Force acting on a particle moving in the x-y plane is `vecF=(y^2hati+xhatj)N`, x and y are in metre. As shown in figure, the particle moves from the origin O to point A `(6m, 6m)`. The figure shows three paths, OLA, OMA, and OA for the motion of the particle from O to A.
Now consider another situation. A force `vecF=(4hati+3hatj)N` acts on a particle of mass `2kg`. The particle under the action of this force moves from the origin to a point A `(4m, -8m)`. Initial speed of the particle, i.e., its speed at the origin is `2sqrt6ms^-1`. Figure shows three paths for the motion of the particle from O to A.

Speed of the particle at A will be nearly

Options

A. (a) `4.0ms^-1`
B. (b) `2.8ms^-1`
C. (c) `3.6ms^-1`
D. (d) `5.6ms^-1`

Answer

A

Explanation

`a=((4hati+3hatj))/(2)`
`v^2=u^2+2ax`
`=4xx6+2((4hati+3hatj))/(2)*(4hati-8hatj)=2A-8=16`
`v=4ms^-1`