## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Multiple-Choice Questions - Page 110: 11

#### Answer

The correct answer is: (a) 1.3 m/s, north

#### Work Step by Step

The correct answer is: (a) 1.3 m/s, north The area under the velocity versus time graph is the displacement. Each block in the graph is $1~m/s \times 120~s = 120~m$ During the entire 30.0 min, there are 26 blocks above the x-axis and 6 blocks below the x-axis. The blocks below the x-axis indicate a negative displacement. 26 blocks - 6 blocks = 20 blocks The displacement is $(20)(120~m) = 2400~m$ A positive displacement means that the jogger moves 2400 m to the north. The average velocity is: $\frac{displacement}{time}$ $v_{ave} = \frac{2400~m}{1800~m} = 1.3~m/s,~north$

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