Question

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:

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

Answer 1

Correct Answer - c
Along OE: `x=ut=3xx4=12m//s`
`bot^(ar)` to OE: `a_(y)=(F)/(m)=(4)/(2)=2m//s^(2)`
`therefore y=u_(y)t+(1)/(2)a_(y)t^(2)`
`=(0)4+(1)/(2)(2)(4)^(2)`
=16,
`therefore` Distance `=sqrt((12)^(2)+(16)^(2))`
`=sqrt(144+256)=sqrt(400)=20m`