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

The correct option is C

20 m

Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds . $\therefore x=ut=3\times 4=12m$ and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds.

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

So its distance from O is given by $d=\sqrt{x^2+y^2}=\sqrt{(12{)}^2+(16{)}^2}$

$\therefore$ d = 20 m