Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 5 - Force and Motion - Exercises and Problems: 10

Answer

The distance traveled in the first four seconds is 8.0 furlongs.

Work Step by Step

The distance traveled is proportional to the square of the time. That is: $d = ct^2$, where $c$ is a constant. An object travels 2.0 furlongs in the first 2.0 seconds. We can use this information to find the constant $c$. $d = ct^2$ $c = \frac{d}{t^2}$ $c = \frac{2.0~furlongs}{(2.0~s)^2}$ $c = 0.5~furlong/s^2$ We can find the distance in the first 4.0 seconds. $d = ct^2$ $d = (0.5~furlong/s^2)(4.0~s)^2$ $d = 8~furlongs$ The distance traveled in the first four seconds is 8.0 furlongs.
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