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 39: 112


$\Delta t=3.75 \times 10^{-3}s$

Work Step by Step

First, find the acceleration. Use the equation $$v_f^2=v_o^2+2a\Delta x$$ Solving for $a$ yields $$a=\frac{v_f^2-v_o^2}{2\Delta x}$$ Using values of $v_o=0.00m/s$, $v_f=640m/s$, and $\Delta x=1.20m$ yields an acceleration of $$a=\frac{(640m/s)^2}{2(1.20m)}=1.71\times 10^5m/s^2$$ This acceleration can be used in the equation $$\Delta x=\frac{1}{2}a\Delta t^2+v_o\Delta t$$ Since $v_o=0.00m/s$, this equation becomes $$\Delta x=\frac{1}{2}a\Delta t^2$$ Solve for $\Delta t$ to get $$\Delta t=\sqrt{\frac{2\Delta x}{a}}$$ Substitute known values of $a=1.71\times 10^5m/s^2$ and $\Delta x=1.20m$ yields a time of $$\Delta t=\sqrt{\frac{2(1.20m)}{1.71\times 10^5m/s^2}}=3.75 \times 10^{-3}s$$
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