Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 45

Answer

(a) The deer is 5.0 meters from the car when the car stops. (b) v = 21.9 m/s

Work Step by Step

(a) It takes 0.50 seconds to apply the brakes. We can find the distance the car moves in that time. $d = v~t = (20~m/s)(0.50~s)$ $d = 10~m$ The car is 25 meters from the deer when the brakes are applied. We can find the distance the car moves after the brakes are applied. $x = \frac{v^2-v_0^2}{2a}$ $x = \frac{0-(20~m/s)^2}{(2)(-10~m/s^2)}$ $x = 20~m$ The deer is 5.0 meters from the car when the car stops. (b) The total distance the car can travel is 35 meters. $(0.50~s)~v + \frac{v^2}{20~m/s^2}=35~m$ $v^2+10~v-700 = 0$ We can use the quadratic formula to find $v$. $v = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $v = \frac{-(10)\pm \sqrt{(10)^2-(4)(1)(-700)}}{(2)(1)}$ $v = -31.9~m/s, 21.9~m/s$ Since the negative value is unphysical, the solution is $v = 21.9~m/s$.
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