The position (x) of a particle of mass 1 kg moving along x-axis at time t is given by `(x=(1)/(2)t^(2))` metre. Find the work done by force acting on it in time interval from t=0 to t=3 s.

Explanation

`x=1/2t^(2)`
`impliesv=(dx)/(dt)=1/2(2t)=t`
` therefore At t=0,v_(1)=0`
`At t =3s, v_(t)=3ms^(-1)`
According to W-E theorem.
`W=DeltaK`
`=K_(t)-K_(i)`
`=1/2mv_(t)^(2)-1/2mv_(t)^(2)`
`=1/2xx1xx3^(2)`
`=4.5J`