Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises: 22

Answer

(a) My average speed is less than 25.0 m/s. More time is spent at 20.0 m/s than at 30.0 m/s. (b) $Average~speed=24.0~m/s$

Work Step by Step

See conversion factors (inside front cover). $1~mi=1609~m$, so $10.0~mi=10.0\times1609~m=16090~m$ $distance_{1}=distance_{2}=16090~m$. $t_{1}=\frac{distance_{1}}{average~speed_{1}}=\frac{16090~m}{20.0m/s}=805~s$ $t_{2}=\frac{distance_{2}}{average~speed_{2}}=\frac{16090~m}{30.0m/s}=536~s$ $Total~time=t_{1}+t_{2}=805~s+536~s=1341~s$ $Total~distance=distance_{1}+distance_{2}=16090~m+16090~m=32180~m$ $Average~speed=\frac{Total~distance}{Total~time}=\frac{32180~m}{1341~s}=24.0~m/s$
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