Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 50: 36

Answer

(a) $a_{av}=2~m/s^{2}$ (b) $a_{av}=0$ (c) $a_{av}=-0.5~m/s^{2}$

Work Step by Step

(a) The segment A begins with $t_{0}=0,~v_{0}=0$. It finishes with $t_{1}=5~s,~v_{1}=10~m/s$ $a_{av}=\frac{\Delta v}{\Delta t}=\frac{v_{1}-v_{0}}{t_{1}-t_{0}}=\frac{10~m/s-0}{5~s-0}=2~m/s^{2}$ (b) The segment B begins with $t_{1}=5~s,~v_{1}=10~m/s$. It finishes with $t_{2}=15~s,~v_{2}=10~m/s$ $a_{av}=\frac{\Delta v}{\Delta t}=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{10~m/s-10~m/s}{15~s-5~s}=0$ (c) The segment C begins with $t_{2}=15~s,~v_{2}=10~m/s$. It finishes with $t_{3}=25~s,~v_{3}=5~m/s$ $a_{av}=\frac{\Delta v}{\Delta t}=\frac{v_{3}-v_{2}}{t_{3}-t_{2}}=\frac{5~m/s-10~m/s}{25~s-15~s}=\frac{-5~m/s}{10~s}=-0.5~m/s^{2}$
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