WEBVTT
00:00:01.230 --> 00:00:13.240
A particle is moving in a straight line such that its displacement from the origin after π‘ seconds is given by π₯ equals one-third cos of two π‘ meters, for π‘ is greater than or equal to zero.
00:00:14.190 --> 00:00:22.980
Find its velocity π£ when π‘ is equal to π by four seconds and its acceleration π when π‘ is equal to π by three seconds.
00:00:23.540 --> 00:00:29.140
In this question, weβve been given an expression for displacement in terms of time π‘.
00:00:29.450 --> 00:00:35.050
Weβre looking to find the velocity and acceleration of the particle, given some information about the time.
00:00:35.400 --> 00:00:40.810
And so, letβs begin by recalling how we link displacement, velocity, and acceleration.
00:00:41.290 --> 00:00:44.970
Velocity is the rate of change of displacement of an object.
00:00:45.350 --> 00:00:48.960
And when we think of rate of change, we think of differentiation.
00:00:49.440 --> 00:00:54.950
So, velocity is the first derivative of displacement π₯ with respect to time.
00:00:55.540 --> 00:01:04.880
Similarly, acceleration is rate of change of velocity, meaning acceleration is the first derivative of velocity with respect to time.
00:01:05.320 --> 00:01:18.850
Now, of course, since velocity is itself the first derivative of π₯ with respect to π‘, we can say that acceleration must therefore be the second derivative of π₯ with respect to π‘, d two π₯ by dπ‘ squared.
00:01:19.340 --> 00:01:26.920
So, weβll begin by differentiating our expression for displacement with respect to time to find an expression for velocity.
00:01:27.240 --> 00:01:30.940
Weβll then differentiate once again to find our expression for acceleration.
00:01:31.530 --> 00:01:40.960
Displacement is given as a third cos of two π‘, so the velocity is the derivative of a third cos of two π‘ with respect to π‘.
00:01:41.270 --> 00:01:48.730
Now, one thing that weβre allowed to do when differentiating is take out constant factors, so we can take out a constant factor of one-third.
00:01:48.730 --> 00:01:54.710
And we can say that the velocity will be one-third times the first derivative of cos of two π‘.
00:01:55.100 --> 00:01:58.950
This isnβt entirely necessary, but it does make the next step a little easier.
00:01:59.270 --> 00:02:05.740
Next, we quote the general result for the derivative of cos of ππ₯ for real constants π.
00:02:06.030 --> 00:02:08.700
Itβs negative π sin of ππ₯.
00:02:08.960 --> 00:02:14.660
And so, this means when we differentiate cos of two π‘, weβll get negative two sin of two π‘.
00:02:14.890 --> 00:02:17.280
And so, our velocity is a third of this.
00:02:17.500 --> 00:02:25.530
Itβs a third times negative two sin of two π‘, which we can, of course, write as negative two-thirds sin of two π‘.
00:02:26.360 --> 00:02:29.090
Weβll next find the expression for the acceleration.
00:02:29.360 --> 00:02:33.510
Then, weβll find the given velocity and acceleration at the times required.
00:02:33.990 --> 00:02:41.010
This time, the acceleration is the first derivative of negative two-thirds of sin two π‘ with respect to π‘.
00:02:41.340 --> 00:02:51.940
Once again, letβs take out that constant factor of negative two-thirds, and we see that weβre going to find the derivative of sin of two π‘ and then multiply this by negative two-thirds.
00:02:52.330 --> 00:02:58.870
And so, letβs quote the general result this time for differentiating sin of ππ₯ for real constants π.
00:02:59.260 --> 00:03:01.310
Itβs π times cos of ππ₯.
00:03:01.550 --> 00:03:07.830
Now, this means that the derivative of sin of two π‘ with respect to π‘ is two cos two π‘.
00:03:08.250 --> 00:03:14.580
And therefore, our expression for acceleration is negative two-thirds times two cos of two π‘.
00:03:14.910 --> 00:03:19.020
That simplifies to negative four-thirds times cos of two π‘.
00:03:19.340 --> 00:03:24.650
Weβre now going to clear some space and evaluate the velocity when π‘ is equal to π by four seconds.
00:03:25.010 --> 00:03:28.300
We could represent that using function notation as shown.
00:03:28.770 --> 00:03:38.070
This indicates to us that to find the velocity when π‘ is equal to π by four, we replace π‘ in our expression for velocity with π by four.
00:03:38.370 --> 00:03:45.420
And so, the velocity at this time must be negative two-thirds times sin of two times π by four.
00:03:45.740 --> 00:03:51.930
Now, two times π by four or two times a quarter π is a half π or π by two.
00:03:52.250 --> 00:03:57.200
So, this simplifies a little to negative two-thirds times sin of π by two.
00:03:57.550 --> 00:04:01.130
But of course, we know that sin of π by two is simply one.
00:04:01.510 --> 00:04:08.110
So, our velocity at this time is just negative two-thirds times one, which is negative two-thirds.
00:04:08.410 --> 00:04:12.730
The displacement is in meters, and our time π‘ is measured in seconds.
00:04:12.950 --> 00:04:19.480
So we can say the velocity when π‘ is equal to π by four seconds is negative two-thirds meters per second.
00:04:20.250 --> 00:04:23.000
Now, we neednβt worry that our velocity is negative.
00:04:23.350 --> 00:04:29.130
We recall that velocity is a vector quantity; it consists of a direction and a magnitude.
00:04:29.470 --> 00:04:33.860
And so, a negative velocity simply tells us the direction in which the particle is traveling.
00:04:34.720 --> 00:04:40.530
Letβs now repeat this process, this time, working out the acceleration when π‘ is equal to π by three.
00:04:41.020 --> 00:04:43.800
If we use function notation, it looks like this.
00:04:44.070 --> 00:04:47.560
And we see that this time weβre going to replace π‘ with π by three.
00:04:48.310 --> 00:04:58.230
Our acceleration is therefore negative four-thirds times cos of two times π by three, which simplifies to negative four-thirds times cos of two π by three.
00:04:58.720 --> 00:05:02.760
But cos of two π by three is negative one-half.
00:05:02.940 --> 00:05:09.110
And so, our acceleration can be calculated by multiplying negative four-thirds by negative one-half.
00:05:09.380 --> 00:05:12.640
We see we can cross cancel by a constant factor of two.
00:05:12.860 --> 00:05:17.180
And this becomes negative two-thirds times negative one over one.
00:05:17.620 --> 00:05:21.380
A negative multiplied by a negative gives us a positive.
00:05:21.720 --> 00:05:25.790
Then, we multiply our numerators and separately multiply our denominators.
00:05:26.050 --> 00:05:36.180
That leaves us with an acceleration of two-thirds or two-thirds meters per square second, which, we can alternatively say is two-thirds meters per second per second.
00:05:36.870 --> 00:05:42.740
When π‘ is equal to π by four seconds, velocity is negative two-thirds meters per second.
00:05:43.110 --> 00:05:48.560
And when π‘ is equal to π by three, acceleration is two-thirds meters per square second.