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 49: 23

Answer

(a) Positive (b) Zero (c) Positive (d) Negative (e) $v_{av}=2~m/s$ (f) $v_{av}=0$ (g) $v_{av}=1~m/s$ (h) $v_{av}=-1.5~m/s$

Work Step by Step

(a) The segment points up (upward progression) (b) Horizontal segment (c) The segment points up (upward progression) (d) The segment points down (downward progression) (e) The segment A Starts at $x_{0}=0,~t_{0}=0$ and finishes at $x_{1}=2~m,~t_{1}=1~s$ $v_{av}=\frac{x_{1}-x_{0}}{t_{1}-t_{0}}=\frac{2~m-0}{1~s-0}=2~m/s$ (f) The segment B Starts at $x_{1}=2~m,~t_{1}=1~s$ and finishes at $x_{2}=2~m,~t_{2}=2~s$ $v_{av}=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}=\frac{2~m-2~m}{2~s-1~s}=0$ (g) The segment C Starts at $x_{2}=2~m,~t_{2}=2~s$ and finishes at $x_{3}=3~m,~t_{3}=3~s$ $v_{av}=\frac{x_{3}-x_{2}}{t_{3}-t_{2}}=\frac{2~m-0}{1~s-0}=1~m/s$ (h) The segment D Starts at $x_{3}=3~m,~t_{3}=3~s$ and finishes at $x_{4}=0,~t_{4}=5~s$ $v_{av}=\frac{x_{4}-x_{3}}{t_{4}-t_{3}}=\frac{0-3~m}{5~s-3~s}=-1.5~m/s$
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