## Calculus (3rd Edition)

$3.675 \ m$
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 $3.675 \ m$