Question

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

• A. 12 m
• B. 20 m
• C. 8 m
• D. 48 m

Answer 1

Answer: (B)

20 m

Answer 2

Displacement of body in 4 sec along OE

$s_{x} = v_{x}t = 3 \times 4 = 12\mspace{6mu} m$

Force along OF (perpendicular to OE) = 4 N

$\therefore$ $a_{y} = \frac{F}{m} = \frac{4}{2} = 2\mspace{6mu} m/s^{2}$

Displacement of body in 4 sec along OF

⇒ $s_{y} = u_{y}t + \frac{1}{2}a_{y}t^{2} = \frac{1}{2} \times 2 \times (4)^{2} = 16\mspace{6mu} m$ [As $u_{y} = 0$]

$\therefore$Net displacement

$s = \sqrt{s_{x}^{2} + s_{y}^{2}}\mspace{6mu} = \sqrt{(12)^{2} + (16)^{2}} = 20\mspace{6mu} m$