Find when velocity is zero

Differentiate $s(t) = 5\cos(t) - t$ to get the velocity function: $v(t) = s'(t) = -5\sin(t) - 1$.

Set $v(t) = 0$: $-5\sin(t) - 1 = 0$, so $\sin(t) = -\dfrac{1}{5}$. On $[0, 2\pi]$, sine is negative in the third and fourth quadrants, giving $t = \pi + \arcsin(1/5) \approx 3.3436$ and $t = 2\pi - \arcsin(1/5) \approx 6.0812$.

Differentiate the velocity function to get acceleration: $a(t) = v'(t) = -5\cos(t)$. Evaluate $\cos(t)$ at each critical time: at $t = \pi + \arcsin(1/5) \approx 3.3436$, $t$ is in the third quadrant so $\cos(t) < 0$, giving $a(t) > 0$; at $t = 2\pi - \arcsin(1/5) \approx 6.0812$, $t$ is in the fourth quadrant so $\cos(t) > 0$, giving $a(t) < 0$.

At $t \approx 3.3436$, we have $v(t) = 0$ and $a(t) > 0$. Since acceleration is positive, $v$ is increasing near this time, so $v$ transitions from negative to positive. Therefore, the particle begins moving in the forward (positive) direction immediately after this time.

At $t \approx 6.0812$, we have $v(t) = 0$ and $a(t) < 0$. Since acceleration is negative, $v$ is decreasing near this time, so $v$ transitions from positive to negative. Therefore, the particle begins moving in the backward (negative) direction immediately after this time.