Question

A body of mass m rests on a horizontal plane with the friction coefficient $k$. At the moment $t=0$ a horizontal force is applied to it which varies with time as $\overrightarrow{F}=\overrightarrow{a}t$, where $\overrightarrow{a}$ is a constant vector. Find the distance traversed by the body during the first $t$ seconds after the force action began.

Answer 1

Since, the applied force is proportional to the time and the frictional force also exists, the motion does not start just after applying the force. The body starts its motion when $F$ equals the limiting friction.
Let the motion start after time $t_0$, then
$F=a{t}_0=kmg$ or, $t_0=\frac{kmg}{a}$
So, for $t\le t_0$, the body remains at rest and for $t>t_0$ obviously
$\frac{mdv}{dt}=a(t-t_0)$ or, $mdv=a(t-t_0)dt$
Integrating, and noting $v=0$ at $t=t_0$, we have for $t>t_0$
${\int }_0^{v}mdv=a{\int }_{t_0}^t(t-t_0)dt$ or $v=\frac{a}{2m}{(t-t_0)}^2$
Thus, $s=\int vdt=\frac{a}{2m}{\int }_{t_0}^t{(t-t_0)}^{2}dt=\frac{a}{6m}{(t-t_0)}^3$