University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 144: 3


a) The body's displacement is $-9m$ and the body's average velocity is $-3m/s$. b) At $t=0$: speed $=3m/s$, acceleration $=6m/s^2$ At $t=3$: speed $=12m/s$, acceleration $=-12m/s^2$ c) The body does not change direction during the given interval.

Work Step by Step

$$s(t)=-t^3+3t^2-3t$$ a) The body's displacement, $\Delta s$, over the interval $[0,3]$ is $$\Delta s=s(3)-s(0)=(-3^3+3\times3^2-3\times3)-(-0^3+3\times0^2-3\times0)$$ $$\Delta s=-9-0=-9m$$ The average velocity, $v_a$, of the body over the interval $[0,3]$ is $$v_a=\frac{\Delta s}{3-0}=\frac{-9}{3}=-3m/s$$ b) We first look for the velocity of the body at time $t$, $v(t)$, and the acceleration of the body at time $t$, $a(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}(-t^3+3t^2-3t)=-3t^2+6t-3$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}(-3t^2+6t-3)=-6t+6$$ We now examine the speed and acceleration at 2 endpoints of $[0,3]$: - At $t=0$: $$v(0)=-3\times0^2+6\times0-3=-3m/s$$ So the speed of the body is $|v(0)|=3m/s$ $$a(0)=-6\times0+6=6m/s^2$$ - At $t=3$: $$v(3)=-3\times3^2+6\times3-3=-27+18-3=-12m/s$$ So the speed of the body is $|v(3)|=12m/s$ $$a(3)=-6\times3+6=-12m/s^2$$ c) We know that if - $v(t)\gt0$: the body moves forward. - $v(t)\lt0$: the body moves backward. So the body will change direction when $v(t)$ changes sign. However, here we have $$v(t)=-3t^2+6t-3=-3(t^2-2t+1)=-3(t-1)^2$$ Since $-3(t-1)^2\le0$ for all $t\in[0,3]$, $v(t)$ cannot change sign in the interval, so as a result, the body will not change direction in the given interval.
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