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: 2.14

Answer

$a_x = 6.42~m/s^2$

Work Step by Step

$v_x(t) = (0.860~m/s^3)~t^2$ We can find the time $t$ when $v_x = 12.0~m/s$ $(0.860~m/s^3)~t^2 = 12.0~m/s$ $t^2 = \frac{12.0~m/s}{0.860~m/s^3}$ $t = \sqrt{\frac{12.0~m/s}{0.860~m/s^3}}$ $t = 3.735~s$ We can use $v_x(t)$ to find an equation for $a_x(t)$. $a_x(t) = \frac{dv}{dt} = 2(0.860~m/s^3)~t$ $a_x(t) = (1.720~m/s^3)~t$ We can find the acceleration when $t = 3.735~s$. $a_x = (1.720~m/s^3)(3.735~s) = 6.42~m/s^2$
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