Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.4 Rates of Change - Exercises - Page 130: 22


(a) $ v(t)=600 -9 t^{2} ~m/s$ (b) $24~m/s$,$ -300~m/s$ (c) $3266~m$

Work Step by Step

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 $$s( \dfrac{10\sqrt{6}}{3})= 3266~m$$
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