Physics: Principles with Applications (7th Edition)

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

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems - Page 46: 57

Answer

(a) At t = 10.0 s, the instantaneous velocity is 0.3 m/s. (b) At t = 30.0 s, the instantaneous velocity is 1.2 m/s. (c) The average velocity is 0.3 m/s (d) The average velocity is 1.4 m/s (e) The average velocity is -1 m/s

Work Step by Step

(a) For the first 15 seconds, the rabbit moves with a constant velocity that is equal to the slope of the graph. We can use the points (0,0) and (10,3) to find the slope. $v = \frac{3~m - 0}{10~s-0} = 3/10 ~m/s = 0.3~m/s$ At t = 10.0 s, the instantaneous velocity is 0.3 m/s. (b) At t = 30.0 s, we can approximate the slope of the graph with the points (25,10) and (35,22). $v = \frac{22~m-10~m}{35~s-25~s} = \frac{12~m}{10~s} = 1.2~m/s$ At t = 30.0 s, the instantaneous velocity is 1.5 m/s. (c) The average velocity is $\frac{\Delta x}{\Delta t}$ $\frac{\Delta x}{\Delta t} = \frac{1.6~m}{5.0~s} = 0.3~m/s$ (d) The average velocity is $\frac{\Delta x}{\Delta t}$ $\frac{\Delta x}{\Delta t} = \frac{16~m - 9~m}{5.0~s} = 1.4~m/s$ (e) The average velocity is $\frac{\Delta x}{\Delta t}$ $\frac{\Delta x}{\Delta t} = \frac{10~m - 20~m}{10.0~s} = -1~m/s$
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