Answer
$3.675 \ m$
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 $3.675 \ m$
