WEBVTT 00:00:03.520 --> 00:00:05.870 Let's now consider two dimensional motion, 00:00:05.870 --> 00:00:08.039 and let's try to analyze how to describe 00:00:08.039 --> 00:00:09.570 the change in velocity. 00:00:09.570 --> 00:00:13.630 So again, let's choose a coordinate system. 00:00:13.630 --> 00:00:17.300 We have an origin plus y plus x. 00:00:17.300 --> 00:00:21.100 And let's draw the trajectory of our object. 00:00:21.100 --> 00:00:24.810 And now let's draw the object at two different times. 00:00:24.810 --> 00:00:31.510 So for instance, if I call this the location at time t1, 00:00:31.510 --> 00:00:33.420 and a little bit later here, this 00:00:33.420 --> 00:00:37.225 is the location of the object at time t2. 00:00:37.225 --> 00:00:41.940 We'll call our unit vectors i hat and j hat. 00:00:44.740 --> 00:00:47.610 We know that the direction of the velocity 00:00:47.610 --> 00:00:50.040 is tangent to this curve. 00:00:50.040 --> 00:00:56.410 So if we draw v at time t1-- and over here, 00:00:56.410 --> 00:01:02.106 notice the direction has changed v at time t2. 00:01:02.106 --> 00:01:05.840 And what we'd like to do now is describe, just as before, 00:01:05.840 --> 00:01:12.760 that our acceleration a of t is the derivative of the velocity 00:01:12.760 --> 00:01:14.530 as a function of time. 00:01:14.530 --> 00:01:17.850 What that means is the limit as delta t goes 00:01:17.850 --> 00:01:22.180 to 0 of delta v over delta t. 00:01:22.180 --> 00:01:27.260 Now, it's much harder to visualize the delta v 00:01:27.260 --> 00:01:28.640 in this drawing. 00:01:28.640 --> 00:01:32.289 And partly, the reason for that is these velocity vectors 00:01:32.289 --> 00:01:34.870 are located at two different points. 00:01:34.870 --> 00:01:39.110 And right now, the backs of these vectors 00:01:39.110 --> 00:01:41.180 have different places in space. 00:01:41.180 --> 00:01:46.890 But remember that delta v is just v, in this case, 00:01:46.890 --> 00:01:51.870 at time t2 minus v at time t1. 00:01:51.870 --> 00:01:54.580 And our principle for subtracting two vectors 00:01:54.580 --> 00:01:57.190 at different locations in space is 00:01:57.190 --> 00:02:01.820 to draw the vectors where we put the tails at the same location. 00:02:01.820 --> 00:02:03.600 So here's a tail at this vector. 00:02:03.600 --> 00:02:07.770 We're just going to translate that vector in space. 00:02:07.770 --> 00:02:10.910 That is still v at time t1. 00:02:10.910 --> 00:02:12.050 These vectors are equal. 00:02:12.050 --> 00:02:16.060 They have the same length, and they have the same direction. 00:02:16.060 --> 00:02:19.050 And so delta v is just the vector 00:02:19.050 --> 00:02:22.360 that connects here to there. 00:02:22.360 --> 00:02:24.880 That's what we mean by delta v. 00:02:24.880 --> 00:02:27.600 And so you can see in this particular case 00:02:27.600 --> 00:02:31.390 that it's not obvious from looking at the orbit what 00:02:31.390 --> 00:02:32.730 the delta v is. 00:02:32.730 --> 00:02:37.720 So what we need to do is just trust our calculus. 00:02:37.720 --> 00:02:47.006 And so when we write the velocity as dx dt i hat plus dy 00:02:47.006 --> 00:02:54.640 dy j hat, and we're now treating each direction independently. 00:02:54.640 --> 00:03:00.520 We call this vx i hat plus vy j hat. 00:03:00.520 --> 00:03:02.420 So that's our velocity vector. 00:03:02.420 --> 00:03:08.130 Then our acceleration is just the derivative of the velocity. 00:03:08.130 --> 00:03:10.190 We take each direction separately, 00:03:10.190 --> 00:03:19.220 so we have dv x dt i hat plus dv y dt j hat. 00:03:19.220 --> 00:03:26.250 Now, again, notice that velocity v of x 00:03:26.250 --> 00:03:33.040 is already the first derivative of the position 00:03:33.040 --> 00:03:35.020 of the exponent function. 00:03:35.020 --> 00:03:40.290 So what we really have here is the second derivative 00:03:40.290 --> 00:03:44.820 of the position function in the i hat 00:03:44.820 --> 00:03:47.390 direction and the second derivative of the component 00:03:47.390 --> 00:03:49.690 function in the y direction. 00:03:49.690 --> 00:03:53.920 And that is what we call the instantaneous acceleration. 00:03:53.920 --> 00:03:57.060 Now, again, this is sometimes awkward to draw, 00:03:57.060 --> 00:04:03.060 but you always must remember that this x component 00:04:03.060 --> 00:04:06.500 of the acceleration by definition 00:04:06.500 --> 00:04:09.640 is the second derivative of the component function 00:04:09.640 --> 00:04:13.350 or the first derivative of the component 00:04:13.350 --> 00:04:15.020 function for the velocity. 00:04:15.020 --> 00:04:21.050 And likewise, the y component of the acceleration ay 00:04:21.050 --> 00:04:24.990 is the second derivative of the component 00:04:24.990 --> 00:04:27.180 function for position. 00:04:27.180 --> 00:04:30.990 And that's also equal, by definition, 00:04:30.990 --> 00:04:34.760 to the first derivative of the component of the velocity 00:04:34.760 --> 00:04:36.350 vector. 00:04:36.350 --> 00:04:40.390 And that's how we describe the acceleration. 00:04:40.390 --> 00:04:44.010 As before, we can talk about the magnitude of a vector. 00:04:44.010 --> 00:04:47.355 And the magnitude of a we'll just write as a. 00:04:47.355 --> 00:04:53.220 It's the components squared, added together, taken 00:04:53.220 --> 00:04:54.740 square root. 00:04:54.740 --> 00:04:58.520 And that's our magnitude. 00:04:58.520 --> 00:05:01.850 And so now we've described all of our kinematic quantities 00:05:01.850 --> 00:05:03.920 in two dimensions-- the position, 00:05:03.920 --> 00:05:06.950 the velocity as the derivative of the position, 00:05:06.950 --> 00:05:10.580 and the acceleration as the derivative of the velocity 00:05:10.580 --> 00:05:15.000 where each direction is treated independently.