Question

At the moment `t=0` a stationary particle of mass m experiences a time-dependent force `F=at(tau-t)`, whera a is a constant vector, `tau` is the time during which the given force acts. Find:
(a) the momentum of the particle when the action of the force discontinued:
(b) the distance covered by the particle while the force acted.

Answer 1

From the equation of the given time dependence force `vecF=vecat(pi-t)` at `t=pi`, the force vanishes,
(a) Thus `Deltavecp=vecp=underset(0)overset(tau)int vecFdt`
or, `vecp=underset(0)overset(tau)int vecat(tau-t)dt(vecatau^3)/(6)`
but `vecp=mvecv` so `vecv=(vecatau^3)/(6m)`
(b) Again from the equation `vecF=mvecw`
`vecat(tau-t)=m(dvecv)/(dt)`
or, `veca(ttau-t^2)dt=mdvecv`
Integrating within the limits for `vecv(t)`,
`underset(0)overset(t)int veca(t tau-t^2)dt=m underset(0)overset(vecv)intdvecv`
or, `vecv=(veca)/(m)((taut^2)/(2)-(t^3)/(3))=(vecat^2)/(m)(tau/2-t/3)`
Thus `v=(at^2)/(m)(tau/2-t/3)` for `tletau`
Hence distance covered during the time interval `t=tau`,
`s=underset(0)overset(tau)int v dt`
`=underset(0)overset(tau)(at^2)/(m)(tau/2-t/3)dt=a/m(tau^4)/(12)`