University Physics with Modern Physics (14th Edition)

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

Chapter 5 - Applying Newton's Laws - Problems - Exercises: 5.35

Answer

(a) The shortest stopping distance is 52.5 meters. (b) We should drive at a speed of 16.0 m/s.

Work Step by Step

(a) We can find the magnitude of deceleration. $F_f = ma$ $mg~\mu_k = ma$ $a = g~\mu_k = (9.80~m/s^2)(0.80)$ $a = 7.84~m/s^2$ We can find the stopping distance. $x = \frac{v^2-v_0^2}{2a} = \frac{0-(28.7~m/s)^2}{(2)(-7.84~m/s^2)}$ $x = 52.5~m$ The shortest stopping distance is 52.5 meters. (b) We can find the magnitude of deceleration. $F_f = ma$ $mg~\mu_k = ma$ $a = g~\mu_k = (9.80~m/s^2)(0.25)$ $a = 2.45~m/s^2$ We can find the initial speed $v_0$. $v^2 = v_0^2 +2ax$ $v_0^2 = 0-2ax = -2ax$ $v_0 = \sqrt{-(2)(-2.45~m/s^2)(52.5~m)}$ $v_0 = 16.0~m/s$ We should drive at a speed of 16.0 m/s.
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