An object of mass 3 kg is at rest. Now a force of `vecF=6t^(2)hati+4thatj` is applied on the object, then the velocity of object at t=3 second is

Options

A. `18hati+3hatj`
B. `18hati+6hatj`
C. `3hati+18hatj`
D. `18hati+4hatj`

Answer

A

Explanation

`F=6t^(2)hati+4thatj`
`F=ma" ":.a=(barF)/(m)=(1)/(3)(6t^(2)hati+4thatj)=2t^(2)hati+(4)/(3)thatj`
`a=(dV)/(dt)" ":.v=int_(0)^(t)adt=int_(0)^(3)2t^(2)hatjdt+int_(0)^(3)(4)/(5)thatjdt=18hati+6hatj`