## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Problems - Page 117: 98

#### Answer

The displacement is $59~mi$ (north) $v_{ave} = 95.7~mi/h$ (north) $a_{ave} = 11.4~mi/h~$ (south)

#### Work Step by Step

The displacement is $185~mi-126~mi = 59~mi$ (north) $v_{ave} = \frac{displacement}{time}$ $v_{ave} = \frac{59~mi}{(37/60)~h}$ $v_{ave} = 95.7~mi/h$ (north) $a_{ave} = \frac{\Delta v}{\Delta t}$ $a_{ave} = \frac{105.0~mi/h-112.0~mi/h}{(37/60)~h}$ $a_{ave} = -11.4~mi/h$ $a_{ave} = 11.4~mi/h~$ (south)

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