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 2 - Kinematics in One Dimension - Exercises and Problems - Page 63: 68

Answer

(a) Tina drives 900 meters. (b) v = 60 m/s

Work Step by Step

(a) We can equate the distance traveled by David and Tina to find the time it takes Tina to pass David; $\frac{1}{2}at^2=v~t$ $t = \frac{2v}{a} = \frac{(2)(30~m/s)}{2.0~m/s^2}$ $t = 30~s$ We can find the distance that Tina travels in 30 seconds. $x = \frac{1}{2}at^2$ $x = \frac{1}{2}(2.0~m/s^2)(30~s)^2$ $x = 900~m$ Tina drives 900 meters. (b) $v = a~t = (2.0~m/s^2)(30~s)$ $v = 60~m/s$
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