A horizontal force `F` of variable magnitude and constant direction acts on a body of mass `m` which is initially at rest at a point `O` on a smooth horizontal surface. The magnitude of `F` is given by `F=beta+alphat` where `t` is the time for which the force has been acting on the distance of the body from `O` at time `t` , then `s` is equal to.

Options

A. `(1)/(2m)(betat+alphat^(2))t`
B. `(1)/(2m)(beta+alphat^(2))`
C. `((beta+alphat)t^(2))/(2m)`
D. `t^(2)/(6m)(3beta+alphat)`

Answer

D

Explanation

`a=(dv)/(dt)=(F)/(m)=(beta+alphat)/(m)`
`int_(0)^(v)dv=int_(0)^(t)((beta+alphat)/(m))dt` implies `v=(betat)/
(m)+(alphat2)/(2m)`
Now `intds=intvdt` implies `int_(0)^(s)ds=int_(0)^(t)((betat)/(m)+(alphat^(2))/(2m))dt`
implies `s=(t^(2))/(6m)(3beta+alphat)`