### Video Transcript

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

In this question, weβve been given
information about the displacement of a particle time π‘. And weβre looking to find
information about the acceleration of that same particle. And so we recall the link between
acceleration and displacement. Acceleration is the rate of change
of velocity of the particle. And the velocity itself is the rate
of change of displacement. So, we differentiate an expression
for displacement with respect to time to get us an expression for velocity. And then we differentiate once more
to get an expression for acceleration.

So, to find an expression for the
acceleration of the particle, weβre going to differentiate negative cos of π‘ with
respect to π‘. In fact, thereβs a cycle that can
help us remember how to differentiate trigonometric functions. The derivative of sin π₯ is cos
π₯. Then, the derivative of cos π₯ is
negative sin π₯. If we differentiate negative sin π₯
with respect to π₯, we get negative cos π₯. Then, if we differentiate negative
cos π₯ with respect to π₯, we get back to sin π₯. So, letβs begin by differentiating
our expression for π₯ with respect to time π‘. This tells us that the velocity is
the derivative of negative cos π‘. We can see from our cycle that the
derivative of negative cos π₯ is sin π₯. So, the derivative of negative cos
π‘ with respect to π‘ is sin π‘.

To find an expression for
acceleration, weβre going to differentiate our expression for velocity with respect
to time. Once again, we see from our cycle
that the derivative of sin π₯ with respect to π₯ is cos π₯. And so this means that the
derivative of sin π‘ with respect to π‘ is cos π‘. So, we have three expressions
describing the motion of the particle. Velocity is sin π‘, acceleration is
cos π‘, and π₯, displacement, is negative cos π‘.

We can see that, in fact, our
expressions for acceleration and displacement look quite similar. However, they are negatives of one
another. So, we can say that π₯ is the
negative of the acceleration or vice versa. π is equal to negative π₯. Going back to the options given to
us in this question, we see that that is equivalent to (D). The answer is (D). It is equal to negative π₯.