A delivery robot moves along a straight corridor…

Step 1. The velocity. $v = \dfrac{ds}{dt} = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)$.

Step 2. The stops. $v = 0$ at $t = 2\ \text{s}$ and $t = 4\ \text{s}$.

Step 3. The acceleration. $a = \dfrac{dv}{dt} = 6t - 18$. At $t = 4\ \text{s}$, $a = 6\ \text{m/s}^2$. At $t = 2\ \text{s}$, $a = -6\ \text{m/s}^2$.

Step 4. The positions. $s(2) = 8 - 36 + 48 = 20\ \text{m}$ and $s(4) = 64 - 144 + 96 = 16\ \text{m}$. The stopping places are $20 - 16 = 4\ \text{m}$ apart.