College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 60: 9

Answer

(a) Trip takes $1.43\,\mathrm{h}$. (b) We save $0.33\,\mathrm{h}$.

Work Step by Step

Conversion - $1\,\mathrm{mile}=1.609\,\mathrm{km}$. (a) Distance traveled, $d=150\,\mathrm{km}$. Speed, $s=65\,\mathrm{mi/h}=104.59\,\mathrm{km/h}$. Time taken, \begin{align*} t&=\frac{d}{s}=\frac{150}{104.59}\\&=1.43\,\mathrm{h} \end{align*}(b) Return speed, $s'=85\,\mathrm{mi/h}=136.76\,\mathrm{km/h}$. Time taken, \begin{align*} t'&=\frac{d}{s'}=\frac{150}{136.76}\\&=1.1\,\mathrm{h} \end{align*}Time saved is $(t'-t)=(1.43-1.1)=0.33\,\mathrm{h}$.
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