A body mass `2kg` has an initial velocity of 3 metre//sec along OE and it is subject to a force of `4N` in a direction perpendicular to OE. The distance of body from `O` after 4 sec will be:

Options

A. `12m`
B. `20m`
C. `8m`
D. `48m`

Answer

B

Explanation

Displacement of body in 4 sec along `OE`
`s_(x)=v_(x)t=3xx4=12m`
Force along `OF` (perpendicular to `OE`) `=4N`
:. `a_(y)=(F)/(m)=(4)/(2)=2m//s^(2)`
Displacement of body in 4 sec along `OF`
implies `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)`
`=(1)/(4)xx2xx(4)^(2)=16m` . `[As u_(y)=0]`
:. Net displacement
`s=sqrt(s_(x)^(2)+s_(y)^(2))=sqrt((12)^(2)+(16)^(2))=20m` .