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: 41

Answer

$v_{f}=10.3~m/s$

Work Step by Step

First, let's find the average acceleration: $a_{av}=\frac{\Delta v}{\Delta t}=\frac{4.7~m/s}{5.0~s}=0.94~m/s^{2}$ If the acceleration remains the constant, we can find the increase in speed in the additional 6.0 s rearranging the equation above: $\Delta v=(a_{av})(\Delta t)=(0.94~m/s^{2})(6.0~s)=5.6~m/s$ But, in this additional 6.0 s, the initial speed was 4.7 m/s: $\Delta v=v_{f}-v_{i}$ $5.6~m/s=v_{f}-4.7~m/s$ $v_{f}=10.3~m/s$
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