Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems: 21d

Answer

$0.006111$ $m/s^2$

Work Step by Step

At time $t = 9$ min, the velocity is $2.2$ $m/s$. At time $t = 3$ min, the velocity is $0$ $m/s$. Thus, the change in velocity over the time interval is $2.2$ $m/s$ - $0$ $m/s$ = $2.2$ $m/s$. The change in time is 9 min - 3 min = 6 min = 360 s. Thus, the average acceleration is given by $a_{avg} = \frac{\Delta v}{\Delta t}$. Thus, $a_{avg}$ = $2.2$ m/s / $360$ s = $0.006111$ $m/s^2$.
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