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 54: 93

Answer

(a) $6.7~m/s$ (b) $-5.9~m/s$ (c) $9.6~m/s$

Work Step by Step

Let the positive direction be upward. (a) $distance~traveled=4.0~m$, $elapsed~time=0.60~s$ $average~speed=\frac{distance}{elapsed~time}=\frac{4.0~m}{0.60~s}=6.7~m/s$ (b) $a=-g=-9.81~m/s^2$, $t=0.60~s$ $\Delta v=at$ $\Delta v=(-9.81~m/s^2)(0.60~s)=-5.886~m/s\approx-5.9~m/s$ (c) $v_{av}=\frac{v_0+v}{2}$. Since the positive direction is upward, $v_{av}=6.7~m/s$. So: $\frac{v_0+v}{2}=6.7~m/s$. Also: $\Delta v=v-v_0=-5.886~m/s$ -> $v=v_0-5.886~m/s$. Replace this information in the above equation: $\frac{v_0+v}{2}=6.7~m/s$ $\frac{v_0+v_0-5.886~m/s}{2}=6.7~m/s$ $2v_0-5.886~m/s=2(6.7~m/s)$ $2v_0=13.4~m/s+5.886~m/s$ $v_0=\frac{19.286~m/s}{2}=9.643~m/s\approx9.6~m/s$
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