## Calculus (3rd Edition)

(a) $v(t)=600 -9 t^{2} ~m/s$ (b) $24~m/s$,$-300~m/s$ (c) $3266~m$
We are given $$s(t)=600 t-3 t^{3} \text { for } 0 \leq t \leq 12$$ (a) We take the derivative of position to find the velocity: \begin{align*} v(t)&=s'(t) \\ &= 600 -9 t^{2} \end{align*} See the plot below. (b) We plug in the $t$ values into velocity: \begin{align*} v(8)&= 600 -9 (8)^{2}=24~m/s\\ v(10)&= 600 -9 (10)^{2}=-300~m/s \end{align*} (c) The maximum height when $v=0$ is at $t= \dfrac{10\sqrt{6}}{3}$, so we have $$s( \dfrac{10\sqrt{6}}{3})= 3266~m$$