Question Video: Determining the Acceleration of a Particle Based on Time and Displacement Mathematics

A particle moves along a straight line. Its displacement at time 𝑑 is π‘₯ = βˆ’sin (𝑑). Which of the following statements about the acceleration of the particle is true? [A] It is equal to the velocity of the particle? [B] It is equal to βˆ’π‘₯. [C] It is equal to π‘₯. [D] it is equal to βˆ’π‘£, where 𝑣 is the velocity of the particle.

02:56

Video Transcript

A particle moves along a straight line. Its displacement at time 𝑑 is π‘₯ equals negative sin of 𝑑. Which of the following statements about the acceleration of the particle is true? Is it (A) it is equal to the velocity of the particle? (B) It is equal to negative π‘₯. (C) It is equal to π‘₯. Or (D) it is equal to negative 𝑣, where 𝑣 is the velocity of the particle.

Here we have an expression for the displacement of the particle π‘₯ given in terms of time 𝑑. And we need to look for some information about the acceleration of that same particle. So let’s begin by recalling the link between acceleration and displacement. Firstly, we begin by defining 𝑣 to be equal to the velocity of the particle. We know that velocity is defined as the rate of change of an object. And when we think about rate of change, we actually think about differentiation.

So velocity is the first derivative of displacement π‘₯ with respect to 𝑑 time. We also know that acceleration is rate of change of velocity. So acceleration is the first derivative of velocity with respect to time. Now, of course, since velocity itself is the first derivative of π‘₯ with respect to 𝑑, we can say that acceleration must be its second derivative. It’s d two π‘₯ by d𝑑 squared.

So we’re going to begin by differentiating our expression for π‘₯ twice. So the velocity dπ‘₯ by d𝑑 will be the first derivative of negative sin 𝑑 with respect to 𝑑. There are a couple of different ways to remember this, but we can quote the general result for the derivative of sin π‘₯. It’s cos π‘₯. This means the derivative of sin 𝑑 with respect to 𝑑 must be cos 𝑑. And so the derivative of negative sin 𝑑 is negative cos 𝑑.

We then need to differentiate this expression with respect to 𝑑. And that will give us the result for acceleration. It’s the first derivative of negative cos 𝑑. This time, we quote the general result for the derivative of cos π‘₯ with respect to π‘₯, and it’s negative sin π‘₯. So differentiating negative cos 𝑑 with respect to 𝑑 gives us negative negative sin 𝑑, which is simply positive sin 𝑑.

If we look at the statements about the acceleration of the particle, we see they are being compared to the velocity and the displacement. So we see π‘₯, the displacement, is negative sin 𝑑; 𝑣, velocity, is negative cos 𝑑; and π‘Ž, acceleration, is sin 𝑑. The main link we can see is between π‘₯ and π‘Ž. π‘Ž is sin 𝑑 and π‘₯ is negative sin 𝑑. So this must mean that π‘Ž is equal to negative π‘₯. And so the answer is (B). The acceleration is equal to negative π‘₯.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.