Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 10

Answer

speeding up: $a = 16~m/s^2$ slowing down: $a = -5.33~m/s^2$

Work Step by Step

The acceleration is equal to the slope of the velocity versus time graph. Therefore, speeding up: $a = \frac{\Delta v}{\Delta t} = \frac{0.8~m/s}{0.05~s}$ $a = 16~m/s^2$ slowing down: $a = \frac{\Delta v}{\Delta t} = \frac{-0.8~m/s}{0.15~s}$ $a = -5.33~m/s^2$
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