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: 74

Answer

(a) $a = \sqrt{\frac{P}{2mt}}$ (b) At t = 2 s: v = 11 m/s At t = 10 s: v = 24.5 m/s (c) At t = 2 s: $a = 2.7~m/s^2$ At t = 10 s: $a = 1.2~m/s^2$

Work Step by Step

(a) $v^2 = \frac{2P}{m}~t$ $v = \sqrt{\frac{2P}{m}~t}$ $a = \frac{dv}{dt}$ $a = \frac{1}{2}(\frac{2P}{m}~t)^{-1/2}~\frac{2P}{m}$ $a = \sqrt{\frac{P}{2mt}}$ (b) At t = 2 s: $v = \sqrt{\frac{2P}{m}~t}$ $v = \sqrt{\frac{(2)(3.6\times 10^4~W)}{1200~kg}(2~s)}$ $v = 11~m/s$ At t = 10 s: $v = \sqrt{\frac{2P}{m}~t}$ $v = \sqrt{\frac{(2)(3.6\times 10^4~W)}{1200~kg}(10~s)}$ $v = 24.5~m/s$ (c) At t = 2 s: $a = \sqrt{\frac{P}{2mt}}$ $a = \sqrt{\frac{3.6\times 10^4~W}{(2)(1200~kg)(2~s)}}$ $a = 2.7~m/s^2$ At t = 10 s: $a = \sqrt{\frac{P}{2mt}}$ $a = \sqrt{\frac{3.6\times 10^4~W}{(2)(1200~kg)(10~s)}}$ $a = 1.2~m/s^2$
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