Question

A particle of mass $2kg$ is travelling at a velocity of $1.5m/s$. A force $t\left(t\right)=3t^2$ (in N) is applied to it in the direction of motion for a duration of $2$ seconds. Where $t$ denotes time in seconds. The velocity (in $m/s$ up to one decimal place) of the particle immediately after the removal of the force is

• A. 5.5

Answer 1

The correct option is A 5.5
$f\left(t\right)=3t^2$
$ma=3t^2$
$m\frac{dv}{dt}=3t^2$
$2{\int }_{1.5}^{v}dv={\int }_0^{2}3t^{2}dt$
$2(v-1.5)=8$
$v=5.5m/s$