Physics Technology Update (4th Edition)

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

Chapter 1 - Introduction to Physics - Problems and Conceptual Exercises - Page 17: 52

Answer

$43~mi/h$

Work Step by Step

$speed_{1}=v$ and $speed_{2}=v+7.9~mi/h$ $time_{1}=t$ and $time_{2}=t-13~s$ $distance_{1}=distance_{2}=1.0~mi$ $time=\frac{distance}{speed}$ $time_{1}=\frac{distance_{1}}{speed_{1}}$ $t=\frac{1.0~mi}{v}$ $time_{2}=\frac{distance_{2}}{speed_{2}}$ $t-13~s=\frac{1.0~mi}{v+7.9~mi/h}$, but $13~s=\frac{13}{3600}~h$ $t=\frac{1.0~mi}{v+7.9~mi/h}+\frac{13}{3600}~h$ Now, we eliminate t: $\frac{1.0~mi}{v}=\frac{1.0~mi}{v+7.9~mi/h}+\frac{13}{3600}~h$ $(3600~mi)(v+7.9~mi/h)=(3600~mi)v+(13~h)(v+7.9~mi/h)v$ $(13~h)(v^{2})+(102.7~mi)v-28440~mi^{2}/h=0$ $v=\frac{-102.7~mi+\sqrt {(102.7~mi)^{2}+1478880~mi^{2}}}{26~h}=43~mi/h$ $v=\frac{-102.7~mi-\sqrt {(102.7~mi)^{2}+1478880~mi^{2}}}{26~h}\lt0$. But, the speed must be greater than zero.
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