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

Options

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

Answer

C

Explanation

Along OE,`x=ut+=3xx4=12m//s`
`_|_` to `OE,a_(y)=F/m=4/2=2m//s^(2)`
`:.y=u_(y)t+1./2a_(y)t^(2)`
`=(0)4+1/2(2)(4)^(2)=16m`
`:.` Distance `=sqrt((12)^(2)+(16)^(2))=sqrt(144+256)=sqrt(400)=20m`