University Physics with Modern Physics (14th Edition)

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

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 59: 2.13

Answer

a) Its acceleration is not constant. b) i) $a_{av} = 12.77 m/s^{2}$ ii) $a_{av} = 3.50 m/s^{2}$ iii) $a_{av} = 0.81 m/s^{2}$

Work Step by Step

a) $a(t) = v'(t)$. If $a$ was constant, the funtion $v(t)$ would be linear. It is not the case, so the acceleration is not constant. b) Let's convert the velocities [$mi/h$] to velocities [$m/s$] : $v(t = 0) = 0 mi/h = 0 m/s$ $v(t = 2.1) = 60 mi/h = 26.82 m/s$ $v(t = 20.0) = 200 mi/h = 89.41 m/s$ $v(t = 53) = 260 mi/h = 116.23 m/s$ i) $a_{av} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(2.1)-v(0)}{2.1 - 0} = \dfrac{26.82 - 0}{2.1 - 0} = 12.77 m/s^{2}$ ii) $a_{av} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(20.0)-v(2.1)}{20.0 - 2.1} = \dfrac{89.41 - 26.82}{20.0 - 2.1} = 3.50 m/s^{2}$ iii) $a_{av} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(53)-v(20.0)}{53 - 20.0} = \dfrac{116.23-89.41}{53 - 20.0} = 0.81 m/s^{2}$ On the graph, we see that the velocity is less and less growing, which means that its derivative (the acceleration) is decreasing. These results also show that the acceleration is decreasing.
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