Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 101: 6

Answer

The person would experience a force of -19,000 N. The distance traveled is 1.1 meters.

Work Step by Step

$a = 30g = (30)(9.80 ~m/s^2) = 294 ~m/s^2$ $F = ma = (65 ~kg)(-294 ~m/s^2) = -19,000 ~N$ The person would experience a force of -19,000 N. Now, we find the distance traveled. The first step in this regard is to find the initial speed: $v_0 = (95 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 26 ~m/s$ We can use the initial velocity to find the distance traveled. $x = \frac{v^2 - v_0^2}{2a} = \frac{0 - (26 ~m/s)^2}{(2)(-294 ~m/s^2)} = 1.1 ~m$
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