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. 12 metres
B. 20 metres
C. 8 metres
D. 48 metres

Answer

B

Explanation

The acceleration of the body perpendicular to `OE` is `a=(F)/(m)=(4)/(2)=2m//s^(2)`
Displacement along `OE` , `s_(1)=vt=3xx4=12m`
Displacement perpendicular to `OE`
`s_(2)=(1)/(2)at^(2)=(1)/(2)xx2xx(4)^(2)=16m`
the resultant displacement
`s=sqrt(s_(1)^(2)+s_(2)^(2))=sqrt(144+256)=sqrt(400)=20m`