Question

Question 33 of 45 You may use your calculator for this question. A particle moves along a line so that at time t where 0 ≤t≤n, its position is given by s(t)=-4 sint- t/2+10. What is the acceleration of the particle the 2 first time its velocity equals zero? ​

Answer 1

The particle's velocity at time $t$ is equal to the first derivative of its position at that time, and acceleration is the second derivative.

We have

$s(t) = -4\sin(t) - \dfrac t2 + 10 \implies s'(t) = -4\cos(t) - \dfrac12 \implies s''(t) = 4\sin(t)$

Find when the velocity is zero:

$s'(t) = -4\cos(t) - \dfrac12 = 0 \implies \cos(t) = -\dfrac18 \implies t = \cos^{-1}\left(-\dfrac18\right) \approx 1.696$

At this time, the acceleration of the particle is approximately

$s''(1.696) \approx \boxed{3.969}$

(B)