Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems - Page 33: 17e


$30.3$ cm/s

Work Step by Step

First, the positions at t=2.00s and at t=3.00 s are $x(2.00)$ and $x(3.00)$, respectively. Let $x_{23}$ be the midpoint of $x(2.00)$ and $x(3.00):$ $x_{23}=\displaystyle \frac{x(2.00)+x(3.00)}{2}$ $=\displaystyle \frac{9.75+1.50(2.00)^{3}+9.75+1.50(3.00)^{3}}{2}=36.0$ cm To find the t when the particle is at $x_{23}=36.0$ cm, we solve for t: $36=9.75+1.50t^{3}$ $1.50t^{3}=36.0-9.75$ $t^{3}=\displaystyle \frac{36.0-9.75}{1.50}$ $t=(\displaystyle \frac{36.0-9.75}{1.50})^{1/3}=2.596$ s So, the instantaneous speed at this time is (speed = $|v(t)|$ We have already stated in previous parts that $v(t)=4.50t^{2}$ (cm/s) $|v(t)|=|4.5(2.596)^{2}|$ cm/s$=30.3$ cm/s
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