A body of mass `2 kg` has an initial velocity of `3 m s^-1` along `OE` and it is subjected to a force of `4 N` in `O F` direction perpendicular to `O E`. Find the distance of the body from `O` after `4 s`.
.

Options

A. `12m`
B. `28m`
C. `20m`
D. `48m`

Answer

c

Explanation

`u_(x)=3ms^(-1), as=0, u_(y)=2m//se^(2) and t=4sec`.
If `s_(x) and s_(y)` be the displacement alont x-axis and y-axis respectively then
`s_(x)=u_(x)t+1/2a_(x)t^(2)`
`=3xx4+1/2xx0xx(4)^(2)=12m`
and `s_(y)=u_(y)t+1/2a_(y)t^(2)`
`=0xx4+1/2xx2xx(4)^(2)=16m`
`s=sqrt(s_(x)^(2)+s_(y)^(2))=sqrt((12)^(2)+(16)^(2))`
`=sqrt(144+256)=sqrt(400)=20m`