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 - Page 49: 12


(a) $Average$ $speed=9.934\times10^{-2}m/s$ (b) $Average$ $speed=0.2222mi/h$

Work Step by Step

$10months=(10 months)(\frac{30days}{1month})(\frac{24h}{1day})=7200h$ $7200h=(7200h)(\frac{3600s}{1h})=2.592\times10^{7}s$ $1600 mi = (1600mi)(\frac{5280ft}{1mi})(\frac{1m}{3.281ft})=2.575\times10^{6}m$ (a) $Average$ $speed=\frac{distance}{time elapsed}=\frac{2.575\times10^{6}}{2.592\times10^{7}}=9.934\times10^{-2}m/s$ (b) $Average$ $speed=\frac{distance}{time elapsed}=\frac{1600mi}{7200h}=0.2222mi/h$
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