Video: Finding the Acceleration of a Particle Moving in a Straight Line given Its Distance-Time Relationship

A particle moves along a straight line. Its displacement at time ๐‘ก is ๐‘ฅ = โˆ’cos (๐‘ก). Which of the following statements about the acceleration of the particle is true? [A] it is equal to ๐‘ฅ [B] it is equal to โˆ’๐‘ฃ, where ๐‘ฃ is the velocity of the particle [C] it is equal to the velocity of the particle [D] it is equal to โˆ’๐‘ฅ

02:41

Video Transcript

A particle moves along a straight line. Its displacement at time ๐‘ก is ๐‘ฅ equals negative cos of ๐‘ก. Which of the following statements about the acceleration of the particle is true? Is it (A) it is equal to ๐‘ฅ? (B) It is equal to negative ๐‘ฃ, where ๐‘ฃ 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 ๐‘ฅ?

In this question, weโ€™ve been given information about the displacement of a particle time ๐‘ก. And weโ€™re 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, weโ€™re going to differentiate negative cos of ๐‘ก with respect to ๐‘ก. In fact, thereโ€™s a cycle that can help us remember how to differentiate trigonometric functions. The derivative of sin ๐‘ฅ is cos ๐‘ฅ. Then, the derivative of cos ๐‘ฅ is negative sin ๐‘ฅ. If we differentiate negative sin ๐‘ฅ with respect to ๐‘ฅ, we get negative cos ๐‘ฅ. Then, if we differentiate negative cos ๐‘ฅ with respect to ๐‘ฅ, we get back to sin ๐‘ฅ. So, letโ€™s begin by differentiating our expression for ๐‘ฅ with respect to time ๐‘ก. This tells us that the velocity is the derivative of negative cos ๐‘ก. We can see from our cycle that the derivative of negative cos ๐‘ฅ is sin ๐‘ฅ. So, the derivative of negative cos ๐‘ก with respect to ๐‘ก is sin ๐‘ก.

To find an expression for acceleration, weโ€™re going to differentiate our expression for velocity with respect to time. Once again, we see from our cycle that the derivative of sin ๐‘ฅ with respect to ๐‘ฅ is cos ๐‘ฅ. And so this means that the derivative of sin ๐‘ก with respect to ๐‘ก is cos ๐‘ก. So, we have three expressions describing the motion of the particle. Velocity is sin ๐‘ก, acceleration is cos ๐‘ก, and ๐‘ฅ, displacement, is negative cos ๐‘ก.

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 ๐‘ฅ is the negative of the acceleration or vice versa. ๐‘Ž is equal to negative ๐‘ฅ. 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 ๐‘ฅ.

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