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: 84a

Answer

$a=29g$

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 1800 km/h to $$\frac{1800km}{1hr} \times \frac{1 hr}{60min} \times \frac{1min}{60s} \times \frac{1.0\times 10^3m}{1.0km}=5.0\times 10^2m/s$$ Substituting this value of $\Delta v = 5.0\times 10^2 m/s$ and $\Delta t=1.8s$ yields an acceleration of $$a=\frac{5.0\times 10^2m/s}{1.8s}=280m/s^2$$ To express in terms of $g$, divide the acceleration by $g$. This gives you a coefficient of $\frac{280m/s^2}{9.8m/s^2}=29$ This means that the acceleration is $29g$.
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