12:19.530 ","End":"12:22.095","Text":"Now let\u0027s take a look at question number 4."},{"Start":"12:22.095 ","End":"12:27.680","Text":"How long must Edna wait after the release of Julie\u0027s rock in order to release"},{"Start":"12:27.680 ","End":"12:30.755","Text":"her rock if she would want both her rock"},{"Start":"12:30.755 ","End":"12:34.235","Text":"and Julie\u0027s rock to hit the ground simultaneously?"},{"Start":"12:34.235 ","End":"12:35.450","Text":"Let\u0027s take a look."},{"Start":"12:35.450 ","End":"12:39.305","Text":"Julie\u0027s rock takes root 6 seconds to hit the ground,"},{"Start":"12:39.305 ","End":"12:44.915","Text":"which means that Edna needs that after root 6 seconds,"},{"Start":"12:44.915 ","End":"12:49.530","Text":"both her rock and Julie\u0027s rock will have reached the ground,"},{"Start":"12:49.530 ","End":"12:51.620","Text":"so how long does Edna have to wait?"},{"Start":"12:51.620 ","End":"12:55.175","Text":"I know that the total time is root 6 seconds,"},{"Start":"12:55.175 ","End":"13:01.685","Text":"and I know that the time taken for Edna\u0027s rock to fall to the ground is root 3 seconds,"},{"Start":"13:01.685 ","End":"13:04.135","Text":"so she has to wait root 6,"},{"Start":"13:04.135 ","End":"13:07.395","Text":"negative root 3 seconds,"},{"Start":"13:07.395 ","End":"13:11.180","Text":"and then both the rocks will fall at the exact same time."},{"Start":"13:11.180 ","End":"13:12.755","Text":"Now we can check this,"},{"Start":"13:12.755 ","End":"13:18.740","Text":"if we look if Julie drops her rock and it takes root 6 seconds to hit the ground,"},{"Start":"13:18.740 ","End":"13:20.540","Text":"and this is our total time needed."},{"Start":"13:20.540 ","End":"13:23.780","Text":"Then Edna waits root 3 seconds,"},{"Start":"13:23.780 ","End":"13:27.075","Text":"so negative root 3,"},{"Start":"13:27.075 ","End":"13:32.625","Text":"and then it takes her rock another root 3 seconds to fall,"},{"Start":"13:32.625 ","End":"13:34.580","Text":"so then these 2 cross out,"},{"Start":"13:34.580 ","End":"13:39.035","Text":"which just leaves us with roots 6 seconds, which as we know,"},{"Start":"13:39.035 ","End":"13:44.295","Text":"is going to be our total time so that\u0027s how we know that it\u0027s correct."},{"Start":"13:44.295 ","End":"13:49.810","Text":"That\u0027s the end of this question and we\u0027re going to move on to harder questions next."}],"ID":9230},{"Watched":false,"Name":"Vertical Trajectory","Duration":"9m 36s","ChapterTopicVideoID":8976,"CourseChapterTopicPlaylistID":5376,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.500 ","End":"00:02.970","Text":"Hello. In this lesson,"},{"Start":"00:02.970 ","End":"00:06.030","Text":"we\u0027re going to be speaking about vertical trajectories."},{"Start":"00:06.030 ","End":"00:10.470","Text":"Now, what does this mean? We saw previously a lesson about free fall."},{"Start":"00:10.470 ","End":"00:14.730","Text":"Now what\u0027s the difference between vertical trajectory and freefall?"},{"Start":"00:14.730 ","End":"00:18.900","Text":"In freefall, we have an object and we drop it and it falls"},{"Start":"00:18.900 ","End":"00:23.985","Text":"down freely according to gravitational acceleration."},{"Start":"00:23.985 ","End":"00:28.545","Text":"That\u0027s freefall. Now vertical trajectory, on the other hand,"},{"Start":"00:28.545 ","End":"00:31.890","Text":"is if there\u0027s an object and we provide"},{"Start":"00:31.890 ","End":"00:37.290","Text":"some force which flings the object upwards or downwards."},{"Start":"00:37.290 ","End":"00:39.220","Text":"What is vertical trajectory?"},{"Start":"00:39.220 ","End":"00:42.535","Text":"When the body has a starting velocity in the y direction."},{"Start":"00:42.535 ","End":"00:49.700","Text":"There\u0027s some force that throws the ball upwards or downwards."},{"Start":"00:49.700 ","End":"00:52.955","Text":"In the y direction means upwards or downwards."},{"Start":"00:52.955 ","End":"01:00.090","Text":"As we know, this is the y-axis and this would be the x-axis."},{"Start":"01:00.560 ","End":"01:05.390","Text":"In vertical trajectories, the acceleration is always downwards."},{"Start":"01:05.390 ","End":"01:07.550","Text":"Even if we throw the object upwards,"},{"Start":"01:07.550 ","End":"01:11.135","Text":"there\u0027s always g or gravitational acceleration,"},{"Start":"01:11.135 ","End":"01:14.510","Text":"which is trying to pull the object down with an acceleration of,"},{"Start":"01:14.510 ","End":"01:18.670","Text":"we saw 9.8 meters per second to 1 decimal point."},{"Start":"01:18.670 ","End":"01:20.525","Text":"When throwing a body downwards,"},{"Start":"01:20.525 ","End":"01:23.450","Text":"we will choose the direction of our axes to be downwards."},{"Start":"01:23.450 ","End":"01:27.725","Text":"When during upwards, we will choose the direction of our axes to be upwards."},{"Start":"01:27.725 ","End":"01:32.900","Text":"Our a or acceleration will equal negative g. If"},{"Start":"01:32.900 ","End":"01:38.074","Text":"there\u0027s someone standing here and she throws the ball upwards,"},{"Start":"01:38.074 ","End":"01:39.800","Text":"then the ball will go upwards,"},{"Start":"01:39.800 ","End":"01:44.840","Text":"but it will decelerate until it reaches a stop at the top."},{"Start":"01:44.840 ","End":"01:49.370","Text":"V final at the top will equal 0 and it will"},{"Start":"01:49.370 ","End":"01:54.035","Text":"experience a deceleration of g. Then it will fall."},{"Start":"01:54.035 ","End":"01:55.805","Text":"When it comes back down."},{"Start":"01:55.805 ","End":"01:57.860","Text":"It will be accelerating."},{"Start":"01:57.860 ","End":"02:03.050","Text":"An acceleration of g when it\u0027s coming back down."},{"Start":"02:03.050 ","End":"02:04.505","Text":"But when it\u0027s going up,"},{"Start":"02:04.505 ","End":"02:07.715","Text":"the acceleration will equal negative g,"},{"Start":"02:07.715 ","End":"02:10.295","Text":"because g will be decelerating."},{"Start":"02:10.295 ","End":"02:16.011","Text":"If this girl were to throw her object downwards,"},{"Start":"02:16.011 ","End":"02:20.690","Text":"then the acceleration would be slightly different because then we would have"},{"Start":"02:20.690 ","End":"02:28.955","Text":"an acceleration of g plus starting force over mass,"},{"Start":"02:28.955 ","End":"02:33.200","Text":"which equals acceleration, if you remember, a equals MF."},{"Start":"02:33.200 ","End":"02:35.975","Text":"Now over here, when throwing upwards,"},{"Start":"02:35.975 ","End":"02:37.865","Text":"when the body reaches its peak height,"},{"Start":"02:37.865 ","End":"02:39.200","Text":"its velocity will equal 0,"},{"Start":"02:39.200 ","End":"02:41.090","Text":"which is what we spoke about."},{"Start":"02:41.090 ","End":"02:44.194","Text":"If someone throws this rock upwards,"},{"Start":"02:44.194 ","End":"02:47.390","Text":"because gravity is working against this object,"},{"Start":"02:47.390 ","End":"02:53.195","Text":"our object will be accelerating at negative g or decelerating,"},{"Start":"02:53.195 ","End":"02:58.470","Text":"and will go up and up and up until it reaches its peak height."},{"Start":"02:58.470 ","End":"03:05.690","Text":"Height max, where its velocity will equal 0,"},{"Start":"03:05.690 ","End":"03:11.870","Text":"and then it will begin dropping down again in this direction at"},{"Start":"03:11.870 ","End":"03:18.440","Text":"an acceleration of g. That\u0027s exactly what we spoke about also on the last page."},{"Start":"03:18.440 ","End":"03:23.750","Text":"Then we can work out the time taken to reach its height by using our velocity equation."},{"Start":"03:23.750 ","End":"03:27.140","Text":"Remember our favorite equation,"},{"Start":"03:27.140 ","End":"03:31.675","Text":"v(t) equals v_0 plus at."},{"Start":"03:31.675 ","End":"03:33.620","Text":"This is our equation."},{"Start":"03:33.620 ","End":"03:38.090","Text":"If we sub in that a is equal to negative g,"},{"Start":"03:38.090 ","End":"03:39.410","Text":"then we get this equation,"},{"Start":"03:39.410 ","End":"03:41.390","Text":"v equals negative gt."},{"Start":"03:41.390 ","End":"03:46.520","Text":"Then if we want to find what our time is to reach maximum height,"},{"Start":"03:46.520 ","End":"03:49.820","Text":"we know that our final velocity will be equal to 0."},{"Start":"03:49.820 ","End":"03:52.670","Text":"Then if we substitute into this equation 0,"},{"Start":"03:52.670 ","End":"03:57.595","Text":"we\u0027ll get 0 equals v_0 minus gt."},{"Start":"03:57.595 ","End":"04:00.680","Text":"Then if we rearrange this equation,"},{"Start":"04:00.680 ","End":"04:03.470","Text":"then we\u0027ll get the t equals v_0 over"},{"Start":"04:03.470 ","End":"04:07.400","Text":"g. To get a little bit of intuition about what this equation"},{"Start":"04:07.400 ","End":"04:15.410","Text":"means is that if I give a higher starting velocity to my object and I\u0027m throwing,"},{"Start":"04:15.410 ","End":"04:21.290","Text":"then the time taken for my object to reach its maximum height will be larger."},{"Start":"04:21.290 ","End":"04:26.165","Text":"Now, in order to calculate the time taken for the body to return to the starting point,"},{"Start":"04:26.165 ","End":"04:31.055","Text":"we will substitute in y equals 0 into the following equation."},{"Start":"04:31.055 ","End":"04:33.350","Text":"We know this equation."},{"Start":"04:33.350 ","End":"04:36.635","Text":"It\u0027s the same as when we had the x(t), remember,"},{"Start":"04:36.635 ","End":"04:46.245","Text":"of the form x(t) equals x_0 plus v_0t plus 1/2at^2."},{"Start":"04:46.245 ","End":"04:47.975","Text":"This is the exact same equation,"},{"Start":"04:47.975 ","End":"04:50.900","Text":"but instead of being for the x-axis,"},{"Start":"04:50.900 ","End":"04:56.850","Text":"we\u0027re doing it for the y-axis because we\u0027re throwing in the direction of the y-axis."},{"Start":"04:56.850 ","End":"04:59.830","Text":"Now we cancel out."},{"Start":"04:59.830 ","End":"05:03.155","Text":"Here, we could say that we have plus y_0."},{"Start":"05:03.155 ","End":"05:08.240","Text":"We cancel out this y_0 because our starting position,"},{"Start":"05:08.240 ","End":"05:10.195","Text":"we\u0027re going to set at the origin"},{"Start":"05:10.195 ","End":"05:14.270","Text":"because we like starting at the origin for our ease of calculation."},{"Start":"05:14.270 ","End":"05:17.000","Text":"This is 0, so we can just rub it out."},{"Start":"05:17.000 ","End":"05:20.240","Text":"We don\u0027t have to take it into account in our calculations."},{"Start":"05:20.240 ","End":"05:23.570","Text":"Then in our equation just like over here,"},{"Start":"05:23.570 ","End":"05:26.000","Text":"we have our starting velocity multiplied by"},{"Start":"05:26.000 ","End":"05:28.700","Text":"t. We know that our starting velocity is not going to"},{"Start":"05:28.700 ","End":"05:34.035","Text":"be equal to 0 because we know that the body is given an initial force,"},{"Start":"05:34.035 ","End":"05:36.690","Text":"an initial starting velocity."},{"Start":"05:36.690 ","End":"05:38.915","Text":"It\u0027s not starting from rest,"},{"Start":"05:38.915 ","End":"05:40.955","Text":"which is also different from freefall,"},{"Start":"05:40.955 ","End":"05:45.215","Text":"because in freefall an object is released from rest and falls downwards."},{"Start":"05:45.215 ","End":"05:49.585","Text":"Here know an object is thrown upwards or downwards."},{"Start":"05:49.585 ","End":"05:52.785","Text":"Then plus 1/2at^2."},{"Start":"05:52.785 ","End":"05:55.740","Text":"Again our a being"},{"Start":"05:55.740 ","End":"05:58.720","Text":"either g or negative g"},{"Start":"05:58.720 ","End":"06:02.735","Text":"depending on whether we\u0027re throwing the object upwards or downwards."},{"Start":"06:02.735 ","End":"06:09.155","Text":"Then if we substitute all of this in and we start with our y equaling 0,"},{"Start":"06:09.155 ","End":"06:12.279","Text":"because we\u0027re starting at the origin as we just said,"},{"Start":"06:12.279 ","End":"06:17.945","Text":"then we\u0027ll get that the time taken for our object to reach"},{"Start":"06:17.945 ","End":"06:24.665","Text":"its maximum height and then come back down is going to be t equals 0,"},{"Start":"06:24.665 ","End":"06:29.900","Text":"because at time 0 our object will be at the origin over here,"},{"Start":"06:29.900 ","End":"06:31.487","Text":"if this is our object."},{"Start":"06:31.487 ","End":"06:33.560","Text":"Then second t,"},{"Start":"06:33.560 ","End":"06:37.580","Text":"our t_2 is going to be once the object has reached"},{"Start":"06:37.580 ","End":"06:43.070","Text":"its maximum height and fallen back down to reach the same point again."},{"Start":"06:43.070 ","End":"06:45.860","Text":"That\u0027s going to be at t_2,"},{"Start":"06:45.860 ","End":"06:47.570","Text":"which is 2v_0,"},{"Start":"06:47.570 ","End":"06:52.860","Text":"our initial velocity divided by g acceleration."},{"Start":"06:53.270 ","End":"06:56.090","Text":"This is something that\u0027s really handy to"},{"Start":"06:56.090 ","End":"06:59.090","Text":"remember because it will make your calculations a lot faster."},{"Start":"06:59.090 ","End":"07:00.710","Text":"If you don\u0027t remember this,"},{"Start":"07:00.710 ","End":"07:06.260","Text":"just remember this equation or this equation and instead of the x\u0027s,"},{"Start":"07:06.260 ","End":"07:08.415","Text":"you substitute in for y."},{"Start":"07:08.415 ","End":"07:10.095","Text":"Instead of your x_0,"},{"Start":"07:10.095 ","End":"07:16.065","Text":"you know that your y_0 it\u0027s going to be 0 because your starting at the origin."},{"Start":"07:16.065 ","End":"07:17.375","Text":"Then again over here,"},{"Start":"07:17.375 ","End":"07:19.670","Text":"your end position will also be 0."},{"Start":"07:19.670 ","End":"07:22.385","Text":"Then you just rearrange to find your 2t\u0027s."},{"Start":"07:22.385 ","End":"07:29.240","Text":"This one representing the starting time when your object is at the origin at t=0,"},{"Start":"07:29.240 ","End":"07:32.180","Text":"and t number 2 is when your object returns."},{"Start":"07:32.180 ","End":"07:34.400","Text":"Notice that the time taken to return to"},{"Start":"07:34.400 ","End":"07:38.570","Text":"the starting position is twice that of reaching the maximum height."},{"Start":"07:38.570 ","End":"07:40.895","Text":"If you have a question that asks you,"},{"Start":"07:40.895 ","End":"07:45.385","Text":"how long will it take for your object to reach maximum height?"},{"Start":"07:45.385 ","End":"07:48.035","Text":"Then you work it out and you figure it out,"},{"Start":"07:48.035 ","End":"07:50.375","Text":"I don\u0027t know, for example, 2 seconds."},{"Start":"07:50.375 ","End":"07:52.865","Text":"Then if in the 2nd question they ask you,"},{"Start":"07:52.865 ","End":"07:58.155","Text":"what will be the time taken for the object to come back to its starting position?"},{"Start":"07:58.155 ","End":"08:01.083","Text":"Then you know that it will just be 2 times this,"},{"Start":"08:01.083 ","End":"08:03.230","Text":"so it\u0027ll be 4 seconds."},{"Start":"08:03.230 ","End":"08:07.070","Text":"Now, we can calculate the maximum height that the body will reach by"},{"Start":"08:07.070 ","End":"08:12.440","Text":"substituting in the time to reach the maximum height into our position equation."},{"Start":"08:12.440 ","End":"08:17.240","Text":"Now, we already saw a few minutes ago that we can figure out the time taken to"},{"Start":"08:17.240 ","End":"08:22.100","Text":"reach our maximum height by substituting in that our velocity will be equal to 0."},{"Start":"08:22.100 ","End":"08:24.710","Text":"Then we saw that we will find that the time taken"},{"Start":"08:24.710 ","End":"08:27.820","Text":"to reach this maximum height will be v_0,"},{"Start":"08:27.820 ","End":"08:30.635","Text":"our starting velocity divided by g,"},{"Start":"08:30.635 ","End":"08:34.580","Text":"which will be a deceleration in this point."},{"Start":"08:34.580 ","End":"08:38.165","Text":"Now if we take this t and substitute this"},{"Start":"08:38.165 ","End":"08:41.525","Text":"into our position equation that we already know,"},{"Start":"08:41.525 ","End":"08:44.885","Text":"then we will get that our maximum height,"},{"Start":"08:44.885 ","End":"08:47.570","Text":"our maximum y-value in this case,"},{"Start":"08:47.570 ","End":"08:52.555","Text":"will equal our starting velocity squared divided by 2g."},{"Start":"08:52.555 ","End":"08:55.520","Text":"Again, this is useful to remember by height."},{"Start":"08:55.520 ","End":"08:59.104","Text":"If you don\u0027t, if you just remember this equation,"},{"Start":"08:59.104 ","End":"09:02.480","Text":"you can derive this solution yourself."},{"Start":"09:02.480 ","End":"09:04.700","Text":"Now a little point. Why is this a minus?"},{"Start":"09:04.700 ","End":"09:07.850","Text":"Because we know that if we throw something in the upwards direction,"},{"Start":"09:07.850 ","End":"09:09.920","Text":"it\u0027s decelerating at a rate of g,"},{"Start":"09:09.920 ","End":"09:15.809","Text":"which means that our acceleration is negative g. That\u0027s why this is a negative here."},{"Start":"09:15.809 ","End":"09:21.500","Text":"We know that our y is 0 because here you would expect to see a y_0."},{"Start":"09:21.500 ","End":"09:26.510","Text":"We know that our y_0 is equal to 0 because we like"},{"Start":"09:26.510 ","End":"09:32.070","Text":"starting at the origin for ease of calculation."},{"Start":"09:32.070 ","End":"09:34.355","Text":"That\u0027s the end of this lesson."},{"Start":"09:34.355 ","End":"09:37.200","Text":"Let\u0027s go and do some examples."}],"ID":9231},{"Watched":false,"Name":"Example- Rock Is Thrown Upwards","Duration":"15m 36s","ChapterTopicVideoID":8977,"CourseChapterTopicPlaylistID":5376,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"Hello. In this question,"},{"Start":"00:02.070 ","End":"00:07.770","Text":"we\u0027re being told that a rock is thrown upwards at a velocity of 40 meters per second."},{"Start":"00:07.770 ","End":"00:10.200","Text":"Let\u0027s draw our information."},{"Start":"00:10.200 ","End":"00:13.515","Text":"We have our y-axis over here,"},{"Start":"00:13.515 ","End":"00:18.165","Text":"and we\u0027ll call our starting point at the origin."},{"Start":"00:18.165 ","End":"00:22.980","Text":"Then we know that we have a rock being thrown upwards at a starting velocity,"},{"Start":"00:22.980 ","End":"00:27.749","Text":"v_0, equals 40 meters per second."},{"Start":"00:27.749 ","End":"00:32.190","Text":"Now, we know that our acceleration is in"},{"Start":"00:32.190 ","End":"00:38.550","Text":"the y direction and equals to negative g. Know why negative g?"},{"Start":"00:38.550 ","End":"00:44.150","Text":"Because the acceleration is working in the opposite direction to our direction of travel."},{"Start":"00:44.150 ","End":"00:47.015","Text":"Then, we can also say because we\u0027re starting at the origin,"},{"Start":"00:47.015 ","End":"00:50.540","Text":"that our y_0 is equal to 0."},{"Start":"00:50.540 ","End":"00:54.380","Text":"Now that we\u0027ve written down all of our basic bits of information,"},{"Start":"00:54.380 ","End":"00:56.930","Text":"we can start answering the questions."},{"Start":"00:56.930 ","End":"01:01.265","Text":"Our first question is, where will the rock be after 3 seconds?"},{"Start":"01:01.265 ","End":"01:04.055","Text":"Now, a quick note before we start answering everything."},{"Start":"01:04.055 ","End":"01:07.190","Text":"The only reason we can use these 2 equations is"},{"Start":"01:07.190 ","End":"01:11.210","Text":"because we know that our acceleration is a constant."},{"Start":"01:11.210 ","End":"01:16.580","Text":"These equations are only valid for a constant acceleration."},{"Start":"01:16.580 ","End":"01:19.665","Text":"Let\u0027s answer question number 1."},{"Start":"01:19.665 ","End":"01:22.515","Text":"Where will the rock be after 3 seconds?"},{"Start":"01:22.515 ","End":"01:27.945","Text":"All we have to do is substitute this information into our position equation."},{"Start":"01:27.945 ","End":"01:32.195","Text":"Remember, we\u0027ve seen this equation just instead of the y_t and y_0,"},{"Start":"01:32.195 ","End":"01:36.210","Text":"we\u0027ve seen it with x(t) and x_0,"},{"Start":"01:36.210 ","End":"01:38.250","Text":"but here it\u0027s the exact same thing,"},{"Start":"01:38.250 ","End":"01:42.670","Text":"just in the y direction because we\u0027re working up and down the y-axis."},{"Start":"01:42.670 ","End":"01:47.255","Text":"We want to find out our position at t equals 3."},{"Start":"01:47.255 ","End":"01:54.065","Text":"Our y at t equals 3 is equal to y_0."},{"Start":"01:54.065 ","End":"01:55.495","Text":"Now, what\u0027s our y_0?"},{"Start":"01:55.495 ","End":"01:57.650","Text":"We said it\u0027s 0 because we\u0027re starting at the origin."},{"Start":"01:57.650 ","End":"02:00.170","Text":"0 plus our starting velocity,"},{"Start":"02:00.170 ","End":"02:03.735","Text":"which is 40, multiplied by t,"},{"Start":"02:03.735 ","End":"02:08.875","Text":"which is 3 plus 1/2 multiplied by our a,"},{"Start":"02:08.875 ","End":"02:13.270","Text":"which is negative g, multiplied by 3^2."},{"Start":"02:13.270 ","End":"02:19.220","Text":"Now, we\u0027re going to say that our g is going to be equal to negative 10,"},{"Start":"02:19.220 ","End":"02:23.458","Text":"even though it\u0027s actually negative 9.81,"},{"Start":"02:23.458 ","End":"02:24.665","Text":"and so on and so forth,"},{"Start":"02:24.665 ","End":"02:26.230","Text":"just for ease of calculation,"},{"Start":"02:26.230 ","End":"02:27.620","Text":"so we\u0027re going to round it."},{"Start":"02:27.620 ","End":"02:31.280","Text":"Then we can say that we have 40 times 3,"},{"Start":"02:31.280 ","End":"02:35.110","Text":"which is 120, plus,"},{"Start":"02:35.110 ","End":"02:38.360","Text":"but then a negative because there\u0027s a negative sign over here,"},{"Start":"02:38.360 ","End":"02:42.200","Text":"10/2, is 5 times 9,"},{"Start":"02:42.200 ","End":"02:47.165","Text":"which equals 120 minus 45,"},{"Start":"02:47.165 ","End":"02:50.720","Text":"which will equal 75 meters."},{"Start":"02:50.720 ","End":"02:53.900","Text":"Where will the rock be after 3 seconds?"},{"Start":"02:53.900 ","End":"02:58.220","Text":"It will be 75 meters up the y-axis."},{"Start":"02:58.220 ","End":"03:00.770","Text":"Now, our second question is asking us,"},{"Start":"03:00.770 ","End":"03:04.210","Text":"what will be its velocity after 4 seconds?"},{"Start":"03:04.210 ","End":"03:05.765","Text":"How do I figure this out?"},{"Start":"03:05.765 ","End":"03:07.640","Text":"I\u0027m going to use this equation."},{"Start":"03:07.640 ","End":"03:14.590","Text":"My velocity is equal to my initial velocity plus my acceleration multiplied by my time."},{"Start":"03:14.590 ","End":"03:18.860","Text":"I can just say that my velocity is equal to my v_0,"},{"Start":"03:18.860 ","End":"03:22.505","Text":"which is 40, plus my acceleration,"},{"Start":"03:22.505 ","End":"03:27.030","Text":"which is negative g. Here we said that g,"},{"Start":"03:27.030 ","End":"03:28.200","Text":"we\u0027re going to say is 10,"},{"Start":"03:28.200 ","End":"03:34.400","Text":"so negative 10 multiplied by t. This is our equation for the velocity."},{"Start":"03:34.400 ","End":"03:38.000","Text":"Now, because I want to find my velocity at t equals"},{"Start":"03:38.000 ","End":"03:42.860","Text":"4 seconds as being requested in the question,"},{"Start":"03:42.860 ","End":"03:49.940","Text":"you can say that 40 minus 10 multiplied by 4,"},{"Start":"03:49.940 ","End":"03:54.685","Text":"which will equal 40 minus 40, which equals to 0."},{"Start":"03:54.685 ","End":"03:56.990","Text":"Why does this equal to 0?"},{"Start":"03:56.990 ","End":"04:00.845","Text":"If you remember, when our velocity is 0,"},{"Start":"04:00.845 ","End":"04:02.765","Text":"when something\u0027s being thrown upwards,"},{"Start":"04:02.765 ","End":"04:06.275","Text":"then it means that the object has reached its maximum height."},{"Start":"04:06.275 ","End":"04:11.165","Text":"It\u0027s decelerated such that its velocity has reached 0."},{"Start":"04:11.165 ","End":"04:15.065","Text":"That means that after t equals 4 seconds,"},{"Start":"04:15.065 ","End":"04:21.730","Text":"our object is going to start returning back downwards to its starting position."},{"Start":"04:21.730 ","End":"04:25.040","Text":"At 4 seconds we\u0027ve reached our maximum height."},{"Start":"04:25.040 ","End":"04:28.250","Text":"Now let\u0027s take a look at question number 3."},{"Start":"04:28.250 ","End":"04:31.615","Text":"How long will the rock be moving in an upwards direction?"},{"Start":"04:31.615 ","End":"04:34.880","Text":"We\u0027ve actually answered that question in question number 2,"},{"Start":"04:34.880 ","End":"04:36.155","Text":"it will take 4 seconds."},{"Start":"04:36.155 ","End":"04:41.105","Text":"Because what we\u0027re trying to find out is when our velocity will equal 0,"},{"Start":"04:41.105 ","End":"04:44.203","Text":"because that means that we\u0027ve reached a maximum height,"},{"Start":"04:44.203 ","End":"04:47.270","Text":"and then we just have to find the time taken to get to"},{"Start":"04:47.270 ","End":"04:51.090","Text":"this position where our velocity is equal to 0."},{"Start":"04:51.090 ","End":"04:56.195","Text":"In the question, we already saw that when t equals 4,"},{"Start":"04:56.195 ","End":"04:57.770","Text":"our velocity equals 0,"},{"Start":"04:57.770 ","End":"04:59.495","Text":"meaning that it took 4 seconds."},{"Start":"04:59.495 ","End":"05:02.795","Text":"However, if I wasn\u0027t asked this question and they didn\u0027t see this,"},{"Start":"05:02.795 ","End":"05:05.225","Text":"I would have to work it out from scratch."},{"Start":"05:05.225 ","End":"05:06.905","Text":"How would I do this?"},{"Start":"05:06.905 ","End":"05:09.740","Text":"I would use my velocity equation."},{"Start":"05:09.740 ","End":"05:11.960","Text":"I would say that my velocity,"},{"Start":"05:11.960 ","End":"05:15.290","Text":"which I want to be 0,"},{"Start":"05:15.290 ","End":"05:17.180","Text":"because when it\u0027s 0 means that"},{"Start":"05:17.180 ","End":"05:21.230","Text":"my upwards motion has stopped and downwards motion is going to begin,"},{"Start":"05:21.230 ","End":"05:24.245","Text":"is equal to my initial velocity,"},{"Start":"05:24.245 ","End":"05:27.305","Text":"which here is 40 meters per second,"},{"Start":"05:27.305 ","End":"05:29.075","Text":"plus my at,"},{"Start":"05:29.075 ","End":"05:33.885","Text":"which is in fact negative 10t."},{"Start":"05:33.885 ","End":"05:36.560","Text":"Negative 10 because that\u0027s my acceleration due to"},{"Start":"05:36.560 ","End":"05:39.890","Text":"gravity multiplied by t. Then if I rearrange this,"},{"Start":"05:39.890 ","End":"05:44.630","Text":"I would get that 40 equals 10t divide both sides by 10,"},{"Start":"05:44.630 ","End":"05:51.005","Text":"and I would get that 4 equals t. This would mean that at t equals 4 seconds,"},{"Start":"05:51.005 ","End":"05:56.135","Text":"I\u0027ve reached my maximum height and then downwards motion will begin."},{"Start":"05:56.135 ","End":"06:03.095","Text":"Now, question number 4 is asking us what the maximum height that the rock will reach is."},{"Start":"06:03.095 ","End":"06:05.930","Text":"We\u0027re going to use our position equation."},{"Start":"06:05.930 ","End":"06:10.790","Text":"We want to find a maximum height at t equals 4 seconds,"},{"Start":"06:10.790 ","End":"06:13.900","Text":"because we\u0027ve seen that at 4 seconds we reach our maximum height."},{"Start":"06:13.900 ","End":"06:19.265","Text":"We\u0027ll say that our height at t equals 4 seconds equals our initial height,"},{"Start":"06:19.265 ","End":"06:21.140","Text":"which is 0, because we\u0027re at the origin,"},{"Start":"06:21.140 ","End":"06:23.074","Text":"plus our initial velocity,"},{"Start":"06:23.074 ","End":"06:25.595","Text":"which is 40, multiplied by our t,"},{"Start":"06:25.595 ","End":"06:33.140","Text":"which is 4, plus half times our acceleration multiplied by t^2."},{"Start":"06:33.140 ","End":"06:35.555","Text":"We\u0027ve said that our acceleration is negative g,"},{"Start":"06:35.555 ","End":"06:41.285","Text":"which here will be negative 10 times 1/2,"},{"Start":"06:41.285 ","End":"06:44.780","Text":"multiplied by t^2,"},{"Start":"06:44.780 ","End":"06:48.795","Text":"so multiplied by 4^2, which is 16."},{"Start":"06:48.795 ","End":"06:54.510","Text":"Then we have 40 times 4 is a 160, negative 10/2,"},{"Start":"06:54.510 ","End":"07:01.470","Text":"which is 5, and 5 times 16 equals 80."},{"Start":"07:01.470 ","End":"07:03.720","Text":"Then we have 160 minus 80,"},{"Start":"07:03.720 ","End":"07:06.255","Text":"which is equal to 80 meters."},{"Start":"07:06.255 ","End":"07:12.184","Text":"Now we know that our maximum height is at 80 meters."},{"Start":"07:12.184 ","End":"07:14.690","Text":"Now let\u0027s take a look at question number 5."},{"Start":"07:14.690 ","End":"07:19.120","Text":"What will be the rock\u0027s velocity when it returns to its starting position?"},{"Start":"07:19.120 ","End":"07:22.790","Text":"Now what we want to see is the rock was thrown upwards,"},{"Start":"07:22.790 ","End":"07:25.580","Text":"it reaches its y max,"},{"Start":"07:25.580 ","End":"07:31.875","Text":"and now it\u0027s falling back down and we want to see what our v_final is equal to."},{"Start":"07:31.875 ","End":"07:36.900","Text":"If we put this into our position equation first,"},{"Start":"07:36.900 ","End":"07:41.195","Text":"first of all, we want to find the time at which this happens."},{"Start":"07:41.195 ","End":"07:45.630","Text":"We can see that our position at t equals something,"},{"Start":"07:45.630 ","End":"07:47.765","Text":"this is the variable which we want to find out,"},{"Start":"07:47.765 ","End":"07:51.830","Text":"is equal to our starting position,"},{"Start":"07:51.830 ","End":"07:54.320","Text":"which as we know,"},{"Start":"07:54.320 ","End":"08:01.595","Text":"is equal to 0 plus our initial velocity,"},{"Start":"08:01.595 ","End":"08:04.025","Text":"which is 40 multiplied by t,"},{"Start":"08:04.025 ","End":"08:05.405","Text":"which is our unknown,"},{"Start":"08:05.405 ","End":"08:08.270","Text":"plus half at^2,"},{"Start":"08:08.270 ","End":"08:12.930","Text":"which as we know is negative 5t^2."},{"Start":"08:14.670 ","End":"08:19.120","Text":"Now we\u0027ll see that we get 2 options for the time at"},{"Start":"08:19.120 ","End":"08:23.410","Text":"which our object will return to its starting position."},{"Start":"08:23.410 ","End":"08:25.705","Text":"Because this equals 0,"},{"Start":"08:25.705 ","End":"08:29.305","Text":"because we want our starting position to equal 0."},{"Start":"08:29.305 ","End":"08:32.440","Text":"Because we want it to return back to the origin."},{"Start":"08:32.440 ","End":"08:35.620","Text":"Now we\u0027re going to see that we can get 2 answers."},{"Start":"08:35.620 ","End":"08:39.925","Text":"How do I do this? I have to put this into a quadratic equation."},{"Start":"08:39.925 ","End":"08:42.520","Text":"I\u0027ll take out my common factors,"},{"Start":"08:42.520 ","End":"08:49.300","Text":"which will be 5 and t. Then I have multiplied by negative t,"},{"Start":"08:49.300 ","End":"08:55.975","Text":"will get me this negative 5t^2 plus 8 will get me this for t is equal to 0."},{"Start":"08:55.975 ","End":"08:58.180","Text":"When does this equation equal to 0?"},{"Start":"08:58.180 ","End":"08:59.665","Text":"At t equals 0."},{"Start":"08:59.665 ","End":"09:01.810","Text":"Because then everything will equal to 0."},{"Start":"09:01.810 ","End":"09:05.905","Text":"Or when what\u0027s inside the brackets will equal to 0."},{"Start":"09:05.905 ","End":"09:09.040","Text":"Or t equals 8."},{"Start":"09:09.040 ","End":"09:12.430","Text":"Because then I\u0027ll have negative 8 plus 8, which will equal 0."},{"Start":"09:12.430 ","End":"09:16.165","Text":"Now, this when t equals 0."},{"Start":"09:16.165 ","End":"09:20.410","Text":"Of course, we\u0027re going to"},{"Start":"09:20.410 ","End":"09:24.265","Text":"be at the origin then because that\u0027s when I started. That\u0027s interesting."},{"Start":"09:24.265 ","End":"09:28.000","Text":"T equals 8 is when it will return."},{"Start":"09:28.000 ","End":"09:29.980","Text":"After 8 seconds,"},{"Start":"09:29.980 ","End":"09:35.590","Text":"object or rock would have reached its maximum height and come back within 8 seconds."},{"Start":"09:35.590 ","End":"09:39.100","Text":"Also, if you remember from 1 of the lectures,"},{"Start":"09:39.100 ","End":"09:45.130","Text":"we say that the time taken for an object to reach its maximum height and then"},{"Start":"09:45.130 ","End":"09:47.845","Text":"return to its starting position is"},{"Start":"09:47.845 ","End":"09:51.385","Text":"2 times the time taken for it to reach its maximum height."},{"Start":"09:51.385 ","End":"09:52.840","Text":"In Question 3,"},{"Start":"09:52.840 ","End":"09:55.825","Text":"we checked how long it would take for it to reach its maximum height,"},{"Start":"09:55.825 ","End":"09:58.420","Text":"which was 4 seconds and 4 times 2,"},{"Start":"09:58.420 ","End":"10:00.280","Text":"lo and behold, is 8."},{"Start":"10:00.280 ","End":"10:02.095","Text":"We can see that that\u0027s correct."},{"Start":"10:02.095 ","End":"10:04.120","Text":"Now we know that after 8 seconds,"},{"Start":"10:04.120 ","End":"10:05.545","Text":"it returns back to the origin."},{"Start":"10:05.545 ","End":"10:07.015","Text":"But what we\u0027re being asked is,"},{"Start":"10:07.015 ","End":"10:11.095","Text":"what will be the rock\u0027s velocity when it returns to the origin?"},{"Start":"10:11.095 ","End":"10:13.195","Text":"Now we\u0027re going to use this equation."},{"Start":"10:13.195 ","End":"10:15.850","Text":"Velocity, this is our unknown,"},{"Start":"10:15.850 ","End":"10:18.684","Text":"is going to be equal our starting velocity,"},{"Start":"10:18.684 ","End":"10:23.065","Text":"which is 40 plus 8t."},{"Start":"10:23.065 ","End":"10:26.080","Text":"Again, acceleration is negative 10,"},{"Start":"10:26.080 ","End":"10:28.525","Text":"so negative 10 multiplied by our t,"},{"Start":"10:28.525 ","End":"10:31.510","Text":"which we know is 8."},{"Start":"10:31.510 ","End":"10:37.480","Text":"Now, we just do 40 negative 80,"},{"Start":"10:37.480 ","End":"10:41.650","Text":"which will equal negative 40 meters per second."},{"Start":"10:41.650 ","End":"10:43.270","Text":"Here\u0027s a negative sign."},{"Start":"10:43.270 ","End":"10:46.985","Text":"Now how do we know that this answer makes sense?"},{"Start":"10:46.985 ","End":"10:48.630","Text":"Later on in the course,"},{"Start":"10:48.630 ","End":"10:51.555","Text":"you\u0027re going to understand energy conservation,"},{"Start":"10:51.555 ","End":"10:54.960","Text":"in which case you\u0027ll know that if something starts off with"},{"Start":"10:54.960 ","End":"10:57.420","Text":"some initial velocity and it\u0027s thrown up in"},{"Start":"10:57.420 ","End":"11:00.425","Text":"the air once it comes down because of energy conservation,"},{"Start":"11:00.425 ","End":"11:04.540","Text":"the size will be exactly the same."},{"Start":"11:04.540 ","End":"11:06.235","Text":"The size is still 40."},{"Start":"11:06.235 ","End":"11:09.910","Text":"Here we have a positive because we know that we\u0027re going in the upwards direction,"},{"Start":"11:09.910 ","End":"11:12.850","Text":"and here we have a negative because we\u0027re going in the downwards direction."},{"Start":"11:12.850 ","End":"11:15.460","Text":"But the speed without"},{"Start":"11:15.460 ","End":"11:17.140","Text":"the direction is equal to"},{"Start":"11:17.140 ","End":"11:20.620","Text":"the same thing and the velocity will just have a change in sign."},{"Start":"11:20.620 ","End":"11:25.135","Text":"We know that every time if something has thrown up at 40 meters per second,"},{"Start":"11:25.135 ","End":"11:28.585","Text":"it will return down and negative 40 meters per second."},{"Start":"11:28.585 ","End":"11:31.315","Text":"If it\u0027s thrown at 2 meters per second,"},{"Start":"11:31.315 ","End":"11:34.975","Text":"it will come back down at 2 meters per second."},{"Start":"11:34.975 ","End":"11:39.505","Text":"This is why this answer makes perfect sense."},{"Start":"11:39.505 ","End":"11:41.680","Text":"Now question number 6,"},{"Start":"11:41.680 ","End":"11:46.450","Text":"how long will it take for the rock to reach 5 meters below its starting point?"},{"Start":"11:46.450 ","End":"11:50.380","Text":"This is easy. Again, we\u0027re going to use our position equation."},{"Start":"11:50.380 ","End":"11:55.975","Text":"We want our final position at t is unknown."},{"Start":"11:55.975 ","End":"11:59.155","Text":"This is a variable we\u0027re trying to find has to equal"},{"Start":"11:59.155 ","End":"12:05.605","Text":"negative 5 meters because we want it to be over here at negative 5."},{"Start":"12:05.605 ","End":"12:07.960","Text":"From 0 to negative 5."},{"Start":"12:07.960 ","End":"12:12.070","Text":"Now we say that this is equal to our starting position,"},{"Start":"12:12.070 ","End":"12:15.985","Text":"which is equal to 0 plus our velocity."},{"Start":"12:15.985 ","End":"12:20.320","Text":"Which will be 40 multiplied by our time t,"},{"Start":"12:20.320 ","End":"12:24.370","Text":"which is unknown, plus half 8t^2,"},{"Start":"12:24.370 ","End":"12:27.490","Text":"which as we know, is negative 5t^2."},{"Start":"12:27.490 ","End":"12:29.710","Text":"Again, we don\u0027t know what this t^2 is."},{"Start":"12:29.710 ","End":"12:34.315","Text":"Now, we can say that we can move this to the other side,"},{"Start":"12:34.315 ","End":"12:35.710","Text":"add 5 to both sides."},{"Start":"12:35.710 ","End":"12:44.355","Text":"Then we\u0027ll have that 0 equals negative 5t^2 plus 40, t plus 5."},{"Start":"12:44.355 ","End":"12:49.980","Text":"Now, we can solve this by saying that t1 and t2 is equal to,"},{"Start":"12:49.980 ","End":"12:52.370","Text":"now remember, this is our a,"},{"Start":"12:52.370 ","End":"12:53.950","Text":"this is our b,"},{"Start":"12:53.950 ","End":"13:00.175","Text":"and this is our c. We know that it\u0027s equal to negative b plus or minus the square root of"},{"Start":"13:00.175 ","End":"13:08.035","Text":"b^2 minus 4ac divided by 2a,"},{"Start":"13:08.035 ","End":"13:15.490","Text":"which here our negative b is negative 40 plus or minus square root of b^2,"},{"Start":"13:15.490 ","End":"13:23.020","Text":"which will be 1,600 minus 4 times negative 5,"},{"Start":"13:23.020 ","End":"13:26.650","Text":"so we\u0027ll have plus 4 times 5,"},{"Start":"13:26.650 ","End":"13:31.205","Text":"which is 20 multiplied by 5,"},{"Start":"13:31.205 ","End":"13:40.200","Text":"so 20 multiplied by 5 is equal to 100 divided by 2 times a,"},{"Start":"13:40.200 ","End":"13:41.820","Text":"which is 2 times negative 5,"},{"Start":"13:41.820 ","End":"13:44.005","Text":"which is negative 10."},{"Start":"13:44.005 ","End":"13:47.980","Text":"Now, all I have to do is I can say that my"},{"Start":"13:47.980 ","End":"13:52.255","Text":"negative 40 divided by negative 10 will equal 4,"},{"Start":"13:52.255 ","End":"13:56.695","Text":"plus or minus my square root of"},{"Start":"13:56.695 ","End":"14:03.325","Text":"1,700 divided by negative 10."},{"Start":"14:03.325 ","End":"14:06.880","Text":"Now the trick here is to notice that I have a plus and a minus over"},{"Start":"14:06.880 ","End":"14:10.840","Text":"here because I have a negative number in my denominator,"},{"Start":"14:10.840 ","End":"14:16.345","Text":"then I know that when I have 4 plus the square root divided by a negative number,"},{"Start":"14:16.345 ","End":"14:22.570","Text":"I get a negative number and 4 minus the square root divided by a negative number,"},{"Start":"14:22.570 ","End":"14:23.785","Text":"I\u0027ll get a positive number."},{"Start":"14:23.785 ","End":"14:27.295","Text":"Now obviously, my time isn\u0027t going to be a negative time,"},{"Start":"14:27.295 ","End":"14:31.240","Text":"so my number can\u0027t be a negative number because that would mean"},{"Start":"14:31.240 ","End":"14:35.350","Text":"before this whole process and this whole system came about,"},{"Start":"14:35.350 ","End":"14:37.150","Text":"I need a positive time."},{"Start":"14:37.150 ","End":"14:40.345","Text":"Therefore, I know that I\u0027m not going to do this plus"},{"Start":"14:40.345 ","End":"14:43.840","Text":"because it will equal a total negative and"},{"Start":"14:43.840 ","End":"14:46.450","Text":"then 4 minus root of"},{"Start":"14:46.450 ","End":"14:50.440","Text":"1,700 divided by 10 is going to be a negative number and that\u0027s not what I want."},{"Start":"14:50.440 ","End":"14:58.750","Text":"I want a plus. I\u0027ll get rid of my t1 and only my second t is valid."},{"Start":"14:58.750 ","End":"15:02.845","Text":"My t will equal to 4 plus,"},{"Start":"15:02.845 ","End":"15:05.290","Text":"because minus and a minus is plus,"},{"Start":"15:05.290 ","End":"15:12.370","Text":"root 1,700 divided by 10."},{"Start":"15:12.370 ","End":"15:18.520","Text":"At this time, our rock will be 5 meters below its start point."},{"Start":"15:18.520 ","End":"15:20.800","Text":"It will be over here."},{"Start":"15:20.800 ","End":"15:23.170","Text":"That\u0027s the end of this lesson."},{"Start":"15:23.170 ","End":"15:26.020","Text":"If there\u0027s anything here that you didn\u0027t understand,"},{"Start":"15:26.020 ","End":"15:30.145","Text":"please go over this because this is the basics and"},{"Start":"15:30.145 ","End":"15:32.950","Text":"the little tricks that we go through over here will also"},{"Start":"15:32.950 ","End":"15:36.830","Text":"help you in the exam to answer questions a lot faster."}],"ID":9232}],"Thumbnail":null,"ID":5376},{"Name":"Projectile Motion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Projectile Motion","Duration":"30m 9s","ChapterTopicVideoID":10259,"CourseChapterTopicPlaylistID":8958,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.104","Text":"Hello. In this lesson,"},{"Start":"00:02.104 ","End":"00:05.924","Text":"we\u0027re going to be speaking about projectiles or projectile motion."},{"Start":"00:05.924 ","End":"00:09.959","Text":"Now, most of you will have studied this in high school and hopefully,"},{"Start":"00:09.959 ","End":"00:14.670","Text":"you remember it, so this lesson is going to be a brief recap."},{"Start":"00:14.670 ","End":"00:19.980","Text":"When we\u0027re dealing with projectile motion we have the classic case where we have"},{"Start":"00:19.980 ","End":"00:25.147","Text":"our ground or our x-axis,"},{"Start":"00:25.147 ","End":"00:29.550","Text":"and then we have some kind of body which is"},{"Start":"00:29.550 ","End":"00:34.928","Text":"shot with some initial velocity v_0 and it has an angle;"},{"Start":"00:34.928 ","End":"00:39.430","Text":"some angle Theta to the x-axis."},{"Start":"00:39.430 ","End":"00:48.375","Text":"We can discuss what the motion of this body will be once it is shot like so,"},{"Start":"00:48.375 ","End":"00:52.745","Text":"then we have horizontal projectile motion."},{"Start":"00:52.745 ","End":"00:57.209","Text":"That\u0027s when we have some height,"},{"Start":"00:57.209 ","End":"00:58.620","Text":"clef or whatever,"},{"Start":"00:58.620 ","End":"01:05.490","Text":"and we have a body which is shot horizontally with v_0."},{"Start":"01:06.080 ","End":"01:09.754","Text":"If we look, if this is our x-axis,"},{"Start":"01:09.754 ","End":"01:14.920","Text":"we can see that our shot is fired parallel to our x-axis,"},{"Start":"01:14.920 ","End":"01:17.620","Text":"so this is horizontal projectile motion."},{"Start":"01:17.620 ","End":"01:25.715","Text":"Then we can see how this body is going to fly or land where on the ground."},{"Start":"01:25.715 ","End":"01:28.692","Text":"Here we know that in the projectile motion"},{"Start":"01:28.692 ","End":"01:32.250","Text":"we\u0027re going to have something that looks like this,"},{"Start":"01:32.250 ","End":"01:38.480","Text":"and here in horizontal projectile motion there\u0027s something that looks like this."},{"Start":"01:39.380 ","End":"01:42.279","Text":"These are the 2 basics and of course,"},{"Start":"01:42.279 ","End":"01:45.610","Text":"we can mix and match between the 2."},{"Start":"01:45.610 ","End":"01:48.430","Text":"We can have some height and we have"},{"Start":"01:48.430 ","End":"01:52.449","Text":"our x-axis over here and we can have our body which is over"},{"Start":"01:52.449 ","End":"02:01.114","Text":"here shots at some angle Theta to our horizontal x-axis."},{"Start":"02:01.114 ","End":"02:06.612","Text":"We can also have that it\u0027s shot downwards at this angle,"},{"Start":"02:06.612 ","End":"02:12.710","Text":"so we can see that here our Theta to angle is some negative angle."},{"Start":"02:12.710 ","End":"02:15.890","Text":"It\u0027s below our x is equal to 0."},{"Start":"02:15.890 ","End":"02:22.230","Text":"We can mix, and match, and play around with it and see what will happen to this body."},{"Start":"02:22.880 ","End":"02:27.499","Text":"What we can see is that this horizontal projectile motion"},{"Start":"02:27.499 ","End":"02:31.975","Text":"is a specific case of regular projectile motion,"},{"Start":"02:31.975 ","End":"02:34.310","Text":"the only thing is that instead of having"},{"Start":"02:34.310 ","End":"02:40.030","Text":"some angle Theta we have that our Theta is equal to 0."},{"Start":"02:41.000 ","End":"02:45.410","Text":"Let\u0027s see our projectile motion and then"},{"Start":"02:45.410 ","End":"02:49.825","Text":"afterwards we\u0027ll speak about the case of horizontal projectile motion."},{"Start":"02:49.825 ","End":"02:52.635","Text":"We have our buddy over here,"},{"Start":"02:52.635 ","End":"02:55.820","Text":"and it doesn\u0027t matter if it\u0027s at a height or on the ground."},{"Start":"02:55.820 ","End":"03:00.560","Text":"We know that in this direction we have our x-axis,"},{"Start":"03:00.560 ","End":"03:02.644","Text":"and in this direction,"},{"Start":"03:02.644 ","End":"03:04.925","Text":"we have our y-axis."},{"Start":"03:04.925 ","End":"03:10.219","Text":"We have been told that our body is being shot at an initial velocity of v_0"},{"Start":"03:10.219 ","End":"03:16.600","Text":"and an angle of Theta relative to the x-axis."},{"Start":"03:17.120 ","End":"03:23.134","Text":"Let\u0027s take a look at what is going on, on our x-axis."},{"Start":"03:23.134 ","End":"03:28.429","Text":"Our velocity in the x-direction is simply"},{"Start":"03:28.429 ","End":"03:33.910","Text":"going to be the projection of our v_0 on our x-axis."},{"Start":"03:33.910 ","End":"03:41.330","Text":"So that\u0027s going to be equal to v_0 cosine of the angle which here is Theta."},{"Start":"03:41.330 ","End":"03:49.105","Text":"What we can see is that our velocity is constant."},{"Start":"03:49.105 ","End":"03:53.299","Text":"That means that we can write this equation for"},{"Start":"03:53.299 ","End":"03:57.960","Text":"the position along the x-axis as a function of time."},{"Start":"03:57.960 ","End":"04:03.010","Text":"That\u0027s going to be our x_0 which is our initial position"},{"Start":"04:03.010 ","End":"04:08.725","Text":"plus our velocity in the x-direction multiplied by time."},{"Start":"04:08.725 ","End":"04:18.260","Text":"Here it\u0027s going to be specifically x_0 plus v_0 cosine Theta multiplied by"},{"Start":"04:18.260 ","End":"04:27.680","Text":"t. Now let\u0027s talk about what\u0027s going on our y-axis in the y-direction."},{"Start":"04:27.680 ","End":"04:33.520","Text":"First of all, we can see that in the y-direction we have acceleration,"},{"Start":"04:33.520 ","End":"04:36.529","Text":"and that is due to gravity."},{"Start":"04:36.529 ","End":"04:40.623","Text":"Now usually, our projectile is going to be shot upwards"},{"Start":"04:40.623 ","End":"04:44.835","Text":"in which case gravity is obviously pulling the body down."},{"Start":"04:44.835 ","End":"04:49.532","Text":"The acceleration is going to be in the negative direction to gravity."},{"Start":"04:49.532 ","End":"04:57.320","Text":"It will be negative g which sometimes happens in 1 of the questions,"},{"Start":"04:57.320 ","End":"05:00.800","Text":"if we\u0027re shooting our projectile in this direction,"},{"Start":"05:00.800 ","End":"05:04.370","Text":"so our projectile is going to be moving in the same direction as"},{"Start":"05:04.370 ","End":"05:08.510","Text":"gravity and then our a_y will be equal"},{"Start":"05:08.510 ","End":"05:16.355","Text":"to positive g. The minute that we see that we have acceleration,"},{"Start":"05:16.355 ","End":"05:20.140","Text":"there are 2 equations that we can work with."},{"Start":"05:20.140 ","End":"05:24.035","Text":"The first equation is going to be"},{"Start":"05:24.035 ","End":"05:31.418","Text":"our velocity on the y-axis as a function of time."},{"Start":"05:31.418 ","End":"05:35.764","Text":"Because our velocity isn\u0027t constant because we have acceleration."},{"Start":"05:35.764 ","End":"05:40.279","Text":"So our velocity in the y-direction as a function of time is going to be equal"},{"Start":"05:40.279 ","End":"05:45.285","Text":"to our initial velocity in the y-direction."},{"Start":"05:45.285 ","End":"05:48.529","Text":"It\u0027s going to be in a similar format to what we have here"},{"Start":"05:48.529 ","End":"05:52.468","Text":"our initial velocity in the x-direction"},{"Start":"05:52.468 ","End":"06:01.069","Text":"plus our acceleration in the y-direction multiplied by t. Here specifically,"},{"Start":"06:01.069 ","End":"06:05.195","Text":"our initial velocity in the y-direction is simply going to be"},{"Start":"06:05.195 ","End":"06:12.539","Text":"v_0 sine Theta because"},{"Start":"06:12.539 ","End":"06:16.439","Text":"we want our projection of v_0 on our y-axis,"},{"Start":"06:16.439 ","End":"06:19.215","Text":"so that\u0027s v_0 sine Theta."},{"Start":"06:19.215 ","End":"06:26.029","Text":"Then plus our acceleration in the y-direction which is negative g multiplied"},{"Start":"06:26.029 ","End":"06:33.127","Text":"by t. What do we have here in our red boxes?"},{"Start":"06:33.127 ","End":"06:39.580","Text":"This is the equation for the position along the x-axis as a function of t,"},{"Start":"06:39.580 ","End":"06:47.029","Text":"and this equation over here is our velocity on the y-axis as a function of"},{"Start":"06:47.029 ","End":"06:56.894","Text":"t. The next equation is our position on the y-axis as a function of time,"},{"Start":"06:56.894 ","End":"07:02.285","Text":"and that\u0027s going to be equal to our initial y coordinate"},{"Start":"07:02.285 ","End":"07:09.695","Text":"plus our initial velocity in the y-direction multiplied by t,"},{"Start":"07:09.695 ","End":"07:19.200","Text":"plus half of the acceleration in the y-direction multiplied by t^2."},{"Start":"07:19.340 ","End":"07:26.938","Text":"Now, once we substitute in all this we\u0027ll get that our y_0 is our initial position."},{"Start":"07:26.938 ","End":"07:31.580","Text":"Usually, our y_0 and our x_0 will be at the origin,"},{"Start":"07:31.580 ","End":"07:33.320","Text":"so they\u0027ll both be equal to 0,"},{"Start":"07:33.320 ","End":"07:35.014","Text":"so we can just ignore them."},{"Start":"07:35.014 ","End":"07:40.435","Text":"But sometimes like in the case where we have some heights and our body is here,"},{"Start":"07:40.435 ","End":"07:44.449","Text":"so sometimes we\u0027ll say that this is the origin which means that"},{"Start":"07:44.449 ","End":"07:50.470","Text":"the x-value will maybe be 0 and the y-value will be some kind of height."},{"Start":"07:50.470 ","End":"07:52.740","Text":"It depends on the question."},{"Start":"07:52.740 ","End":"07:56.344","Text":"We have our initial y position plus"},{"Start":"07:56.344 ","End":"08:01.579","Text":"our initial velocity in the y-direction which we said was"},{"Start":"08:01.579 ","End":"08:08.730","Text":"v_0 sine of Theta multiplied by t and then we\u0027re going to"},{"Start":"08:08.730 ","End":"08:16.820","Text":"have plus half our acceleration in the y-direction which is negative g,"},{"Start":"08:16.820 ","End":"08:24.360","Text":"so let\u0027s put negative 1/2g multiplied by t^2."},{"Start":"08:24.650 ","End":"08:28.460","Text":"These are the 3 most important equations"},{"Start":"08:28.460 ","End":"08:31.775","Text":"you need to remember when dealing with projectile motion."},{"Start":"08:31.775 ","End":"08:35.569","Text":"We have our position along the x-axis as"},{"Start":"08:35.569 ","End":"08:39.710","Text":"a function of time where our x_0 is our initial position and"},{"Start":"08:39.710 ","End":"08:44.134","Text":"our v_0 is our initial velocity then we have"},{"Start":"08:44.134 ","End":"08:49.115","Text":"our velocity in the y-direction as a function of time."},{"Start":"08:49.115 ","End":"08:53.111","Text":"We have our v_0 which is our initial velocity"},{"Start":"08:53.111 ","End":"08:57.169","Text":"and when we\u0027re working against the direction of gravity,"},{"Start":"08:57.169 ","End":"09:00.214","Text":"so we\u0027ll have negative g. Then,"},{"Start":"09:00.214 ","End":"09:03.790","Text":"we have our position along the y-axis as a function of time"},{"Start":"09:03.790 ","End":"09:08.127","Text":"which will be our initial y-position."},{"Start":"09:08.127 ","End":"09:11.475","Text":"This v_0 is our initial velocity,"},{"Start":"09:11.475 ","End":"09:15.184","Text":"and again a negative over here if we\u0027re working"},{"Start":"09:15.184 ","End":"09:19.015","Text":"against the direction of gravity such as in these examples."},{"Start":"09:19.015 ","End":"09:23.479","Text":"We\u0027ll have a positive over here and here"},{"Start":"09:23.479 ","End":"09:31.440","Text":"if we for instance shield our projector in this direction or in this direction."},{"Start":"09:31.850 ","End":"09:35.900","Text":"These are the 3 most important equations."},{"Start":"09:35.900 ","End":"09:38.329","Text":"Now let\u0027s see the type of questions that they can ask"},{"Start":"09:38.329 ","End":"09:41.735","Text":"you when dealing with projectile motion."},{"Start":"09:41.735 ","End":"09:44.825","Text":"Now, there are a few equations that maybe you can"},{"Start":"09:44.825 ","End":"09:48.274","Text":"automatically use depending on the question, however,"},{"Start":"09:48.274 ","End":"09:52.670","Text":"these 3 equations will bring you to those specific equations,"},{"Start":"09:52.670 ","End":"09:56.755","Text":"so these are the ones to really remember and learn."},{"Start":"09:56.755 ","End":"10:07.625","Text":"Let\u0027s take a look. We have our body which is located at the origin,"},{"Start":"10:07.625 ","End":"10:13.095","Text":"and here we have our x-axis and here we have our y-axis."},{"Start":"10:13.095 ","End":"10:16.640","Text":"Our body is shot with an initial velocity of v_0"},{"Start":"10:16.640 ","End":"10:20.855","Text":"and an angle Theta relative to the x-axis."},{"Start":"10:20.855 ","End":"10:24.230","Text":"Now we know that the projectile motion for something like"},{"Start":"10:24.230 ","End":"10:29.884","Text":"this will go something along these lines."},{"Start":"10:29.884 ","End":"10:32.674","Text":"It will look something like this."},{"Start":"10:32.674 ","End":"10:36.919","Text":"Now we know that it will have a maximum height over"},{"Start":"10:36.919 ","End":"10:41.154","Text":"here right at the center of its projectile motion,"},{"Start":"10:41.154 ","End":"10:45.990","Text":"and this is called H_max; its maximum height."},{"Start":"10:47.250 ","End":"10:53.394","Text":"Let\u0027s discuss how we can find our value for H max."},{"Start":"10:53.394 ","End":"10:59.000","Text":"How are we going to find the maximum height that our projectile will reach?"},{"Start":"10:59.220 ","End":"11:04.224","Text":"The condition for us reaching maximum height is that"},{"Start":"11:04.224 ","End":"11:12.280","Text":"our velocity in the y-direction will be equal to 0."},{"Start":"11:12.280 ","End":"11:17.949","Text":"Through this and substituting into these different equations,"},{"Start":"11:17.949 ","End":"11:23.694","Text":"we can also find the time at which our projectile will reach its maximum height."},{"Start":"11:23.694 ","End":"11:30.014","Text":"Let\u0027s see how we find this."},{"Start":"11:30.014 ","End":"11:33.555","Text":"From this condition that we have over here,"},{"Start":"11:33.555 ","End":"11:36.734","Text":"that our velocity in the y direction has to be equal to 0,"},{"Start":"11:36.734 ","End":"11:40.159","Text":"we\u0027re going to find the time at which this happens."},{"Start":"11:40.159 ","End":"11:42.879","Text":"We\u0027ll substitute in this over here."},{"Start":"11:42.879 ","End":"11:46.779","Text":"We\u0027ll have therefore that our 0 is equal to"},{"Start":"11:46.779 ","End":"11:54.080","Text":"v_0 sine of Theta minus gt."},{"Start":"11:54.120 ","End":"12:02.799","Text":"That means that the time that we get to H max is going to be equal to,"},{"Start":"12:02.799 ","End":"12:04.480","Text":"and we just isolate this out."},{"Start":"12:04.480 ","End":"12:14.650","Text":"v_0 sine Theta divided by g. Now we have our time to our H max,"},{"Start":"12:14.650 ","End":"12:20.679","Text":"and now we want to find what position this is in the y-direction,"},{"Start":"12:20.679 ","End":"12:25.449","Text":"because the position in the y-direction is going to be the maximum height."},{"Start":"12:25.449 ","End":"12:29.950","Text":"What we\u0027re going to do is we\u0027re going to then substitute in this time into"},{"Start":"12:29.950 ","End":"12:35.005","Text":"our equation for the position as a function of time in the y direction."},{"Start":"12:35.005 ","End":"12:39.354","Text":"That means that our position as a function"},{"Start":"12:39.354 ","End":"12:48.475","Text":"of time at our time is equal to tH max,"},{"Start":"12:48.475 ","End":"12:53.694","Text":"kt is equal to t. At H max is going to be equal to,"},{"Start":"12:53.694 ","End":"12:56.809","Text":"so our y_0 position,"},{"Start":"12:58.260 ","End":"13:02.319","Text":"it will be 0, but let\u0027s just for the sake of it write this n,"},{"Start":"13:02.319 ","End":"13:07.060","Text":"plus our v_0 sine of"},{"Start":"13:07.060 ","End":"13:08.689","Text":"Theta"},{"Start":"13:15.350 ","End":"13:17.159","Text":"multiplied"},{"Start":"13:17.159 ","End":"13:18.450","Text":"by our t,"},{"Start":"13:18.450 ","End":"13:20.040","Text":"where our t is our tH max,"},{"Start":"13:20.040 ","End":"13:21.179","Text":"which is this over here,"},{"Start":"13:21.179 ","End":"13:28.874","Text":"so multiplied by v_0 sine Theta divided by"},{"Start":"13:28.874 ","End":"13:39.339","Text":"g. Then we have minus 1/2g multiplied by our t squared,"},{"Start":"13:39.339 ","End":"13:42.159","Text":"so our tH max squared,"},{"Start":"13:42.159 ","End":"13:51.319","Text":"which will be v_0 squared sine squared Theta divided by g squared."},{"Start":"13:52.170 ","End":"13:54.895","Text":"Now let\u0027s simplify it."},{"Start":"13:54.895 ","End":"13:58.089","Text":"Our y_0, because we said here specifically we\u0027re"},{"Start":"13:58.089 ","End":"14:01.450","Text":"dealing with our body beginning at the origin,"},{"Start":"14:01.450 ","End":"14:05.019","Text":"so our initial y value is equal to 0."},{"Start":"14:05.019 ","End":"14:07.930","Text":"Then this g can cross out with the square,"},{"Start":"14:07.930 ","End":"14:10.764","Text":"so we have just g over here."},{"Start":"14:10.764 ","End":"14:16.885","Text":"Here we have v_0 sine of Theta multiplied by v_0 sine of Theta."},{"Start":"14:16.885 ","End":"14:20.830","Text":"That\u0027s just going to be v_0 squared sine squared Theta,"},{"Start":"14:20.830 ","End":"14:23.109","Text":"which is the same as over here."},{"Start":"14:23.109 ","End":"14:26.155","Text":"That means that what we\u0027ll have is"},{"Start":"14:26.155 ","End":"14:32.739","Text":"0 plus v_0 squared sine squared Theta divided by g minus"},{"Start":"14:32.739 ","End":"14:39.924","Text":"half v_0 squared sine squared Theta divided by g. Our final answer will simply be"},{"Start":"14:39.924 ","End":"14:48.500","Text":"v_0 squared sine squared Theta divided by 2g."},{"Start":"14:49.650 ","End":"14:57.145","Text":"Then of course this is equal to height H max."},{"Start":"14:57.145 ","End":"15:00.909","Text":"Now as long as you understand that this"},{"Start":"15:00.909 ","End":"15:04.854","Text":"is the most important condition in order to find your H max,"},{"Start":"15:04.854 ","End":"15:10.690","Text":"and then you just substitute it into these 2 equations to find the maximum height,"},{"Start":"15:10.690 ","End":"15:13.540","Text":"which will be your maximum y is a function of"},{"Start":"15:13.540 ","End":"15:17.230","Text":"t. Then you don\u0027t need to remember this equation."},{"Start":"15:17.230 ","End":"15:22.030","Text":"But a lot of the time people say that the equation in order to find the H max,"},{"Start":"15:22.030 ","End":"15:26.955","Text":"instead of doing these steps is simply this over here."},{"Start":"15:26.955 ","End":"15:30.524","Text":"But if you understand how to substitute n,"},{"Start":"15:30.524 ","End":"15:35.410","Text":"then you don\u0027t have to remember this specific equation."},{"Start":"15:36.300 ","End":"15:42.620","Text":"The next thing that we\u0027re going to be speaking about is our displacement."},{"Start":"15:43.070 ","End":"15:47.430","Text":"When we\u0027re dealing with displacement,"},{"Start":"15:47.430 ","End":"15:53.015","Text":"sometimes we also call this the range."},{"Start":"15:53.015 ","End":"15:57.685","Text":"Sometimes it\u0027s denoted by either the letter D or the letter I."},{"Start":"15:57.685 ","End":"15:59.215","Text":"What does that mean?"},{"Start":"15:59.215 ","End":"16:04.360","Text":"It\u0027s the maximum distance that our body will reach."},{"Start":"16:04.360 ","End":"16:09.009","Text":"That means that if our body is shot with our initial velocity v_0,"},{"Start":"16:09.009 ","End":"16:10.645","Text":"and at this angle of Theta,"},{"Start":"16:10.645 ","End":"16:17.860","Text":"our range or our displacement in this example will be this over here."},{"Start":"16:17.860 ","End":"16:21.295","Text":"This is our total displacement."},{"Start":"16:21.295 ","End":"16:25.840","Text":"The maximum distance traveled by this projectile."},{"Start":"16:25.840 ","End":"16:30.144","Text":"Now specifically in our example that we\u0027re giving,"},{"Start":"16:30.144 ","End":"16:33.329","Text":"we\u0027re dealing with normal projectile motion,"},{"Start":"16:33.329 ","End":"16:36.280","Text":"where we have our body which is shot from the ground,"},{"Start":"16:36.280 ","End":"16:41.830","Text":"and it\u0027s landing on the ground at the same height from where it was shot."},{"Start":"16:41.830 ","End":"16:46.959","Text":"Now obviously these equations and these calculations will be different."},{"Start":"16:46.959 ","End":"16:51.599","Text":"If we\u0027re shooting our body from some height above the ground,"},{"Start":"16:51.599 ","End":"16:56.320","Text":"or it\u0027s landing somewhere to different height where we initially shot the body out,"},{"Start":"16:56.320 ","End":"16:59.545","Text":"for instance at lands slightly lower or higher."},{"Start":"16:59.545 ","End":"17:04.330","Text":"We\u0027re dealing with specifically this example over here."},{"Start":"17:04.330 ","End":"17:10.129","Text":"Our displacement is a maximum value for x."},{"Start":"17:10.920 ","End":"17:14.920","Text":"What we say in order to find this maximum displacement,"},{"Start":"17:14.920 ","End":"17:16.794","Text":"specifically for this case,"},{"Start":"17:16.794 ","End":"17:20.904","Text":"where our body is shot from the same height at which it will land,"},{"Start":"17:20.904 ","End":"17:31.310","Text":"is the condition is that our position in the y-axis as a function of t is equal to 0."},{"Start":"17:31.890 ","End":"17:38.109","Text":"This is because when our position in the y-axis as a function of time is equal to 0,"},{"Start":"17:38.109 ","End":"17:41.484","Text":"as in it\u0027s hit the floor."},{"Start":"17:41.484 ","End":"17:44.020","Text":"Our projectile has hit the ground,"},{"Start":"17:44.020 ","End":"17:47.170","Text":"so it\u0027s not going to carry on traveling."},{"Start":"17:47.170 ","End":"17:49.600","Text":"We know that if it hit the ground,"},{"Start":"17:49.600 ","End":"17:51.084","Text":"and it stopped traveling,"},{"Start":"17:51.084 ","End":"17:55.524","Text":"that it\u0027s going to have also reached its maximum x value."},{"Start":"17:55.524 ","End":"18:01.554","Text":"Then we say that our position on the y-axis is equal to 0."},{"Start":"18:01.554 ","End":"18:06.760","Text":"Usually, we\u0027re dealing with our y_0 is equal to 0,"},{"Start":"18:06.760 ","End":"18:09.925","Text":"because our initial position is also the origin,"},{"Start":"18:09.925 ","End":"18:11.889","Text":"which is what we can see over here."},{"Start":"18:11.889 ","End":"18:17.845","Text":"We\u0027ll have 0 plus our v_0 sine of Theta"},{"Start":"18:17.845 ","End":"18:26.139","Text":"multiplied by t minus 1/2g multiplied by t squared."},{"Start":"18:26.139 ","End":"18:32.643","Text":"Now what we can do is we can divide both sides by t,"},{"Start":"18:32.643 ","End":"18:36.939","Text":"and then what we want to do is we want to find the time at which this happens."},{"Start":"18:36.939 ","End":"18:39.459","Text":"We\u0027re going to isolate out our t,"},{"Start":"18:39.459 ","End":"18:43.554","Text":"so the time when our projectile hits the ground,"},{"Start":"18:43.554 ","End":"18:52.630","Text":"again is simply going to be equal to 2v_0 sine of Theta divided by"},{"Start":"18:52.630 ","End":"19:02.499","Text":"g. This t over here is the t at our maximum displacement."},{"Start":"19:02.499 ","End":"19:05.470","Text":"When we reach maximum displacement is after this time."},{"Start":"19:05.470 ","End":"19:13.479","Text":"Now we\u0027re going to substitute in this t into this equation over here."},{"Start":"19:13.479 ","End":"19:19.659","Text":"Our maximum displacement D is going to be equal to"},{"Start":"19:19.659 ","End":"19:27.110","Text":"our position as a function of time when our time is equal to our td,"},{"Start":"19:27.300 ","End":"19:31.360","Text":"which is simply going to be equal to our x_0,"},{"Start":"19:31.360 ","End":"19:34.930","Text":"so we said we started at the origin, so our 00."},{"Start":"19:34.930 ","End":"19:42.520","Text":"Plus v_0 cosine Theta multiplied by td,"},{"Start":"19:42.520 ","End":"19:50.405","Text":"so multiply it by 2v_0 sine Theta divided by g,"},{"Start":"19:50.405 ","End":"19:52.319","Text":"which is simply equal to,"},{"Start":"19:52.319 ","End":"19:53.850","Text":"if we just simplify this,"},{"Start":"19:53.850 ","End":"20:00.225","Text":"v_0 squared sine 2 Theta divided by"},{"Start":"20:00.225 ","End":"20:09.280","Text":"g. This is the maximum range or the maximum displacement of our projectile."},{"Start":"20:09.280 ","End":"20:14.215","Text":"Again, you can simply use this equation"},{"Start":"20:14.215 ","End":"20:19.134","Text":"if you\u0027re being asked to find maximum displacement of your projectile."},{"Start":"20:19.134 ","End":"20:26.469","Text":"However, if you understand how to use these equations in order to get your displacement,"},{"Start":"20:26.469 ","End":"20:29.739","Text":"so you don\u0027t have to remember this equation."},{"Start":"20:29.739 ","End":"20:33.030","Text":"The most important thing to remember is that"},{"Start":"20:33.030 ","End":"20:36.210","Text":"in order to find your maximum displacement or your range,"},{"Start":"20:36.210 ","End":"20:39.360","Text":"your condition is that your position in"},{"Start":"20:39.360 ","End":"20:43.710","Text":"the y-direction as a function of time is equal to 0."},{"Start":"20:43.710 ","End":"20:47.199","Text":"Then you just play around with the equations from there."},{"Start":"20:47.400 ","End":"20:50.064","Text":"Also, a brief reminder,"},{"Start":"20:50.064 ","End":"20:54.519","Text":"if you do want to use this equation in order to find the maximum displacement."},{"Start":"20:54.519 ","End":"20:59.830","Text":"Remember that this equation is only correct if our projectile is"},{"Start":"20:59.830 ","End":"21:06.775","Text":"shot from the origin and it lands at the same height from which it was shot."},{"Start":"21:06.775 ","End":"21:13.689","Text":"It\u0027s shot from y is equal to 0 and it lands at the ground at y is equal to 0,"},{"Start":"21:13.689 ","End":"21:15.894","Text":"and stops its motion there."},{"Start":"21:15.894 ","End":"21:18.805","Text":"Otherwise, this equation isn\u0027t correct,"},{"Start":"21:18.805 ","End":"21:24.050","Text":"and you\u0027re anyway going to have to do all the stages that we did here."},{"Start":"21:24.570 ","End":"21:34.219","Text":"Now what we\u0027re going to be doing is we\u0027re going to be speaking about our track equation."},{"Start":"21:35.130 ","End":"21:37.690","Text":"What is our track equation?"},{"Start":"21:37.690 ","End":"21:45.085","Text":"This time, it\u0027s going to be our possession on the y-axis as a function of x."},{"Start":"21:45.085 ","End":"21:47.229","Text":"Instead of as a function of time,"},{"Start":"21:47.229 ","End":"21:52.010","Text":"it\u0027s as a function of what\u0027s happening on the x-axis."},{"Start":"21:52.140 ","End":"21:55.839","Text":"We\u0027ll see that once we write out this equation,"},{"Start":"21:55.839 ","End":"22:04.015","Text":"we\u0027ll really get the root or the track with which our projectile follows,"},{"Start":"22:04.015 ","End":"22:07.310","Text":"we\u0027ll see the trajectory."},{"Start":"22:08.040 ","End":"22:13.330","Text":"How are we going to find our track equation?"},{"Start":"22:13.330 ","End":"22:16.930","Text":"The first thing that we\u0027re going to do is we\u0027re going to use"},{"Start":"22:16.930 ","End":"22:21.055","Text":"our equation for our position in the x-axis as a function of time."},{"Start":"22:21.055 ","End":"22:29.650","Text":"What we want to do is we want to isolate out our t. We\u0027ll get that our t is equal"},{"Start":"22:29.650 ","End":"22:39.669","Text":"to x minus x_0 divided by v_0 cosine of Theta."},{"Start":"22:39.669 ","End":"22:46.270","Text":"A lot of the time they call x minus x_0 delta x,"},{"Start":"22:46.270 ","End":"22:47.950","Text":"and it makes no difference."},{"Start":"22:47.950 ","End":"22:50.980","Text":"Also, a lot of the time we say that our x_0,"},{"Start":"22:50.980 ","End":"22:53.230","Text":"such as in this example over here,"},{"Start":"22:53.230 ","End":"22:56.590","Text":"that our x_0 is equal to 0 in which case we can just"},{"Start":"22:56.590 ","End":"23:00.985","Text":"write x divided by v_0 cosine of Theta."},{"Start":"23:00.985 ","End":"23:03.520","Text":"They\u0027re all variations of the same thing"},{"Start":"23:03.520 ","End":"23:07.990","Text":"depending on how you want to work this out and how you want to label it,"},{"Start":"23:07.990 ","End":"23:11.540","Text":"and also your initial positions."},{"Start":"23:12.030 ","End":"23:14.559","Text":"Now that we have our time,"},{"Start":"23:14.559 ","End":"23:17.829","Text":"we\u0027re going to substitute into"},{"Start":"23:17.829 ","End":"23:23.380","Text":"our position as a function of time but this time on the y-axis."},{"Start":"23:23.380 ","End":"23:25.435","Text":"Every time we see a t,"},{"Start":"23:25.435 ","End":"23:31.210","Text":"we\u0027re going to substitute in this and then we\u0027ll see that we\u0027ll have our y as a function"},{"Start":"23:31.210 ","End":"23:38.755","Text":"of only x and Theta rather than of t. Let\u0027s see this."},{"Start":"23:38.755 ","End":"23:43.959","Text":"Then we\u0027ll have that our y as a function of"},{"Start":"23:43.959 ","End":"23:49.405","Text":"x is going to be equal to our initial y position plus"},{"Start":"23:49.405 ","End":"23:58.180","Text":"v_0 sine of Theta multiplied by t. Let\u0027s just call that dx divided"},{"Start":"23:58.180 ","End":"24:08.275","Text":"by v_0 cosine of Theta minus 1/2g multiplied by t^2."},{"Start":"24:08.275 ","End":"24:18.320","Text":"Let\u0027s call that dx^2 divided by v_0^2 cosine squared Theta."},{"Start":"24:18.930 ","End":"24:22.450","Text":"Now what we can do is we can simplify this equation,"},{"Start":"24:22.450 ","End":"24:25.495","Text":"make it look a little bit nicer so a v_0,"},{"Start":"24:25.495 ","End":"24:27.550","Text":"here cancel out,"},{"Start":"24:27.550 ","End":"24:31.090","Text":"and then we\u0027ll have that our y as a function of x,"},{"Start":"24:31.090 ","End":"24:35.514","Text":"our track equation will be equal to our initial position in y"},{"Start":"24:35.514 ","End":"24:41.455","Text":"plus delta x tan of Theta."},{"Start":"24:41.455 ","End":"24:48.535","Text":"Sine of Theta divided by cosine Theta is tan Theta minus g delta x"},{"Start":"24:48.535 ","End":"24:58.010","Text":"squared divided by 2v_0^2 cosine squared Theta."},{"Start":"24:59.100 ","End":"25:03.490","Text":"What we can see is that our y 0 is a constant,"},{"Start":"25:03.490 ","End":"25:05.544","Text":"our Theta is a constant,"},{"Start":"25:05.544 ","End":"25:08.350","Text":"and our v_0 is a constant and so as r,"},{"Start":"25:08.350 ","End":"25:10.840","Text":"g. These are constants."},{"Start":"25:10.840 ","End":"25:14.110","Text":"We can see that our y is a function of x,"},{"Start":"25:14.110 ","End":"25:16.945","Text":"is as a function of Delta x squared over here."},{"Start":"25:16.945 ","End":"25:20.720","Text":"This is a square equation."},{"Start":"25:21.060 ","End":"25:27.520","Text":"A lot of the time people like to remember this equation and just by"},{"Start":"25:27.520 ","End":"25:30.369","Text":"default use this as the track equation instead of doing"},{"Start":"25:30.369 ","End":"25:34.930","Text":"all the steps that we did up until now in order to get this equation."},{"Start":"25:34.930 ","End":"25:42.609","Text":"This equation is a very general equation because we didn\u0027t use this example specifically,"},{"Start":"25:42.609 ","End":"25:44.545","Text":"we kept it general."},{"Start":"25:44.545 ","End":"25:52.970","Text":"We said that our Delta x is equal to something just in case our x_0 is not at the origin."},{"Start":"25:52.980 ","End":"25:58.390","Text":"We also said that our y_0 can also not be at the origin,"},{"Start":"25:58.390 ","End":"26:00.430","Text":"so we can have some height."},{"Start":"26:00.430 ","End":"26:07.750","Text":"This equation is general and you can use this for any case of projectile motion."},{"Start":"26:07.750 ","End":"26:13.584","Text":"You can remember this or you can do these simple stages in order to get to this equation."},{"Start":"26:13.584 ","End":"26:17.349","Text":"Now what we\u0027re going to speak about are some of"},{"Start":"26:17.349 ","End":"26:21.800","Text":"the basic characteristics of projectile motion."},{"Start":"26:22.620 ","End":"26:27.475","Text":"The basic characteristics of this classic projectile motion,"},{"Start":"26:27.475 ","End":"26:31.855","Text":"specifically this case where our body is shot from the origin,"},{"Start":"26:31.855 ","End":"26:33.339","Text":"where y is equal to 0,"},{"Start":"26:33.339 ","End":"26:39.985","Text":"x is equal to 0 and it lands or so at y is equal to 0."},{"Start":"26:39.985 ","End":"26:43.894","Text":"We can say that our h max,"},{"Start":"26:43.894 ","End":"26:50.414","Text":"so the x position of our h max is half of the displacement,"},{"Start":"26:50.414 ","End":"26:52.590","Text":"half of the range."},{"Start":"26:52.590 ","End":"27:00.750","Text":"Number 1, our x position at a maximum height."},{"Start":"27:00.750 ","End":"27:09.760","Text":"When dealing with this specific example is going to be at half D. I just"},{"Start":"27:09.760 ","End":"27:12.820","Text":"wrote over here so that you are"},{"Start":"27:12.820 ","End":"27:18.580","Text":"reminded that this is only correct when our initial x position,"},{"Start":"27:18.580 ","End":"27:24.385","Text":"it doesn\u0027t really matter, but our initial y position is equal to 0."},{"Start":"27:24.385 ","End":"27:29.840","Text":"We\u0027re also landing our final y is also equal to 0."},{"Start":"27:30.120 ","End":"27:34.315","Text":"Now, for the second thing."},{"Start":"27:34.315 ","End":"27:43.315","Text":"If our projectile shot out at v_0 and it lands with some vf."},{"Start":"27:43.315 ","End":"27:53.810","Text":"The size of our initial velocity is going to be equal to the size of our final velocity."},{"Start":"27:59.010 ","End":"28:04.329","Text":"Similarly, if our projectile,"},{"Start":"28:04.329 ","End":"28:07.269","Text":"it\u0027s shot at an angle of Theta, initially,"},{"Start":"28:07.269 ","End":"28:12.415","Text":"it\u0027s going to land at the exact same angle Theta."},{"Start":"28:12.415 ","End":"28:16.149","Text":"If this is theta i and this is theta f,"},{"Start":"28:16.149 ","End":"28:21.820","Text":"we can say that our Theta I is equal to Theta"},{"Start":"28:21.820 ","End":"28:29.350","Text":"f. Our initial Theta o"},{"Start":"28:29.350 ","End":"28:32.155","Text":"is going to be equal to I Theta final."},{"Start":"28:32.155 ","End":"28:37.549","Text":"Also, this angle here is Theta final."},{"Start":"28:38.490 ","End":"28:42.789","Text":"The third and final thing is"},{"Start":"28:42.789 ","End":"28:46.450","Text":"something that you\u0027ll be asked relatively often in the questions."},{"Start":"28:46.450 ","End":"28:52.000","Text":"That\u0027s to find the magnitude of the velocity at some time."},{"Start":"28:52.000 ","End":"28:55.600","Text":"The magnitude of the velocity at a certain time is"},{"Start":"28:55.600 ","End":"28:58.929","Text":"simply going to be equal to the square root of"},{"Start":"28:58.929 ","End":"29:06.370","Text":"vx^2 squared plus rvy^2."},{"Start":"29:06.370 ","End":"29:10.810","Text":"Our vx^2 is simply going to be this value squared."},{"Start":"29:10.810 ","End":"29:13.690","Text":"We can see that it\u0027s a constant throughout the motion."},{"Start":"29:13.690 ","End":"29:15.534","Text":"It\u0027s just this squared."},{"Start":"29:15.534 ","End":"29:18.955","Text":"Then we have our vy^2,"},{"Start":"29:18.955 ","End":"29:22.330","Text":"which is going to be this squared."},{"Start":"29:22.330 ","End":"29:28.780","Text":"Now notice that our vy is as a function of t. We\u0027re going to have different values"},{"Start":"29:28.780 ","End":"29:35.470","Text":"for our velocity in the y-direction depending on the time into our motion."},{"Start":"29:35.470 ","End":"29:38.440","Text":"All we will have to do is we just have to know"},{"Start":"29:38.440 ","End":"29:45.384","Text":"the time at which we want to know what our velocity in the y-direction is."},{"Start":"29:45.384 ","End":"29:50.829","Text":"Substitute that time into this equation and then square that,"},{"Start":"29:50.829 ","End":"29:54.219","Text":"and then plug it in over here and square root and then"},{"Start":"29:54.219 ","End":"29:58.520","Text":"we have the magnitude of our velocity."},{"Start":"29:59.340 ","End":"30:02.034","Text":"That\u0027s the end of this lesson."},{"Start":"30:02.034 ","End":"30:04.764","Text":"Now this was a relatively brief recap."},{"Start":"30:04.764 ","End":"30:09.230","Text":"I hope that it was clear and that you know the basics now."}],"ID":10595}],"Thumbnail":null,"ID":8958},{"Name":"Track Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Trajectory Equations, Explanation and Example","Duration":"5m 3s","ChapterTopicVideoID":8978,"CourseChapterTopicPlaylistID":5377,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.295","Text":"Hello. In this video,"},{"Start":"00:02.295 ","End":"00:04.815","Text":"we\u0027re going to talk about trajectory equations."},{"Start":"00:04.815 ","End":"00:09.880","Text":"Let\u0027s assume that we have some object that\u0027s moving along the plane."},{"Start":"00:09.920 ","End":"00:12.495","Text":"You can see its trajectory here."},{"Start":"00:12.495 ","End":"00:14.000","Text":"If we make our axis,"},{"Start":"00:14.000 ","End":"00:18.410","Text":"the x-axis is going horizontally and the y-axis is going vertically."},{"Start":"00:18.410 ","End":"00:23.105","Text":"We can describe any given position based on x and y coordinates."},{"Start":"00:23.105 ","End":"00:26.110","Text":"We can also describe its position based on x"},{"Start":"00:26.110 ","End":"00:30.020","Text":"given as a function of t equaling some equation or formula."},{"Start":"00:30.020 ","End":"00:34.980","Text":"Y as a function of t equaling some other equation or formula."},{"Start":"00:35.090 ","End":"00:38.315","Text":"If we want to find the coordinates that way,"},{"Start":"00:38.315 ","End":"00:40.685","Text":"what we do is set."},{"Start":"00:40.685 ","End":"00:47.765","Text":"For example, here we could say x as a function of t=a constant"},{"Start":"00:47.765 ","End":"00:56.210","Text":"a times t. We could also set y as a function of t equal to a different constant,"},{"Start":"00:56.210 ","End":"01:00.660","Text":"let\u0027s say b multiplied by t^2."},{"Start":"01:01.610 ","End":"01:04.490","Text":"To find my trajectory equation,"},{"Start":"01:04.490 ","End":"01:08.450","Text":"what I want to do is take instead of y as a function of t,"},{"Start":"01:08.450 ","End":"01:10.430","Text":"I want to find y as a function of x."},{"Start":"01:10.430 ","End":"01:14.795","Text":"The way that we can do that is by inverting"},{"Start":"01:14.795 ","End":"01:19.820","Text":"the x as a function of t equation set it in terms of t. Instead of x,"},{"Start":"01:19.820 ","End":"01:21.620","Text":"a function of t=at,"},{"Start":"01:21.620 ","End":"01:25.945","Text":"we get t=x over a."},{"Start":"01:25.945 ","End":"01:27.980","Text":"Then we can input that into"},{"Start":"01:27.980 ","End":"01:31.805","Text":"our initial y as a function of t equation in place with a t there."},{"Start":"01:31.805 ","End":"01:36.830","Text":"What you would get as a result is y as a function of"},{"Start":"01:36.830 ","End":"01:42.455","Text":"x=b our constant multiplied by instead of t^2,"},{"Start":"01:42.455 ","End":"01:45.280","Text":"you get x over a^2."},{"Start":"01:45.280 ","End":"01:48.215","Text":"You can write that, in other words,"},{"Start":"01:48.215 ","End":"01:51.005","Text":"as x^2 over a^2."},{"Start":"01:51.005 ","End":"01:54.815","Text":"What happens is if I draw out the function I just created,"},{"Start":"01:54.815 ","End":"02:01.065","Text":"my trajectory equation, I get the exact trajectory that I wanted. As you can see here."},{"Start":"02:01.065 ","End":"02:03.935","Text":"What we got in this example is a parabola."},{"Start":"02:03.935 ","End":"02:10.010","Text":"You would get something along the lines of the following drawing."},{"Start":"02:10.010 ","End":"02:12.005","Text":"Now, with this drawing,"},{"Start":"02:12.005 ","End":"02:14.870","Text":"I\u0027m not going to know exactly where the object is at any given moment,"},{"Start":"02:14.870 ","End":"02:18.480","Text":"but I know all the points that the object must cross."},{"Start":"02:18.770 ","End":"02:21.785","Text":"Let\u0027s take a pretty basic example."},{"Start":"02:21.785 ","End":"02:23.840","Text":"We\u0027re going to use a projectile motion equation,"},{"Start":"02:23.840 ","End":"02:26.240","Text":"and that\u0027s what you\u0027re going to use if you throw something up into the air,"},{"Start":"02:26.240 ","End":"02:29.555","Text":"gravity is acting on it as well as the velocity with which you through it."},{"Start":"02:29.555 ","End":"02:31.340","Text":"The function is as follows."},{"Start":"02:31.340 ","End":"02:37.055","Text":"You take x as a function of time t=v_0,"},{"Start":"02:37.055 ","End":"02:39.365","Text":"the initial velocity times the cosine of Theta."},{"Start":"02:39.365 ","End":"02:42.860","Text":"Theta is the angle at which you through the object multiplied by"},{"Start":"02:42.860 ","End":"02:49.790","Text":"t. Our y is a function of t. We\u0027re given as v_0,"},{"Start":"02:49.790 ","End":"02:59.145","Text":"your initial velocity multiplied by the sine of Theta times t minus 1.5 of gt^2."},{"Start":"02:59.145 ","End":"03:02.375","Text":"If we try to simplify this into our trajectory equation,"},{"Start":"03:02.375 ","End":"03:05.645","Text":"what we have to do is just like in the last example,"},{"Start":"03:05.645 ","End":"03:08.450","Text":"simplify our t and our x."},{"Start":"03:08.450 ","End":"03:13.195","Text":"We find that t=x over v_0 cosine Theta."},{"Start":"03:13.195 ","End":"03:16.190","Text":"If we plug that into our y of t equation,"},{"Start":"03:16.190 ","End":"03:24.590","Text":"we get y as a function of x=v_0 sine Theta multiplied by"},{"Start":"03:24.590 ","End":"03:32.925","Text":"x over v_0 cosine Theta"},{"Start":"03:32.925 ","End":"03:35.360","Text":"minus 1.5 g. Again,"},{"Start":"03:35.360 ","End":"03:46.240","Text":"instead of t, we put in x^2 over v_0^2 times cosine^2 of Theta."},{"Start":"03:46.610 ","End":"03:48.890","Text":"If we simplify this a little bit,"},{"Start":"03:48.890 ","End":"03:56.975","Text":"we can eliminate our v_0\u0027s and we end up with y as a function of x=x"},{"Start":"03:56.975 ","End":"04:06.365","Text":"tangent Theta minus 1.5 g x^2 over v^2 cosine^2 Theta."},{"Start":"04:06.365 ","End":"04:09.935","Text":"Now that we have our function as y in terms of x,"},{"Start":"04:09.935 ","End":"04:11.840","Text":"and it\u0027s all with constants,"},{"Start":"04:11.840 ","End":"04:17.735","Text":"we can actually map it out and see because our negative 1.5 constants times x^2,"},{"Start":"04:17.735 ","End":"04:20.480","Text":"we have some degrading parabola."},{"Start":"04:20.480 ","End":"04:22.160","Text":"Because of our x in the beginning,"},{"Start":"04:22.160 ","End":"04:26.990","Text":"there\u0027s something that\u0027s going to push our object upwards at the beginning."},{"Start":"04:26.990 ","End":"04:30.110","Text":"In fact, its maximum height is going to be some in the middle,"},{"Start":"04:30.110 ","End":"04:34.745","Text":"as you can see there, we have some a parabola that\u0027s degrading over time."},{"Start":"04:34.745 ","End":"04:37.010","Text":"To summarize this in short,"},{"Start":"04:37.010 ","End":"04:40.430","Text":"what we\u0027re doing is using some basic algebra to simplify"},{"Start":"04:40.430 ","End":"04:42.770","Text":"our 2 equations of x as a function of t"},{"Start":"04:42.770 ","End":"04:45.620","Text":"and y is a function of t to find our y as a function of x."},{"Start":"04:45.620 ","End":"04:49.310","Text":"Once we have that, we can really plot out the path of"},{"Start":"04:49.310 ","End":"04:56.930","Text":"our projectile and see every point at which that projectile will cross along its path."},{"Start":"04:56.930 ","End":"04:59.245","Text":"Really, it\u0027s just some basic algebra"},{"Start":"04:59.245 ","End":"05:02.430","Text":"and you can have everything figured out pretty quickly."}],"ID":9233}],"Thumbnail":null,"ID":5377},{"Name":"Normal and Tangential Acceleration","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Describing Types of Acceleration","Duration":"4m 45s","ChapterTopicVideoID":8979,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.910","Text":"Hello. In this video,"},{"Start":"00:02.910 ","End":"00:07.425","Text":"we\u0027re going to talk about your normal acceleration and tangential acceleration."},{"Start":"00:07.425 ","End":"00:09.255","Text":"First, let start with some symbols."},{"Start":"00:09.255 ","End":"00:18.700","Text":"Normal acceleration is a_n and tangential acceleration is symbolized by a_t."},{"Start":"00:19.220 ","End":"00:24.270","Text":"At this early point,"},{"Start":"00:24.270 ","End":"00:32.070","Text":"it\u0027s really important to just note that your normal acceleration is not the same thing as"},{"Start":"00:32.070 ","End":"00:35.970","Text":"radial acceleration and that your tangential acceleration is"},{"Start":"00:35.970 ","End":"00:40.440","Text":"not the same as acceleration in the direction of Theta acceleration."},{"Start":"00:40.440 ","End":"00:42.620","Text":"They are similar, there are some shared properties,"},{"Start":"00:42.620 ","End":"00:46.430","Text":"but a lot of people make this mistake and I want to make sure at the first point,"},{"Start":"00:46.430 ","End":"00:49.780","Text":"at the get-go, to make sure they\u0027re not the same thing."},{"Start":"00:49.780 ","End":"00:52.205","Text":"With that out of the way, let\u0027s begin."},{"Start":"00:52.205 ","End":"00:57.230","Text":"Imagine that your origin is a point here and that you have"},{"Start":"00:57.230 ","End":"01:03.240","Text":"an object that\u0027s moving in a curved path along your Cartesian coordinates like so."},{"Start":"01:03.460 ","End":"01:06.365","Text":"This is just an arbitrary graph,"},{"Start":"01:06.365 ","End":"01:09.680","Text":"but let\u0027s say we want to find any given point on this path,"},{"Start":"01:09.680 ","End":"01:12.410","Text":"an arbitrary point again, let\u0027s say this point."},{"Start":"01:12.410 ","End":"01:15.665","Text":"I want to know some of the properties of the movement here."},{"Start":"01:15.665 ","End":"01:22.250","Text":"First of all, I know that my velocity is in the tangential direction like this."},{"Start":"01:22.250 ","End":"01:26.270","Text":"In fact, people say oftentimes tangential velocity,"},{"Start":"01:26.270 ","End":"01:29.420","Text":"but really all velocity is tangential in the sense that it\u0027s going,"},{"Start":"01:29.420 ","End":"01:31.910","Text":"as you can see here, not exactly along the path,"},{"Start":"01:31.910 ","End":"01:35.000","Text":"but continuing in a tangential direction,"},{"Start":"01:35.000 ","End":"01:37.310","Text":"which is tangential to the main path."},{"Start":"01:37.310 ","End":"01:39.320","Text":"It could be tangential in either direction,"},{"Start":"01:39.320 ","End":"01:41.705","Text":"but assuming that we\u0027re moving from left to right here,"},{"Start":"01:41.705 ","End":"01:46.570","Text":"it\u0027s always going to be pointing towards the right and tangential to the path."},{"Start":"01:46.570 ","End":"01:52.820","Text":"We know our velocity at any point here, at least the direction of it."},{"Start":"01:52.820 ","End":"01:54.545","Text":"If we\u0027re talking about our acceleration,"},{"Start":"01:54.545 ","End":"01:58.640","Text":"it can really go in any direction with the caveat that it\u0027s always"},{"Start":"01:58.640 ","End":"02:03.260","Text":"going to be going towards the inside of a curve, the concave part."},{"Start":"02:03.260 ","End":"02:10.087","Text":"In this example, at the point where our velocity is symbolized by that black arrow,"},{"Start":"02:10.087 ","End":"02:14.045","Text":"in red, you see some potential acceleration vectors."},{"Start":"02:14.045 ","End":"02:15.800","Text":"It cannot be that top line,"},{"Start":"02:15.800 ","End":"02:19.345","Text":"it has to be 1 of those ones going in towards the inside of a curve."},{"Start":"02:19.345 ","End":"02:21.860","Text":"Let\u0027s say it\u0027s that one, so that means that if we"},{"Start":"02:21.860 ","End":"02:24.110","Text":"choose the second where the curve changes,"},{"Start":"02:24.110 ","End":"02:26.945","Text":"our acceleration is going to be going up on our screen here."},{"Start":"02:26.945 ","End":"02:30.695","Text":"Again, if we go back to the end of the graph where the curves back down,"},{"Start":"02:30.695 ","End":"02:32.915","Text":"you\u0027re going to see the red line going in."},{"Start":"02:32.915 ","End":"02:34.280","Text":"Again, it\u0027s always concave,"},{"Start":"02:34.280 ","End":"02:39.050","Text":"and if you notice the velocity is actually always going towards the convex side."},{"Start":"02:39.050 ","End":"02:44.135","Text":"These 3 red vectors we just drew are all general acceleration vector,"},{"Start":"02:44.135 ","End":"02:46.775","Text":"they describe our total acceleration."},{"Start":"02:46.775 ","End":"02:52.265","Text":"If we want to find our normal acceleration and our tangential acceleration,"},{"Start":"02:52.265 ","End":"02:56.545","Text":"what we need to do is basically break this general vector down into 2 bits."},{"Start":"02:56.545 ","End":"03:00.665","Text":"Now we already know that you can break down your acceleration into"},{"Start":"03:00.665 ","End":"03:05.420","Text":"a y acceleration and x acceleration based on the x and y axes."},{"Start":"03:05.420 ","End":"03:07.760","Text":"Your x acceleration will look something like this,"},{"Start":"03:07.760 ","End":"03:09.335","Text":"parallel to the x-axis,"},{"Start":"03:09.335 ","End":"03:13.025","Text":"and your y acceleration will be parallel to the y-axis as well."},{"Start":"03:13.025 ","End":"03:16.850","Text":"Now what we can do is we can break down our normal and"},{"Start":"03:16.850 ","End":"03:20.855","Text":"our tangential acceleration using just a different set of axes."},{"Start":"03:20.855 ","End":"03:23.720","Text":"Instead of going based on the x and the y,"},{"Start":"03:23.720 ","End":"03:27.600","Text":"we\u0027re actually going to draw a different axis,"},{"Start":"03:27.600 ","End":"03:31.835","Text":"1 that is based on the tangent that we looked at before,"},{"Start":"03:31.835 ","End":"03:33.500","Text":"and the other that\u0027s orthogonal,"},{"Start":"03:33.500 ","End":"03:35.645","Text":"that\u0027s perpendicular to that."},{"Start":"03:35.645 ","End":"03:38.345","Text":"This is the orthogonal axis,"},{"Start":"03:38.345 ","End":"03:41.240","Text":"the perpendicular axis if you will and"},{"Start":"03:41.240 ","End":"03:46.250","Text":"our initial tangent from the velocity is our tangential axis."},{"Start":"03:46.250 ","End":"03:49.775","Text":"We\u0027re going to take each of these components and give them a new name,"},{"Start":"03:49.775 ","End":"03:52.880","Text":"the perpendicular or orthogonal line,"},{"Start":"03:52.880 ","End":"03:54.815","Text":"which is in this case more vertical,"},{"Start":"03:54.815 ","End":"03:58.080","Text":"is going to be our normal acceleration."},{"Start":"03:58.270 ","End":"04:03.260","Text":"The other line which is more parallel to"},{"Start":"04:03.260 ","End":"04:10.380","Text":"our path is the vector of your tangential acceleration."},{"Start":"04:10.380 ","End":"04:12.870","Text":"These are our 2 components."},{"Start":"04:12.870 ","End":"04:17.730","Text":"They are again broken down from the general acceleration and there are different than"},{"Start":"04:17.730 ","End":"04:19.980","Text":"the y acceleration and"},{"Start":"04:19.980 ","End":"04:24.425","Text":"the x acceleration in the sense of the orientation of the axes being different."},{"Start":"04:24.425 ","End":"04:27.560","Text":"But there\u0027s a reason we do this that we\u0027ll talk about soon."},{"Start":"04:27.560 ","End":"04:29.420","Text":"Now we\u0027ve described this graphically,"},{"Start":"04:29.420 ","End":"04:31.910","Text":"but the next thing I want to do is explore this algebraically,"},{"Start":"04:31.910 ","End":"04:35.600","Text":"how we can use the vector of velocity and"},{"Start":"04:35.600 ","End":"04:40.890","Text":"the vector of your acceleration to find a_n and a_t."}],"ID":9234},{"Watched":false,"Name":"Finding the Tangential Acceleration","Duration":"2m 58s","ChapterTopicVideoID":8980,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"The way is as follows."},{"Start":"00:03.480 ","End":"00:05.940","Text":"What I have is the vector a,"},{"Start":"00:05.940 ","End":"00:10.440","Text":"and I want to use that to find the vector a_t,"},{"Start":"00:10.440 ","End":"00:14.400","Text":"which is the component of the vector a going in the direction of"},{"Start":"00:14.400 ","End":"00:19.520","Text":"v. We\u0027ve done some practice on this before and"},{"Start":"00:19.520 ","End":"00:23.160","Text":"what we\u0027re going to do to get 1 vector that\u0027s going to be in"},{"Start":"00:23.160 ","End":"00:27.845","Text":"the direction of another vector is we take a vector"},{"Start":"00:27.845 ","End":"00:37.915","Text":"times v vector over the absolute value of v^2 multiplied by v vector."},{"Start":"00:37.915 ","End":"00:41.345","Text":"Your scalar multiplication on top will give you a row number."},{"Start":"00:41.345 ","End":"00:43.580","Text":"Your absolute value, and the bottom will give you a number in which you"},{"Start":"00:43.580 ","End":"00:45.830","Text":"can end up as some number multiplied by the vector"},{"Start":"00:45.830 ","End":"00:49.040","Text":"v. You can also write this as a formula and I"},{"Start":"00:49.040 ","End":"00:52.910","Text":"actually want to stop here and give an explanation of why this formula works."},{"Start":"00:52.910 ","End":"00:59.720","Text":"If you remember when we do a scalar multiplication of the vectors a and"},{"Start":"00:59.720 ","End":"01:02.600","Text":"v. What you\u0027re going to get as"},{"Start":"01:02.600 ","End":"01:09.095","Text":"a result is the absolute value of a times the absolute value of v,"},{"Start":"01:09.095 ","End":"01:12.950","Text":"times the cosine of the angle between the 2 of them."},{"Start":"01:12.950 ","End":"01:16.490","Text":"Now the angle between the 2 of them is, as you can see here."},{"Start":"01:16.490 ","End":"01:21.590","Text":"We\u0027re using Alpha to symbolize it and it\u0027s that angle."},{"Start":"01:21.590 ","End":"01:26.660","Text":"Now what happens if we divide the whole thing"},{"Start":"01:26.660 ","End":"01:31.655","Text":"by the absolute value of v ones is we actually get a simplification here,"},{"Start":"01:31.655 ","End":"01:36.230","Text":"the v\u0027s dropout and we end up with is the absolute value of a times"},{"Start":"01:36.230 ","End":"01:38.690","Text":"the cosine of the angle between them and that"},{"Start":"01:38.690 ","End":"01:41.825","Text":"equals the length of the vector for the tangent line a_t."},{"Start":"01:41.825 ","End":"01:43.940","Text":"Now there are times when you\u0027re interested in only"},{"Start":"01:43.940 ","End":"01:46.280","Text":"knowing the length or the value of that line,"},{"Start":"01:46.280 ","End":"01:47.450","Text":"not necessarily the direction,"},{"Start":"01:47.450 ","End":"01:52.870","Text":"so what you can do is you can write a formula that negates the vector lines."},{"Start":"01:52.870 ","End":"01:58.430","Text":"Instead of vector a_t is just a_t and it equals the scalar multiplication of vector"},{"Start":"01:58.430 ","End":"02:01.370","Text":"a and vector v over the absolute value of"},{"Start":"02:01.370 ","End":"02:04.985","Text":"v. If you want to be more clear to make sure your notes just giving you a value."},{"Start":"02:04.985 ","End":"02:09.080","Text":"If you want, you can write in those 2 lines as well."},{"Start":"02:09.080 ","End":"02:11.960","Text":"This will just give us the absolute value,"},{"Start":"02:11.960 ","End":"02:14.255","Text":"the size of our vector."},{"Start":"02:14.255 ","End":"02:15.980","Text":"If we want to find the direction,"},{"Start":"02:15.980 ","End":"02:17.780","Text":"we can actually add that bit in as"},{"Start":"02:17.780 ","End":"02:20.950","Text":"well and it will reveal to us how the whole formula works."},{"Start":"02:20.950 ","End":"02:23.030","Text":"Remember, if we want to get the direction of vector,"},{"Start":"02:23.030 ","End":"02:25.865","Text":"we take vector a_t for the vector we\u0027re trying to find"},{"Start":"02:25.865 ","End":"02:31.645","Text":"equals a_t absolute value times v hat,"},{"Start":"02:31.645 ","End":"02:35.510","Text":"which is for that vector we\u0027re measuring there."},{"Start":"02:35.510 ","End":"02:38.900","Text":"Now remember that v hat equals and we\u0027re going"},{"Start":"02:38.900 ","End":"02:42.005","Text":"to add in our a_t to make sure our equations are equal v hat equals"},{"Start":"02:42.005 ","End":"02:45.125","Text":"vector v over absolute value of v. Now if you take"},{"Start":"02:45.125 ","End":"02:48.830","Text":"that bit here for the direction and add it into our original formula,"},{"Start":"02:48.830 ","End":"02:52.730","Text":"you end up with the formula on top that we originally started with."},{"Start":"02:52.730 ","End":"02:54.050","Text":"As you can see,"},{"Start":"02:54.050 ","End":"02:56.585","Text":"this will give us both the size and the direction of a vector."},{"Start":"02:56.585 ","End":"02:58.920","Text":"Now let\u0027s move on to a_n."}],"ID":9235},{"Watched":false,"Name":"Finding the Normal Acceleration","Duration":"2m 43s","ChapterTopicVideoID":8981,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"If we want to find the a_n vector,"},{"Start":"00:03.390 ","End":"00:05.625","Text":"there\u0027s really only 1 way to do it."},{"Start":"00:05.625 ","End":"00:09.840","Text":"What we need to do is take the total of a vector, if you recall,"},{"Start":"00:09.840 ","End":"00:13.305","Text":"that\u0027s our line that\u0027s going at about a 45-degree angle here,"},{"Start":"00:13.305 ","End":"00:16.785","Text":"that line, and we need to subtract from it the a_t vector."},{"Start":"00:16.785 ","End":"00:19.680","Text":"Why is that? Because if you remember,"},{"Start":"00:19.680 ","End":"00:24.195","Text":"the way we get the total vector is a_t plus a_n."},{"Start":"00:24.195 ","End":"00:28.875","Text":"If we subtract a_t from the a vector,"},{"Start":"00:28.875 ","End":"00:31.710","Text":"what we\u0027re left with is a_n."},{"Start":"00:31.710 ","End":"00:36.285","Text":"We have to find the value of our a_t vector first."},{"Start":"00:36.285 ","End":"00:39.890","Text":"This is how you find the entire vector."},{"Start":"00:39.890 ","End":"00:43.625","Text":"Sometimes you\u0027re interested in just the magnitude."},{"Start":"00:43.625 ","End":"00:45.560","Text":"If you\u0027re interested just the magnitude,"},{"Start":"00:45.560 ","End":"00:46.850","Text":"there\u0027s a couple of ways to find that."},{"Start":"00:46.850 ","End":"00:49.760","Text":"First, what you can do is take the same equation and just write"},{"Start":"00:49.760 ","End":"00:53.395","Text":"it in terms of magnitude instead of in terms of the entire vector."},{"Start":"00:53.395 ","End":"00:55.890","Text":"Again, you put your bars around"},{"Start":"00:55.890 ","End":"01:00.055","Text":"a_n and you\u0027re going to put your bars around a minus a_t,"},{"Start":"01:00.055 ","End":"01:01.940","Text":"and you\u0027ll find your absolute values,"},{"Start":"01:01.940 ","End":"01:03.545","Text":"your magnitudes that way."},{"Start":"01:03.545 ","End":"01:05.420","Text":"This is a little bit complicated though,"},{"Start":"01:05.420 ","End":"01:08.030","Text":"because the equation is referring to the equation we"},{"Start":"01:08.030 ","End":"01:11.800","Text":"just wrote above is a little complicated in itself."},{"Start":"01:11.800 ","End":"01:14.035","Text":"Shortcut is to do as follows;"},{"Start":"01:14.035 ","End":"01:16.655","Text":"to find the magnitude of a_n,"},{"Start":"01:16.655 ","End":"01:22.220","Text":"you\u0027re going to take the magnitude of the vector a,"},{"Start":"01:22.220 ","End":"01:26.855","Text":"the total vector, multiplied by the sine of the angle Alpha,"},{"Start":"01:26.855 ","End":"01:28.750","Text":"which is between the 2 of them."},{"Start":"01:28.750 ","End":"01:33.575","Text":"The way to find the value of sin(Alpha) is through a cross product or vector product."},{"Start":"01:33.575 ","End":"01:40.570","Text":"You\u0027re going to do a cross-product of the a vector and the v vector."},{"Start":"01:40.570 ","End":"01:44.495","Text":"What that\u0027s going to equal is the magnitude of"},{"Start":"01:44.495 ","End":"01:50.170","Text":"a times the magnitude of v times the sin(Alpha)."},{"Start":"01:50.170 ","End":"01:53.240","Text":"You want to make sure to put your lines there"},{"Start":"01:53.240 ","End":"01:55.880","Text":"to make sure you\u0027re finding a magnitude and not just a vector."},{"Start":"01:55.880 ","End":"02:00.870","Text":"If you notice, the 2nd equation we just wrote is very similar to the 1 above it."},{"Start":"02:00.880 ","End":"02:06.245","Text":"What you\u0027re going to do to find the value of a_n, your magnitude,"},{"Start":"02:06.245 ","End":"02:12.845","Text":"is you\u0027re going to take the cross-product of"},{"Start":"02:12.845 ","End":"02:20.615","Text":"a and v and divide that by the magnitude of the vector v,"},{"Start":"02:20.615 ","End":"02:21.710","Text":"and that\u0027ll give you your magnitude."},{"Start":"02:21.710 ","End":"02:23.915","Text":"It won\u0027t give you direction. It won\u0027t give you the angle,"},{"Start":"02:23.915 ","End":"02:26.710","Text":"but it\u0027ll give you the magnitude if that\u0027s what you\u0027re interested in."},{"Start":"02:26.710 ","End":"02:29.330","Text":"Of course, if you want to find the direction,"},{"Start":"02:29.330 ","End":"02:31.460","Text":"you need to go with the prior formula but we"},{"Start":"02:31.460 ","End":"02:34.595","Text":"can also write this out as a nice formula to use later."},{"Start":"02:34.595 ","End":"02:37.640","Text":"Before figuring out exactly what this is telling us,"},{"Start":"02:37.640 ","End":"02:41.790","Text":"I just want to go through and do an example real quick. Let\u0027s go for it."}],"ID":9236},{"Watched":false,"Name":"Using the Tangential Acceleration","Duration":"4m 42s","ChapterTopicVideoID":8982,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.130","Text":"Now the real question is,"},{"Start":"00:02.130 ","End":"00:04.290","Text":"what do you do with this information?"},{"Start":"00:04.290 ","End":"00:10.365","Text":"How do you use your tangential and acceleration."},{"Start":"00:10.365 ","End":"00:13.215","Text":"Let\u0027s start with the tangential acceleration."},{"Start":"00:13.215 ","End":"00:16.780","Text":"The tangential acceleration is in the direction of your velocity."},{"Start":"00:16.780 ","End":"00:18.390","Text":"Let\u0027s suppose for a second you only have"},{"Start":"00:18.390 ","End":"00:21.820","Text":"your tangential acceleration at and it\u0027s only in direction of your velocity."},{"Start":"00:21.820 ","End":"00:24.900","Text":"What you\u0027re going to do is it\u0027s actually going to change"},{"Start":"00:24.900 ","End":"00:28.170","Text":"the magnitude or the length of your vector,"},{"Start":"00:28.170 ","End":"00:29.595","Text":"but not the direction."},{"Start":"00:29.595 ","End":"00:35.650","Text":"Let\u0027s assume that you have a velocity at any given time really,"},{"Start":"00:35.650 ","End":"00:38.125","Text":"and it\u0027s in this direction."},{"Start":"00:38.125 ","End":"00:41.210","Text":"It\u0027s just like in the picture more or less."},{"Start":"00:41.210 ","End":"00:45.230","Text":"What we do is we take our acceleration and add it onto that."},{"Start":"00:45.230 ","End":"00:49.640","Text":"Now acceleration is the difference in velocity over the difference in time."},{"Start":"00:49.640 ","End":"00:52.190","Text":"Meaning that our difference in velocity is the"},{"Start":"00:52.190 ","End":"00:54.740","Text":"same as acceleration times the difference in time."},{"Start":"00:54.740 ","End":"00:58.760","Text":"You take that acceleration vector, that at vector,"},{"Start":"00:58.760 ","End":"01:01.145","Text":"and we multiply it by a tiny amount of time,"},{"Start":"01:01.145 ","End":"01:04.075","Text":"and it\u0027s going to give you a tiny increase in your velocity."},{"Start":"01:04.075 ","End":"01:06.890","Text":"It\u0027s going to be in that same direction because as we said,"},{"Start":"01:06.890 ","End":"01:08.300","Text":"the vector is in that same direction,"},{"Start":"01:08.300 ","End":"01:10.190","Text":"so it\u0027s only in that direction."},{"Start":"01:10.190 ","End":"01:13.070","Text":"You\u0027re going to get a slightly larger velocity."},{"Start":"01:13.070 ","End":"01:15.290","Text":"Now, if this is confusing,"},{"Start":"01:15.290 ","End":"01:17.705","Text":"there\u0027s a lot of differentials here, a lot of algebra."},{"Start":"01:17.705 ","End":"01:23.180","Text":"Just remember the basic principle that your tangential acceleration,"},{"Start":"01:23.180 ","End":"01:25.055","Text":"if it is truly tangential,"},{"Start":"01:25.055 ","End":"01:27.830","Text":"is only going to increase the magnitude or"},{"Start":"01:27.830 ","End":"01:32.135","Text":"affect the magnitude of your velocity and not the direction."},{"Start":"01:32.135 ","End":"01:35.130","Text":"In fact, we can even write that down as a formula."},{"Start":"01:35.130 ","End":"01:36.485","Text":"What you need to remember is,"},{"Start":"01:36.485 ","End":"01:37.760","Text":"I\u0027ll write it out here a_t,"},{"Start":"01:37.760 ","End":"01:41.300","Text":"your tangential acceleration is going to give you the difference in"},{"Start":"01:41.300 ","End":"01:46.840","Text":"the magnitude of your velocity over dt."},{"Start":"01:46.840 ","End":"01:51.995","Text":"Here it is in a more clean formula typed out for you."},{"Start":"01:51.995 ","End":"01:56.830","Text":"Again, changing only the magnitude of the velocity."},{"Start":"01:56.830 ","End":"02:01.970","Text":"If we\u0027re talking about our normal acceleration as opposed to the tangential acceleration,"},{"Start":"02:01.970 ","End":"02:05.915","Text":"it\u0027s only going to affect the direction of the velocity,"},{"Start":"02:05.915 ","End":"02:07.790","Text":"not the magnitude, but the direction."},{"Start":"02:07.790 ","End":"02:09.170","Text":"Is it in fact does the opposite thing,"},{"Start":"02:09.170 ","End":"02:13.460","Text":"so I\u0027ll try to explain here why the normal acceleration only affects"},{"Start":"02:13.460 ","End":"02:19.075","Text":"the direction and not the magnitude of the velocity."},{"Start":"02:19.075 ","End":"02:23.140","Text":"The best way to do that is through a hypothetical example here."},{"Start":"02:23.140 ","End":"02:26.375","Text":"Let\u0027s say there\u0027s an object that\u0027s traveling on a path."},{"Start":"02:26.375 ","End":"02:30.725","Text":"Let\u0027s say this is the path, and at some time t,"},{"Start":"02:30.725 ","End":"02:35.165","Text":"the velocity is tangential to the path as follows."},{"Start":"02:35.165 ","End":"02:42.920","Text":"We\u0027re going to call this V at t. Let\u0027s say at some other time, a second later,"},{"Start":"02:42.920 ","End":"02:45.455","Text":"millisecond later even less,"},{"Start":"02:45.455 ","End":"02:50.885","Text":"the velocity we know is going to be tangential in that direction that I just drew there."},{"Start":"02:50.885 ","End":"02:52.580","Text":"Again, our top 1 is Vt,"},{"Start":"02:52.580 ","End":"02:55.300","Text":"or second is Vt plus dt."},{"Start":"02:55.300 ","End":"02:58.010","Text":"Ideally, this would be right next to each other,"},{"Start":"02:58.010 ","End":"02:59.360","Text":"but for the sake of elucidation,"},{"Start":"02:59.360 ","End":"03:01.300","Text":"it\u0027s easier to draw them a little farther apart,"},{"Start":"03:01.300 ","End":"03:04.200","Text":"so you can actually see the difference with your eyes."},{"Start":"03:04.250 ","End":"03:11.010","Text":"What we can do is take the vector V at t and write it out below,"},{"Start":"03:11.260 ","End":"03:18.320","Text":"and then drag the Vt plus dt vector and put it right alongside there,"},{"Start":"03:18.320 ","End":"03:20.005","Text":"so you can see the difference."},{"Start":"03:20.005 ","End":"03:23.540","Text":"When we compare them, and I have to say with the caveat"},{"Start":"03:23.540 ","End":"03:27.170","Text":"that we\u0027re talking about a velocity whose magnitude stays constant."},{"Start":"03:27.170 ","End":"03:28.730","Text":"We\u0027re only talking about a difference in direction,"},{"Start":"03:28.730 ","End":"03:32.310","Text":"not the speed or the velocity."},{"Start":"03:32.320 ","End":"03:35.390","Text":"If we overlay them and see the difference,"},{"Start":"03:35.390 ","End":"03:39.500","Text":"we\u0027re going to find a slight change in angle,"},{"Start":"03:39.500 ","End":"03:44.045","Text":"and that\u0027s going to be what\u0027s caused only by the normal acceleration,"},{"Start":"03:44.045 ","End":"03:46.735","Text":"not by the tangential acceleration."},{"Start":"03:46.735 ","End":"03:51.830","Text":"If we imagine something like infinitesimally small as our d/t,"},{"Start":"03:51.830 ","End":"03:54.200","Text":"imagine that this is a nanosecond afterwards,"},{"Start":"03:54.200 ","End":"03:58.670","Text":"what we\u0027d find is 2 lines that are almost touching,"},{"Start":"03:58.670 ","End":"04:00.770","Text":"they\u0027re pretty much plumped 1 with the other."},{"Start":"04:00.770 ","End":"04:05.730","Text":"That would be orthogonal angle, your normal acceleration."},{"Start":"04:05.750 ","End":"04:07.860","Text":"If we draw it out here,"},{"Start":"04:07.860 ","End":"04:15.995","Text":"you can see a little better maybe that if Vt and Vdt are infinitesimally different,"},{"Start":"04:15.995 ","End":"04:19.130","Text":"that you would find only a slight change in angle,"},{"Start":"04:19.130 ","End":"04:23.985","Text":"no change in the velocity itself at all."},{"Start":"04:23.985 ","End":"04:28.070","Text":"Hopefully that helps explain to you why the normal acceleration changes"},{"Start":"04:28.070 ","End":"04:33.040","Text":"only the angle and not the magnitude of the velocity."},{"Start":"04:33.040 ","End":"04:38.160","Text":"In the next video, we\u0027ll see the formula for the normal acceleration."}],"ID":9237},{"Watched":false,"Name":"Curvature Radius","Duration":"3m 52s","ChapterTopicVideoID":8983,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:04.470","Text":"In this video, we\u0027re going to talk about the normal acceleration"},{"Start":"00:04.470 ","End":"00:09.120","Text":"and how you use that to find the radius of the curvature."},{"Start":"00:09.120 ","End":"00:12.600","Text":"The first thing to do is explain what is the radius of the curvature."},{"Start":"00:12.600 ","End":"00:14.055","Text":"In a basic sense,"},{"Start":"00:14.055 ","End":"00:16.230","Text":"on any curved slope,"},{"Start":"00:16.230 ","End":"00:19.155","Text":"and any curved path of an object like this one,"},{"Start":"00:19.155 ","End":"00:21.075","Text":"at any given point,"},{"Start":"00:21.075 ","End":"00:23.910","Text":"your object is on a slight arc, on a slight curve."},{"Start":"00:23.910 ","End":"00:26.385","Text":"It\u0027s not traveling exactly straight."},{"Start":"00:26.385 ","End":"00:29.235","Text":"I\u0027ll actually draw this out in blue,"},{"Start":"00:29.235 ","End":"00:32.015","Text":"but imagine that little curve there,"},{"Start":"00:32.015 ","End":"00:34.160","Text":"and it could really be Infinitesimally small."},{"Start":"00:34.160 ","End":"00:35.735","Text":"We\u0027re talking about points at the end of the day,"},{"Start":"00:35.735 ","End":"00:37.895","Text":"not really long lines."},{"Start":"00:37.895 ","End":"00:40.460","Text":"But that curve for the sake of seeing it,"},{"Start":"00:40.460 ","End":"00:43.525","Text":"It\u0027s a little larger, is on an arc."},{"Start":"00:43.525 ","End":"00:48.990","Text":"That arc that it\u0027s on can really be extended into a full circle if you want it,"},{"Start":"00:49.120 ","End":"00:54.200","Text":"so here\u0027s an extension into a circle just so you can visualize it a little better."},{"Start":"00:54.200 ","End":"00:56.310","Text":"Didn\u0027t come out so bad."},{"Start":"00:56.310 ","End":"01:00.020","Text":"Now, this is actually part of a circle if you think about it in that way,"},{"Start":"01:00.020 ","End":"01:04.085","Text":"and if you go to a different point on that same slope on that same path,"},{"Start":"01:04.085 ","End":"01:06.995","Text":"you\u0027ll find that each point as part of a different circle."},{"Start":"01:06.995 ","End":"01:09.590","Text":"At the top there you can see another potential circle."},{"Start":"01:09.590 ","End":"01:11.120","Text":"This one didn\u0027t come out perfectly,"},{"Start":"01:11.120 ","End":"01:13.000","Text":"but you get the idea."},{"Start":"01:13.000 ","End":"01:15.800","Text":"What I can say is that any given point here is part"},{"Start":"01:15.800 ","End":"01:18.320","Text":"of an arc or part of a larger circle really."},{"Start":"01:18.320 ","End":"01:20.825","Text":"As I changed around, change the direction,"},{"Start":"01:20.825 ","End":"01:25.240","Text":"the circle that is a part of indirection of it and the size of it is going to change."},{"Start":"01:25.240 ","End":"01:28.520","Text":"Ideally, what we have here is that our normal acceleration,"},{"Start":"01:28.520 ","End":"01:31.640","Text":"that orthogonal line, is going to lead me"},{"Start":"01:31.640 ","End":"01:35.785","Text":"right to the center of the circle of which that point is a part of."},{"Start":"01:35.785 ","End":"01:39.455","Text":"Then we can make a line for the radius of that curvature,"},{"Start":"01:39.455 ","End":"01:40.760","Text":"the radius of that circle."},{"Start":"01:40.760 ","End":"01:44.240","Text":"I\u0027ll draw it out for you here so you get a sense of what that might be."},{"Start":"01:44.240 ","End":"01:48.595","Text":"This line going to the right."},{"Start":"01:48.595 ","End":"01:52.010","Text":"Any given point here is going to be part of a twisting motion,"},{"Start":"01:52.010 ","End":"01:55.610","Text":"part of a torque, and it\u0027s going to have its own circle with its own radius."},{"Start":"01:55.610 ","End":"02:00.080","Text":"This one has a larger radius compared to the second I drew earlier on the curve,"},{"Start":"02:00.080 ","End":"02:01.790","Text":"which has a smaller radius."},{"Start":"02:01.790 ","End":"02:05.255","Text":"Actually, a smaller radius indicates a more intense curve,"},{"Start":"02:05.255 ","End":"02:12.480","Text":"whereas a larger radius indicates a more moderate curve."},{"Start":"02:13.730 ","End":"02:19.145","Text":"Now you know what we\u0027re talking about when we\u0027re saying the radius of the curvature."},{"Start":"02:19.145 ","End":"02:21.860","Text":"The next thing to do is figure out how to calculate it."},{"Start":"02:21.860 ","End":"02:23.480","Text":"If you\u0027re thinking about the radius of the curvature,"},{"Start":"02:23.480 ","End":"02:25.880","Text":"you actually think about the physics of the curve,"},{"Start":"02:25.880 ","End":"02:29.375","Text":"not of the body and motion of the object in motion."},{"Start":"02:29.375 ","End":"02:32.660","Text":"We can kind of forget larger curve and just focus on this one second,"},{"Start":"02:32.660 ","End":"02:34.990","Text":"this one little moment of time."},{"Start":"02:34.990 ","End":"02:36.830","Text":"We know that if it\u0027s moving on a curve,"},{"Start":"02:36.830 ","End":"02:38.960","Text":"we can use the formula for curvature motion,"},{"Start":"02:38.960 ","End":"02:42.290","Text":"which is the radial acceleration is equal"},{"Start":"02:42.290 ","End":"02:48.750","Text":"to the velocity squared over the radius of the curvature."},{"Start":"02:49.610 ","End":"02:54.230","Text":"That means that the radius of the curvature is equal to velocity"},{"Start":"02:54.230 ","End":"02:58.960","Text":"squared over the radial acceleration."},{"Start":"02:58.960 ","End":"03:01.415","Text":"In this very instant,"},{"Start":"03:01.415 ","End":"03:05.480","Text":"the normal acceleration is actually going to equal the radial acceleration,"},{"Start":"03:05.480 ","End":"03:08.210","Text":"but it\u0027s only in this instant, only in this case."},{"Start":"03:08.210 ","End":"03:10.775","Text":"Again, we said it\u0027s not the exact same thing."},{"Start":"03:10.775 ","End":"03:12.650","Text":"In this case, they are equal."},{"Start":"03:12.650 ","End":"03:17.020","Text":"We end up with this formula that R,"},{"Start":"03:17.020 ","End":"03:21.770","Text":"the radius of the curvature equals v^2 over an,"},{"Start":"03:21.770 ","End":"03:26.420","Text":"meaning the velocity squared over the normal acceleration."},{"Start":"03:26.420 ","End":"03:30.060","Text":"Now we never divide by a vector."},{"Start":"03:30.060 ","End":"03:32.570","Text":"We know that this is just a magnitude we\u0027re talking about."},{"Start":"03:32.570 ","End":"03:34.160","Text":"But again, because it\u0027s v^2,"},{"Start":"03:34.160 ","End":"03:36.520","Text":"we also can derive that."},{"Start":"03:36.520 ","End":"03:39.710","Text":"Now the only thing that remains is to explain how"},{"Start":"03:39.710 ","End":"03:43.250","Text":"the radial acceleration is not the same as the normal acceleration,"},{"Start":"03:43.250 ","End":"03:46.400","Text":"and how the tangential acceleration is"},{"Start":"03:46.400 ","End":"03:50.760","Text":"not the same as the acceleration in the direction of Theta."}],"ID":9620},{"Watched":false,"Name":"The Difference in Relation to Radial Acceleration","Duration":"3m 41s","ChapterTopicVideoID":8984,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.350","Text":"In this example,"},{"Start":"00:01.350 ","End":"00:04.970","Text":"we\u0027ve discussed here tangential acceleration in your normal acceleration are here,"},{"Start":"00:04.970 ","End":"00:06.945","Text":"and here on these two lines."},{"Start":"00:06.945 ","End":"00:11.955","Text":"If we\u0027re talking about radial acceleration and Theta acceleration,"},{"Start":"00:11.955 ","End":"00:15.270","Text":"you\u0027re talking about two different lines based on a different set of coordinates"},{"Start":"00:15.270 ","End":"00:19.170","Text":"called your polar coordinates something we can talk about more later."},{"Start":"00:19.170 ","End":"00:22.560","Text":"But they\u0027re basically a curved set"},{"Start":"00:22.560 ","End":"00:25.260","Text":"of coordinates that can be an alternative to your x, and your y,"},{"Start":"00:25.260 ","End":"00:28.950","Text":"so your acceleration vector"},{"Start":"00:28.950 ","End":"00:32.565","Text":"for radial acceleration is going to look like that coming into the origin,"},{"Start":"00:32.565 ","End":"00:35.155","Text":"and it\u0027s going to point towards your object."},{"Start":"00:35.155 ","End":"00:38.225","Text":"I didn\u0027t do it perfectly here, but I think you get the idea."},{"Start":"00:38.225 ","End":"00:42.840","Text":"This element can be your AR vector,"},{"Start":"00:42.840 ","End":"00:45.500","Text":"and perpendicular to that,"},{"Start":"00:45.500 ","End":"00:48.680","Text":"you\u0027re going to have a vector for your acceleration in"},{"Start":"00:48.680 ","End":"00:51.920","Text":"the direction of Theta, and actually,"},{"Start":"00:51.920 ","End":"00:55.830","Text":"these two can ultimately serve as an alternative to your x,"},{"Start":"00:55.830 ","End":"00:59.330","Text":"and y axes because they\u0027re 45-degree angle,"},{"Start":"00:59.330 ","End":"01:02.700","Text":"and can be useful for other kinds of measurements."},{"Start":"01:03.950 ","End":"01:07.050","Text":"That\u0027s a Theta."},{"Start":"01:07.050 ","End":"01:09.530","Text":"What you\u0027re going to see is at our object here,"},{"Start":"01:09.530 ","End":"01:13.490","Text":"it looks like AR is parallel to a vector."},{"Start":"01:13.490 ","End":"01:14.940","Text":"In fact, that\u0027s a coincidence,"},{"Start":"01:14.940 ","End":"01:18.120","Text":"that\u0027s not really the case so just for the sake"},{"Start":"01:18.120 ","End":"01:22.780","Text":"of elucidation let\u0027s draw it a little off-centered from a,"},{"Start":"01:22.780 ","End":"01:24.730","Text":"and then at a 45-degree angle from that,"},{"Start":"01:24.730 ","End":"01:27.440","Text":"you\u0027ll have your a Theta vector."},{"Start":"01:27.440 ","End":"01:31.270","Text":"What you\u0027ll notice is that AR,"},{"Start":"01:31.270 ","End":"01:34.885","Text":"your radial acceleration is not the same as your normal acceleration,"},{"Start":"01:34.885 ","End":"01:36.845","Text":"neither in magnitude nor direction,"},{"Start":"01:36.845 ","End":"01:41.530","Text":"and that a Theta is not the same as your tangential acceleration neither,"},{"Start":"01:41.530 ","End":"01:43.490","Text":"and magnitude nor direction."},{"Start":"01:43.490 ","End":"01:46.165","Text":"These are definitely two different things."},{"Start":"01:46.165 ","End":"01:47.740","Text":"You can see that an example here,"},{"Start":"01:47.740 ","End":"01:52.945","Text":"but the reason people often get confused is an example that we use actually"},{"Start":"01:52.945 ","End":"01:58.375","Text":"rather regularly when you\u0027re moving in a circular motion around the origin,"},{"Start":"01:58.375 ","End":"02:00.765","Text":"a circular trajectory that is."},{"Start":"02:00.765 ","End":"02:04.590","Text":"For example, if we take this graph,"},{"Start":"02:04.590 ","End":"02:07.585","Text":"and move in a circular trajectory around the origin,"},{"Start":"02:07.585 ","End":"02:14.410","Text":"let\u0027s say that our motion is going counterclockwise so if that\u0027s our object,"},{"Start":"02:14.410 ","End":"02:20.755","Text":"our velocity has to be tangential to the trajectory so that\u0027s our velocity line,"},{"Start":"02:20.755 ","End":"02:24.280","Text":"and that means that our acceleration has to"},{"Start":"02:24.280 ","End":"02:28.135","Text":"be going inwards from the curvature toward the concave side."},{"Start":"02:28.135 ","End":"02:30.925","Text":"Let\u0027s draw it as that for example,"},{"Start":"02:30.925 ","End":"02:33.950","Text":"now if we want to find our normal,"},{"Start":"02:33.950 ","End":"02:38.275","Text":"and our tangential acceleration for our tangential acceleration,"},{"Start":"02:38.275 ","End":"02:41.920","Text":"we know that it has to be parallel to the velocity,"},{"Start":"02:41.920 ","End":"02:43.785","Text":"has to be tangential to the curve."},{"Start":"02:43.785 ","End":"02:46.200","Text":"That\u0027s your 80 right there,"},{"Start":"02:46.200 ","End":"02:49.430","Text":"and that perpendicular to that is going to be your a,"},{"Start":"02:49.430 ","End":"02:51.620","Text":"and your normal acceleration,"},{"Start":"02:51.620 ","End":"02:55.915","Text":"and the third red line there is your an."},{"Start":"02:55.915 ","End":"02:59.250","Text":"Now if you notice, because at is tangential to"},{"Start":"02:59.250 ","End":"03:02.465","Text":"the curve an is going to be pointing towards the origin."},{"Start":"03:02.465 ","End":"03:05.060","Text":"This particular example, I didn\u0027t draw it perfectly,"},{"Start":"03:05.060 ","End":"03:07.730","Text":"but I think you understand the idea."},{"Start":"03:07.730 ","End":"03:12.770","Text":"But an is pointing towards the origin,"},{"Start":"03:12.770 ","End":"03:17.020","Text":"which actually makes it the same as your radial acceleration,"},{"Start":"03:17.020 ","End":"03:19.790","Text":"so normal acceleration is the same as radial acceleration here."},{"Start":"03:19.790 ","End":"03:23.000","Text":"Tangential acceleration is the same as Theta acceleration again,"},{"Start":"03:23.000 ","End":"03:25.895","Text":"because of the perpendicularity between those two sets,"},{"Start":"03:25.895 ","End":"03:27.410","Text":"and for that reason,"},{"Start":"03:27.410 ","End":"03:31.620","Text":"people often will confuse the two because in this example they really are the same,"},{"Start":"03:31.620 ","End":"03:33.605","Text":"and you can talk about them the same way."},{"Start":"03:33.605 ","End":"03:35.540","Text":"Of course, with any other trajectory,"},{"Start":"03:35.540 ","End":"03:37.310","Text":"they are totally different,"},{"Start":"03:37.310 ","End":"03:41.460","Text":"and that ends our lecture. Thanks for listening."}],"ID":9239},{"Watched":false,"Name":"Example, Part 1","Duration":"6m 22s","ChapterTopicVideoID":9287,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"In this example, we have the velocity given to us in the 2-dimensions."},{"Start":"00:04.980 ","End":"00:12.690","Text":"We have 9t squared plus 2 on the x-axis and 5t^3 on the y-axis."},{"Start":"00:12.690 ","End":"00:18.840","Text":"What we want to find is our tangential acceleration at time 2."},{"Start":"00:18.840 ","End":"00:20.280","Text":"The first thing we need to do is find"},{"Start":"00:20.280 ","End":"00:24.960","Text":"the our vector for overall acceleration and then we can break it down from there."},{"Start":"00:24.960 ","End":"00:28.170","Text":"Our a vector, the way we\u0027re going to find that of course,"},{"Start":"00:28.170 ","End":"00:31.130","Text":"is through a derivative of the v vector."},{"Start":"00:31.130 ","End":"00:34.370","Text":"The derivative of the v vector is going"},{"Start":"00:34.370 ","End":"00:37.730","Text":"to equal and we\u0027re going to have to derive each bit on its own,"},{"Start":"00:37.730 ","End":"00:45.210","Text":"so 9t squared plus 2 becomes 9 times 2t that\u0027s our x-axis."},{"Start":"00:45.210 ","End":"00:46.560","Text":"In addition to that,"},{"Start":"00:46.560 ","End":"00:52.300","Text":"we\u0027ll have 15t squared on our y-axis."},{"Start":"00:52.340 ","End":"00:58.925","Text":"Now we have our acceleration vector and if we want to find our a_t vector,"},{"Start":"00:58.925 ","End":"01:03.140","Text":"we\u0027re going to need to use our formula from before."},{"Start":"01:03.140 ","End":"01:06.920","Text":"If you recall, that formula is as follows,"},{"Start":"01:06.920 ","End":"01:11.360","Text":"a_t equals a dot v a scalar multiplication"},{"Start":"01:11.360 ","End":"01:17.040","Text":"over the magnitude of v squared times the vector v. Let\u0027s start,"},{"Start":"01:17.040 ","End":"01:22.190","Text":"a dot v a scalar multiplication we\u0027re going"},{"Start":"01:22.190 ","End":"01:27.545","Text":"to take the values from our a vector and we\u0027ll just use 18 instead of 9 times 2,"},{"Start":"01:27.545 ","End":"01:29.255","Text":"simplify a little bit here."},{"Start":"01:29.255 ","End":"01:33.980","Text":"We\u0027re going to scalar and then multiply them by our v vector so"},{"Start":"01:33.980 ","End":"01:41.303","Text":"18t times in parentheses 9t squared plus 2,"},{"Start":"01:41.303 ","End":"01:46.605","Text":"and that\u0027s our x and our 2x\u0027s drop away and we\u0027ll add to that"},{"Start":"01:46.605 ","End":"01:56.085","Text":"15t squared times 5t^3."},{"Start":"01:56.085 ","End":"01:57.620","Text":"If we multiply that out,"},{"Start":"01:57.620 ","End":"02:02.000","Text":"what we\u0027re going to get is 18 times 9 is"},{"Start":"02:02.000 ","End":"02:10.470","Text":"162t^3 plus another 36t"},{"Start":"02:12.370 ","End":"02:15.650","Text":"and then when we do our y side,"},{"Start":"02:15.650 ","End":"02:21.180","Text":"we\u0027re going to get 75t^5."},{"Start":"02:21.710 ","End":"02:25.415","Text":"As a result of this, we get a value,"},{"Start":"02:25.415 ","End":"02:26.608","Text":"not an angle,"},{"Start":"02:26.608 ","End":"02:28.760","Text":"not a scalar a value."},{"Start":"02:28.760 ","End":"02:31.775","Text":"Now if we want to find the magnitude of v,"},{"Start":"02:31.775 ","End":"02:36.110","Text":"the way we do that again is going to take the values above of v"},{"Start":"02:36.110 ","End":"02:41.450","Text":"and square them and then do a square root but because we want magnitude of v squared,"},{"Start":"02:41.450 ","End":"02:43.640","Text":"we don\u0027t need to do the square root right now."},{"Start":"02:43.640 ","End":"02:46.460","Text":"What we\u0027re going to do is in parentheses,"},{"Start":"02:46.460 ","End":"02:49.025","Text":"9t squared plus 2,"},{"Start":"02:49.025 ","End":"02:51.320","Text":"and that whole figure squared,"},{"Start":"02:51.320 ","End":"02:56.070","Text":"plus 5t^3 also squared."},{"Start":"02:56.170 ","End":"02:59.810","Text":"Again, if we had done the square root here,"},{"Start":"02:59.810 ","End":"03:01.610","Text":"we would get the absolute value of v,"},{"Start":"03:01.610 ","End":"03:06.750","Text":"which would give us the magnitude of a_t but we want the full value,"},{"Start":"03:06.750 ","End":"03:10.685","Text":"we want the angle as well so we\u0027re going to keep it as is."},{"Start":"03:10.685 ","End":"03:14.845","Text":"Now let\u0027s put it back into our formula."},{"Start":"03:14.845 ","End":"03:18.110","Text":"To make things simpler before we actually plug the whole thing in let\u0027s also put"},{"Start":"03:18.110 ","End":"03:22.340","Text":"in our value for time t is 2 in this case."},{"Start":"03:22.340 ","End":"03:30.135","Text":"What will get if we take a dot v at t equals 2 is 162"},{"Start":"03:30.135 ","End":"03:39.400","Text":"times 8 plus 36 times 2 plus 75 times 2^5."},{"Start":"03:39.400 ","End":"03:48.100","Text":"Now I\u0027ll save you a little time dealing with the calculator and your result is 3,768."},{"Start":"03:48.580 ","End":"03:54.770","Text":"If we plug in our time to the magnitude of v squared,"},{"Start":"03:54.770 ","End":"04:04.345","Text":"what we\u0027re going to get again at t equals 2 is 9 times 4 plus 2,"},{"Start":"04:04.345 ","End":"04:06.716","Text":"all of that in parentheses squared,"},{"Start":"04:06.716 ","End":"04:11.610","Text":"plus 5 times 8, again squared."},{"Start":"04:11.610 ","End":"04:12.750","Text":"When you calculate that out,"},{"Start":"04:12.750 ","End":"04:17.580","Text":"what you end up with is 3,044."},{"Start":"04:17.580 ","End":"04:21.680","Text":"When you take your 2 elements and divide them one by the other,"},{"Start":"04:21.680 ","End":"04:26.180","Text":"you\u0027re a dot v divided by the absolute value,"},{"Start":"04:26.180 ","End":"04:28.145","Text":"the magnitude of v squared,"},{"Start":"04:28.145 ","End":"04:34.370","Text":"you\u0027re going to take 3,768 and divide it by 3,044"},{"Start":"04:34.370 ","End":"04:41.420","Text":"and what you end up with is approximately 1.24."},{"Start":"04:41.420 ","End":"04:42.680","Text":"What we can do is take that,"},{"Start":"04:42.680 ","End":"04:45.140","Text":"plug it into the formula and we still need to multiply that"},{"Start":"04:45.140 ","End":"04:47.990","Text":"by the v vector and it\u0027s going"},{"Start":"04:47.990 ","End":"04:50.600","Text":"to be easier for us again to plug in t equals 2 to"},{"Start":"04:50.600 ","End":"04:53.430","Text":"the v vector before doing this calculation."},{"Start":"04:53.430 ","End":"04:56.650","Text":"If we plug in t equals 2,"},{"Start":"04:56.650 ","End":"04:59.265","Text":"we get 9 times 4,"},{"Start":"04:59.265 ","End":"05:05.925","Text":"2 squared plus 2 as our x factor and 5 times 8,"},{"Start":"05:05.925 ","End":"05:08.470","Text":"2^3 as our y factor."},{"Start":"05:09.020 ","End":"05:17.570","Text":"To write this simplified you get 38 in the x direction and 40 in the direction of y."},{"Start":"05:17.570 ","End":"05:22.625","Text":"You should note that I could have done this all without inserting t equals 2."},{"Start":"05:22.625 ","End":"05:25.550","Text":"I could have gotten it all in terms of"},{"Start":"05:25.550 ","End":"05:28.895","Text":"t and then you can use it for any time that you\u0027d like."},{"Start":"05:28.895 ","End":"05:31.640","Text":"It makes a lot easier if you\u0027re looking for multiple times on"},{"Start":"05:31.640 ","End":"05:34.280","Text":"a particular path of an object however,"},{"Start":"05:34.280 ","End":"05:35.690","Text":"because we\u0027re looking at t equals 2,"},{"Start":"05:35.690 ","End":"05:37.520","Text":"it\u0027s easiest to do it this way because it saves"},{"Start":"05:37.520 ","End":"05:40.285","Text":"a lot of writing and it makes a little bit shorter."},{"Start":"05:40.285 ","End":"05:43.490","Text":"When we want to plug all of this back in,"},{"Start":"05:43.490 ","End":"05:48.410","Text":"we want to find a_t what we\u0027re going to do is take that value 38 for x and 40 for y,"},{"Start":"05:48.410 ","End":"05:51.325","Text":"and multiply each of those by 1.24."},{"Start":"05:51.325 ","End":"05:53.390","Text":"What we\u0027re going to get is,"},{"Start":"05:53.390 ","End":"05:54.500","Text":"and you can just write it this way,"},{"Start":"05:54.500 ","End":"05:57.755","Text":"1.24 times 38,"},{"Start":"05:57.755 ","End":"06:01.625","Text":"40 for scalar multiplication,"},{"Start":"06:01.625 ","End":"06:11.505","Text":"and when you calculate that out you\u0027ll end up getting 47.12 for x and 49.6 for y."},{"Start":"06:11.505 ","End":"06:14.060","Text":"Again, that\u0027s at t equals 2."},{"Start":"06:14.060 ","End":"06:16.790","Text":"It\u0027s not all that difficult to find it as a function of"},{"Start":"06:16.790 ","End":"06:20.150","Text":"time for your tangential acceleration."},{"Start":"06:20.150 ","End":"06:22.770","Text":"Now we can find the normal acceleration."}],"ID":9599},{"Watched":false,"Name":"Example, Part 2","Duration":"5m 43s","ChapterTopicVideoID":9288,"CourseChapterTopicPlaylistID":5378,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"In part B of the problem,"},{"Start":"00:02.130 ","End":"00:06.840","Text":"they\u0027re asking us to find the normal acceleration,"},{"Start":"00:06.840 ","End":"00:10.890","Text":"and to remind you the way we do that is take the total acceleration,"},{"Start":"00:10.890 ","End":"00:16.815","Text":"the a vector and subtract from it the at vector, the tangential acceleration."},{"Start":"00:16.815 ","End":"00:21.410","Text":"Because this is asked in a vector in terms of both length and direction,"},{"Start":"00:21.410 ","End":"00:25.830","Text":"we can\u0027t do the absolute values and everything, we can\u0027t just do the magnitude."},{"Start":"00:25.830 ","End":"00:28.330","Text":"We have to use this formula."},{"Start":"00:28.580 ","End":"00:34.260","Text":"The way we\u0027re going to do it is take our a vector at t equals 2,"},{"Start":"00:34.260 ","End":"00:35.910","Text":"which we can already write out from before,"},{"Start":"00:35.910 ","End":"00:39.740","Text":"which is 18 times"},{"Start":"00:39.740 ","End":"00:47.970","Text":"2 as the X factor and 15 times 4 as our Y factor,"},{"Start":"00:47.970 ","End":"00:51.135","Text":"which is 36 and 60,"},{"Start":"00:51.135 ","End":"00:56.990","Text":"and we\u0027re going to subtract from it the at values that we got above."},{"Start":"00:56.990 ","End":"00:58.340","Text":"As you can see right above us,"},{"Start":"00:58.340 ","End":"01:00.950","Text":"it\u0027s 47.12 and 49.6."},{"Start":"01:00.950 ","End":"01:10.515","Text":"We can write it as follows, 36 minus 47.12, 60 minus 49.6."},{"Start":"01:10.515 ","End":"01:12.645","Text":"That\u0027s your solution right there really,"},{"Start":"01:12.645 ","End":"01:14.360","Text":"It\u0027s not all that difficult."},{"Start":"01:14.360 ","End":"01:15.800","Text":"Now in this situation,"},{"Start":"01:15.800 ","End":"01:17.630","Text":"I plugged it into t equals 2."},{"Start":"01:17.630 ","End":"01:19.700","Text":"I could have kept everything in terms of t,"},{"Start":"01:19.700 ","End":"01:22.835","Text":"but it\u0027s easier this way for this particular problem,"},{"Start":"01:22.835 ","End":"01:25.325","Text":"and I can also just get the magnitude if I want."},{"Start":"01:25.325 ","End":"01:30.260","Text":"I can go for magnitude of vector a_n average price"},{"Start":"01:30.260 ","End":"01:35.785","Text":"simplify those values at this point it\u0027s negative 11.12 and 10.4,"},{"Start":"01:35.785 ","End":"01:41.390","Text":"and we can write them out and square and square root them,"},{"Start":"01:41.390 ","End":"01:43.625","Text":"so that would look like this."},{"Start":"01:43.625 ","End":"01:49.610","Text":"We take 11.12^2 plus"},{"Start":"01:49.610 ","End":"01:55.120","Text":"10.4^2 and take that on a square root and we have our magnitude."},{"Start":"01:55.120 ","End":"01:58.700","Text":"Now we can use the other formula as well."},{"Start":"01:58.700 ","End":"02:00.290","Text":"It\u0027s just really a thought exercise."},{"Start":"02:00.290 ","End":"02:01.700","Text":"It\u0027s not necessary at this point,"},{"Start":"02:01.700 ","End":"02:03.380","Text":"but as long as we\u0027re practicing,"},{"Start":"02:03.380 ","End":"02:06.260","Text":"I figure it\u0027s a good way to get some work in."},{"Start":"02:06.260 ","End":"02:10.385","Text":"If at any point you feel like you get it and you\u0027ve already had enough,"},{"Start":"02:10.385 ","End":"02:12.285","Text":"don\u0027t feel like you have to continue,"},{"Start":"02:12.285 ","End":"02:15.845","Text":"but for the sake of the process here, let\u0027s write it out."},{"Start":"02:15.845 ","End":"02:21.425","Text":"We can also find our magnitude of a_n with the formula a cross-product"},{"Start":"02:21.425 ","End":"02:28.145","Text":"of v over v magnitude."},{"Start":"02:28.145 ","End":"02:30.739","Text":"Because we\u0027re only in 2 dimensions,"},{"Start":"02:30.739 ","End":"02:33.370","Text":"this isn\u0027t so difficult to write out."},{"Start":"02:33.370 ","End":"02:39.750","Text":"What we\u0027re going to get here is our a value,"},{"Start":"02:40.010 ","End":"02:42.680","Text":"and again, for the cross multiplication,"},{"Start":"02:42.680 ","End":"02:47.960","Text":"what we\u0027re going to do is take a(x) times"},{"Start":"02:47.960 ","End":"02:54.830","Text":"v(y) and subtract from that a(y) and v(x)."},{"Start":"02:54.830 ","End":"02:56.870","Text":"If we had 3-dimensional, this would be more complicated."},{"Start":"02:56.870 ","End":"03:00.740","Text":"We have to deal with all 3 dimensions and it would get longer,"},{"Start":"03:00.740 ","End":"03:02.480","Text":"but for our purposes,"},{"Start":"03:02.480 ","End":"03:07.100","Text":"this is what we\u0027re using and we\u0027d multiply that by the third vector,"},{"Start":"03:07.100 ","End":"03:09.145","Text":"the z or z-dimension as well,"},{"Start":"03:09.145 ","End":"03:11.090","Text":"but at the moment we don\u0027t really have to deal with that,"},{"Start":"03:11.090 ","End":"03:13.730","Text":"so we\u0027re fortunate and it\u0027s going to take a little less time."},{"Start":"03:13.730 ","End":"03:15.430","Text":"We\u0027re just looking for the magnitude."},{"Start":"03:15.430 ","End":"03:17.240","Text":"We\u0027re going to do is put our lines into mixture."},{"Start":"03:17.240 ","End":"03:21.625","Text":"We know that that\u0027s what we\u0027re looking for as follows,"},{"Start":"03:21.625 ","End":"03:27.155","Text":"and what we get when we write that out is take our ax,"},{"Start":"03:27.155 ","End":"03:31.665","Text":"which is 18 times t,"},{"Start":"03:31.665 ","End":"03:34.170","Text":"and we\u0027re going to multiply that by v y,"},{"Start":"03:34.170 ","End":"03:36.495","Text":"which is 5t^3,"},{"Start":"03:36.495 ","End":"03:37.740","Text":"and subtract from that ay,"},{"Start":"03:37.740 ","End":"03:41.220","Text":"which is 15t^2 times vx,"},{"Start":"03:41.220 ","End":"03:46.340","Text":"which is (9t^2 plus 2)."},{"Start":"03:46.340 ","End":"03:53.105","Text":"What we can do now is plug in t equals 2 and simplify the whole thing."},{"Start":"03:53.105 ","End":"04:01.230","Text":"What we\u0027re going to end up with is 18 times 2 times 5t^3."},{"Start":"04:01.230 ","End":"04:03.035","Text":"If you calculate that all out,"},{"Start":"04:03.035 ","End":"04:09.040","Text":"what you get is 1,440,"},{"Start":"04:09.040 ","End":"04:16.310","Text":"and we\u0027re going to subtract from that 15 times 2^2 times (9 2^2 plus 2) in parentheses,"},{"Start":"04:16.310 ","End":"04:20.370","Text":"and that value is 2,280."},{"Start":"04:20.510 ","End":"04:23.850","Text":"If we subtract out those 2 values,"},{"Start":"04:23.850 ","End":"04:26.555","Text":"we get negative 840."},{"Start":"04:26.555 ","End":"04:28.670","Text":"But again, because we\u0027re just looking for a magnitude,"},{"Start":"04:28.670 ","End":"04:30.545","Text":"we\u0027re looking for the absolute value here."},{"Start":"04:30.545 ","End":"04:34.130","Text":"We may as well write that with a plus sign and put"},{"Start":"04:34.130 ","End":"04:36.200","Text":"our 2 lines there to make sure that we know why we\u0027re getting"},{"Start":"04:36.200 ","End":"04:39.125","Text":"a positive value and a negative value."},{"Start":"04:39.125 ","End":"04:41.690","Text":"We need to subtract that from"},{"Start":"04:41.690 ","End":"04:45.500","Text":"the absolute value of v. The easiest way we can do that is if you see above,"},{"Start":"04:45.500 ","End":"04:48.005","Text":"we have absolute value of v^2, the magnitude of v^2."},{"Start":"04:48.005 ","End":"04:49.700","Text":"If we just take a square root of that,"},{"Start":"04:49.700 ","End":"04:54.215","Text":"which is 3,044, we\u0027re going to get our absolute value of v,"},{"Start":"04:54.215 ","End":"04:55.775","Text":"or magnitude of v,"},{"Start":"04:55.775 ","End":"05:01.415","Text":"and we end up coming up with 55.17 approximately."},{"Start":"05:01.415 ","End":"05:04.610","Text":"If we plug that whole thing into our equation,"},{"Start":"05:04.610 ","End":"05:13.665","Text":"we get 840 divided by 55.17 equaling our magnitude of an,"},{"Start":"05:13.665 ","End":"05:16.830","Text":"and if you calculate that out,"},{"Start":"05:16.830 ","End":"05:19.695","Text":"what you\u0027ll get is 15.22,"},{"Start":"05:19.695 ","End":"05:23.330","Text":"and if you go to the other side of our box here and calculate"},{"Start":"05:23.330 ","End":"05:27.410","Text":"out this 11.12^2 plus 10.4^2 square root,"},{"Start":"05:27.410 ","End":"05:28.950","Text":"you\u0027ll also get the exact same thing,"},{"Start":"05:28.950 ","End":"05:32.195","Text":"15.22, and that\u0027s it."},{"Start":"05:32.195 ","End":"05:34.610","Text":"We\u0027ve explored how to get your tangential acceleration,"},{"Start":"05:34.610 ","End":"05:36.680","Text":"how to get your normal acceleration in both ways to"},{"Start":"05:36.680 ","End":"05:38.660","Text":"get just the magnitude of the normal acceleration."},{"Start":"05:38.660 ","End":"05:42.090","Text":"I hope that clears up any problems. Thanks."}],"ID":9600}],"Thumbnail":null,"ID":5378}]

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