Physics - Kinematics Question with Solution | TestHub

The acceleration is given by $a=\frac{dv}{dt}=3t^2-2t$. Integrating with respect to time, $v=\int (3t^2-2t)dt=t^3-t^2+C$. Since the particle starts from rest, its initial velocity $v(0)=0$, which implies $C=0$. Therefore, the velocity function is $v(t)=t^3-t^2$. At $t=2\text{s}$, the velocity is $v(2)=(2{)}^3-(2{)}^2=8-4=4\text{m/s}$.