College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 62: 34

Answer

t=0 to 1 s, acceleration =$0 m/s^{2}$ t=1 to 3 s, acceleration =$4 m/s^{2}$ t=3 to 8 s, acceleration =$-4 m/s^{2}$ t=8 to 9 s, acceleration =$8 m/s^{2}$ t=9 to 11 s, acceleration =$0 m/s^{2}$ During the last segment, the object moves with constant velocity, i.e. zero acceleration.

Work Step by Step

t=0 to 1 s, acceleration =$\frac{final \,velocity - initial\, velocity}{time \,difference}=\frac{0-0}{1-0} = 0 m/s^{2}$ t=1 to 3 s, acceleration =$\frac{final \,velocity - initial\, velocity}{time \,difference}=\frac{8-0}{3-1} = 4 m/s^{2}$ t=3 to 8 s, acceleration =$\frac{final \,velocity - initial\, velocity}{time \,difference}=\frac{-12-(-8)}{8-3} = -4 m/s^{2}$ t=8 to 9 s, acceleration =$\frac{final \,velocity - initial\, velocity}{time \,difference}=\frac{-4-(-12)}{9-8} = 8 m/s^{2}$ t=9 to 11 s, acceleration =$\frac{final \,velocity - initial\, velocity}{time \,difference}=\frac{-4-(-4)}{11-9} = 0 m/s^{2}$ During the last segment, the object moves with constant velocity, i.e. zero acceleration.
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