Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.6 Net Change as the Integral of a Rate of Change - Exercises - Page 268: 7


$3.675 \ m$

Work Step by Step

Let $x$ be the displacement of the cat. Therefore, $v(t)=\dfrac{dx}{dt}=-9.8t$ Integrate the above equation to obtain the displacement of the particle over the time interval $[1,6]$: $\int_{0.5}^{1} dx=\int_{0.5}^{1} |-9.8t| \ dt$ or, $ =[4.9 t^2]_{0.5}^1 $ or, $=[4.9 (1)^2]-[4.9(0.5)^2]$ or, $=4.9-1.225 $ or, $=3.675 \ m$ Therefore, the total displacement the cat falls during the time interval $[0.5,1]$ is $3.675 \ m$
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