Question

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

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

Answer 1

Answer: (C)

20 m

Answer 2

Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds ∴ $x = ut = 3 \times 4 = 12m$

and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds.

∴ $y = ut + \frac{1}{2}(a)t^{2}$ $= 0 + \frac{1}{2}\left( \frac{F}{m} \right) ⥂ t^{2}$ $= \frac{1}{2}\left( \frac{4}{2} \right)4^{2} = 16m$

So its distance from O is given by

$d = \sqrt{x^{2} + y^{2}} = \sqrt{(12)^{2} + (16)^{2}}$

∴ $d = 20m$