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: 65a

Answer

$v_{max}=2.25m/s$

Work Step by Step

Remember that velocity and acceleration are related using $$v(t)=\int a(t) dt$$ Since the area under the curve can be expressed as a triangle with base $\Delta t=0.050s$ and $a=90.m/s^2$, the area of a triangle is $$A=\frac{1}{2}bh$$ Substituting known values of $a=90.m/s^2$ and $\Delta t=0.050s$ yields a velocity of $$v=A=\frac{1}{2}(0.050s)(90.m/s^2)=2.25m/s$$
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