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`.
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Explanation

Body moves horizontally with constant initial velocity `3 ms^-1` upto `4 s`.
`:. x = ut = 3 xx 4 = 12 m`
And in perpendicular direction it moves under the effect of constant force with zero initial velocity upto `4 s`.
`:. y = u t + (1)/(2) (a) t^2 = 0 + (1)/(2)((F)/(m)) t^2 = (1)/(2)((4)/(2)) 4^2 = 16 m`
So its distance from `O` is given by
`d = sqrt(x^2 + y^2) = sqrt((12)^2 + (16)^2) = 20 m`.