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: 43

Answer

$3.53~m/s^{2}$, north.

Work Step by Step

Let north be the positive direction: $v_{0}=-81.9~m/s$ and $v_{f}=0$ $v_{av}=\frac{1}{2}(v_{f}+v_{i})=\frac{1}{2}[0+(-81.9~m/s)]=-40.95~m/s$ $x=x_{0}+(v_{av})t$. But, $x_{0}=0$ and $x=-949~m$ (origin in the landing position): $-949~m=(-40.95~m/s)t$ $t=\frac{-949~m}{-40.95~m/s}=23.17~s$. Finally: $a_{av}=a=\frac{\Delta v}{\Delta t}=\frac{v_{f}-v_{i}}{t}=\frac{0-(-81.9~m/s)}{23.17~s}=+3.53~m/s^{2}$ + means north.
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