Answer
$33 \ m$
Work Step by Step
Let $x$ be the displacement of the particle. Therefore,
$v(t)=\dfrac{dx}{dt}=4t-3$
Integrate the above equation to obtain the displacement of the particle over the time interval $[2,5]$:
$\int_2^{5} dx=\int_2^{5} (4t-3) \ dt$
or, $ =[2t^2-3t]_2^{5} $
or, $=[2(5)^2-3(5)]-[2(2)^2-3(2)]$
or, $=35-2 $
or, $=33 \ m$
Therefore, the displacement of the particle over the time interval $[2,5]$ is $33 \ m$
