Finding velocity and acceleration functions

Differentiate $s(t) = \sin(2t)$ using the chain rule: the derivative of $\sin(u)$ is $\cos(u) \cdot u'$ where $u = 2t$, so $v(t) = s'(t) = 2\cos(2t)$.

Differentiate $v(t) = 2\cos(2t)$ again using the chain rule: the derivative of $\cos(u)$ is $-\sin(u) \cdot u'$, so $a(t) = v'(t) = -4\sin(2t)$.

Set $v(t) = 2\cos(2t) = 0$, so $\cos(2t) = 0$. On $[0, \pi]$, $2t$ ranges over $[0, 2\pi]$, and $\cos(2t) = 0$ when $2t = \frac{\pi}{2}, \frac{3\pi}{2}$, giving $t = \frac{\pi}{4}, \frac{3\pi}{4}$.

Check the sign of $v(t) = 2\cos(2t)$ around each candidate: near $t = \frac{\pi}{4}$, $v$ goes from positive to negative, and near $t = \frac{3\pi}{4}$, $v$ goes from negative to positive. Since $v$ changes sign at both points, the particle reverses direction at both $t = \frac{\pi}{4}$ and $t = \frac{3\pi}{4}$.