Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.4 - The Derivative as a Rate of Change - Exercises 3.4 - Page 133: 1


a. $-2m$, $-1m/s$. b. $3m/s$ and $1m/s$. $2m/s^2$, $2m/s^2$. c. $1.5s$

Work Step by Step

a. Given $s(t)=t^2-3t+2$ and $0\leq t\leq2$, we have $s(0)=2$ and $s(2)=2^2-3(2)+2=0$. Thus the body’s displacement is $\Delta s=s(2)-s(0)=0-2=-2m$ and average velocity for the given time interval is $\bar v=\Delta s/\Delta t=-2/2=-1m/s$. b. $v(t)=s'(t)=2t-3$ thus speeds at endpoints $|v(0)|=3m/s$ and $|v(2)|=4-3=1m/s$. $a(t)=v'(t)=2$ thus $a(0)=a(2)=2m/s^2$. c. When the body changes its direction, its velocity will pass $0$ in either direction; letting $v(t)=0$, we have $2t-3=0$ and $t=3/2=1.5s$. Thus, the body changed its direction at $t=1.5s$
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