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

The correct option is 20 m.

Step-by-step Explanation:

Let us choose the direction $OE$ as the x-axis and the perpendicular direction $OF$ as the y-axis, with the origin at the starting point $O$.

1. Motion along the $OE$ direction (x-axis):
The body has an initial velocity along $OE$, and there is no force acting in this direction. Therefore, the acceleration along the x-axis is zero.
Given:
Initial velocity along the x-axis, $u_x=3\text{m/s}$
Acceleration along the x-axis, $a_x=0{\text{m/s}}^2$
Time, $t=4\text{s}$

The displacement $x$ along $OE$ after 4 seconds is given by:
$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$x=\left(3\right)\left(4\right)+0=12\text{m}$

2. Motion along the $OF$ direction (y-axis):
The body starts with no initial velocity along $OF$, but is subjected to a constant perpendicular force of 4 N.
Given:
Initial velocity along the y-axis, $u_y=0\text{m/s}$
Force along the y-axis, $F_y=4\text{N}$
Mass of the body, $m=2\text{kg}$

First, we find the acceleration along the y-axis using Newton's second law:
$a_y=\frac{F_y}{m}$

The displacement $y$ along $OF$ after 4 seconds is given by:
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$y=0+\frac{1}{2}\left(2\right){\left(4\right)}^2=16\text{m}$

3. Total distance from the starting point O:
Since the two displacements $x$ and $y$ are perpendicular to each other, the net distance $d$ of the body from the origin $O$ is calculated using the Pythagorean theorem:
$d=\sqrt{x^2+y^2}$
$d=\sqrt{{12}^2+{16}^2}$
$d=\sqrt{144+256}$
$d=\sqrt{400}=20\text{m}$

Thus, the distance of the body from O after 4 seconds is 20 m.