University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises: 2.23

Answer

The stopping distance must be at least 1.71 meters.

Work Step by Step

$v = (105~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 29.2~m/s$ $x = \frac{v^2-v_0^2}{2a} = \frac{0-(29.2~m/s)^2}{(2)(-250~m/s^2)}$ $x = 1.71~m$ The stopping distance must be at least 1.71 meters.
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