## Physics (10th Edition)

Published by Wiley

# Chapter 2 - Kinematics in One Dimension - Problems: 38

#### Answer

$\frac{385}{324}\approx1.2$

#### Work Step by Step

Let d = distance between 55 and 35 mi/h signs = distance between 35 and 25 mi/h signs First option: $time=\frac{distance}{speed}$ $t_{A}=\frac{d}{55} + \frac{d}{35}=\frac{18d}{385}$ Second option: $x=\frac{1}{2}(u+v)t$ $t=\frac{2x}{u+v}$ Between 55 and 35 mi/h signs: $t=\frac{2d}{55+35}=\frac{d}{45}$ Between 35 and 25 mi/h signs: $t=\frac{2d}{35+25}=\frac{d}{30}$ $t_{B}= \frac{d}{45} + \frac{d}{30} = \frac{d}{18}$ $t_{B}/t_{A}=(\frac{d}{18})/(\frac{18d}{385})=\frac{385}{324}\approx1.2$

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