A time varying force `F=6t-2t^(2)N`, at t=0 starts acting on a body of mass 2kg initially at rest, where t is in second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at `t=0` the force `F=0`. Now answer the following:
Find the duration for which the force acts on the body.

Options

A. 2s
B. 3s
C. 3.5s
D. 4.5s

Answer

D

Explanation

Initial velocity: `u=0`
`I= int F dt implies m(v-u)=int _(0)^(t) (6t-2t^(2))dt`
`implies 2v=3t^(2) - 2/3 t^(3)`
Put `v=0`, we get `4.5s`.