Video: Finding the Velocity and Acceleration of a Particle from the Expression of Its Displacement in Time

A particle is moving in a straight line such that its displacement from the origin after 𝑑 seconds is given by π‘₯ = ((1/3) cos 2𝑑) m, 𝑑 β‰₯ 0. Find its velocity 𝑣 when 𝑑 = πœ‹/4 s and its acceleration π‘Ž when 𝑑 = πœ‹/3 s.

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Video Transcript

A particle is moving in a straight line such that its displacement from the origin after 𝑑 seconds is given by π‘₯ equals one-third cos of two 𝑑 meters, for 𝑑 is greater than or equal to zero. Find its velocity 𝑣 when 𝑑 is equal to πœ‹ by four seconds and its acceleration π‘Ž when 𝑑 is equal to πœ‹ by three seconds.

In this question, we’ve been given an expression for displacement in terms of time 𝑑. We’re looking to find the velocity and acceleration of the particle, given some information about the time. And so, let’s begin by recalling how we link displacement, velocity, and acceleration. Velocity is the rate of change of displacement of an object. And when we think of rate of change, we think of differentiation. So, velocity is the first derivative of displacement π‘₯ with respect to time. Similarly, acceleration is rate of change of velocity, meaning acceleration is the first derivative of velocity with respect to time.

Now, of course, since velocity is itself the first derivative of π‘₯ with respect to 𝑑, we can say that acceleration must therefore be the second derivative of π‘₯ with respect to 𝑑, d two π‘₯ by d𝑑 squared. So, we’ll begin by differentiating our expression for displacement with respect to time to find an expression for velocity. We’ll then differentiate once again to find our expression for acceleration. Displacement is given as a third cos of two 𝑑, so the velocity is the derivative of a third cos of two 𝑑 with respect to 𝑑.

Now, one thing that we’re allowed to do when differentiating is take out constant factors, so we can take out a constant factor of one-third. And we can say that the velocity will be one-third times the first derivative of cos of two 𝑑. This isn’t entirely necessary, but it does make the next step a little easier. Next, we quote the general result for the derivative of cos of π‘Žπ‘₯ for real constants π‘Ž. It’s negative π‘Ž sin of π‘Žπ‘₯. And so, this means when we differentiate cos of two 𝑑, we’ll get negative two sin of two 𝑑. And so, our velocity is a third of this. It’s a third times negative two sin of two 𝑑, which we can, of course, write as negative two-thirds sin of two 𝑑.

We’ll next find the expression for the acceleration. Then, we’ll find the given velocity and acceleration at the times required. This time, the acceleration is the first derivative of negative two-thirds of sin two 𝑑 with respect to 𝑑. Once again, let’s take out that constant factor of negative two-thirds, and we see that we’re going to find the derivative of sin of two 𝑑 and then multiply this by negative two-thirds. And so, let’s quote the general result this time for differentiating sin of π‘Žπ‘₯ for real constants π‘Ž. It’s π‘Ž times cos of π‘Žπ‘₯. Now, this means that the derivative of sin of two 𝑑 with respect to 𝑑 is two cos two 𝑑. And therefore, our expression for acceleration is negative two-thirds times two cos of two 𝑑. That simplifies to negative four-thirds times cos of two 𝑑.

We’re now going to clear some space and evaluate the velocity when 𝑑 is equal to πœ‹ by four seconds. We could represent that using function notation as shown. This indicates to us that to find the velocity when 𝑑 is equal to πœ‹ by four, we replace 𝑑 in our expression for velocity with πœ‹ by four. And so, the velocity at this time must be negative two-thirds times sin of two times πœ‹ by four. Now, two times πœ‹ by four or two times a quarter πœ‹ is a half πœ‹ or πœ‹ by two. So, this simplifies a little to negative two-thirds times sin of πœ‹ by two. But of course, we know that sin of πœ‹ by two is simply one.

So, our velocity at this time is just negative two-thirds times one, which is negative two-thirds. The displacement is in meters, and our time 𝑑 is measured in seconds. So we can say the velocity when 𝑑 is equal to πœ‹ by four seconds is negative two-thirds meters per second. Now, we needn’t worry that our velocity is negative. We recall that velocity is a vector quantity; it consists of a direction and a magnitude. And so, a negative velocity simply tells us the direction in which the particle is traveling. Let’s now repeat this process, this time, working out the acceleration when 𝑑 is equal to πœ‹ by three.

If we use function notation, it looks like this. And we see that this time we’re going to replace 𝑑 with πœ‹ by three. Our acceleration is therefore negative four-thirds times cos of two times πœ‹ by three, which simplifies to negative four-thirds times cos of two πœ‹ by three. But cos of two πœ‹ by three is negative one-half. And so, our acceleration can be calculated by multiplying negative four-thirds by negative one-half. We see we can cross cancel by a constant factor of two. And this becomes negative two-thirds times negative one over one. A negative multiplied by a negative gives us a positive. Then, we multiply our numerators and separately multiply our denominators.

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. When 𝑑 is equal to πœ‹ by four seconds, velocity is negative two-thirds meters per second. And when 𝑑 is equal to πœ‹ by three, acceleration is two-thirds meters per square second.

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