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 38: 84b


$\Delta x = 390m$

Work Step by Step

To find acceleration using velocity and change in time, use the equation $$a=\frac{\Delta v}{\Delta t}$$ Velocity must be in meters per second. Converting using dimensional analysis converts 1600 km/h to $$\frac{1600km}{1hr} \times \frac{1 hr}{60min} \times \frac{1min}{60s} \times \frac{1.0\times 10^3m}{1.0km}=440m/s$$ Substituting this value of $\Delta v = 440m/s$ and $\Delta t=1.8s$ yields an acceleration of $$a=\frac{440m/s}{1.8s}=240m/s^2$$ To find distance traveled, use the equation relating initial velocity, displacement, acceleration, and time, which is $$\Delta x=\frac{1}{2}a\Delta t^2+v_o\Delta t$$ Since the object starts from rest, $v_o=0.00m/s$. Substituting known values of $a=240m/s^2$ and $\Delta t=1.8s$ yields a distance traveled of $$\Delta x =\frac{1}{2}(240m/s^2)(1.8s)^2=390m$$
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