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 $4\, N$ in a direction perpendicular to $O E$. The distance of body from $O$ after $4 s$ will be :

Updated On: Jan 18, 2023
  • 12 m
  • 20 m
  • 8 m
  • 48 m
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The Correct Option is B

Solution and Explanation

The acceleration of the body perpendicular to $OE$ is :
$\alpha=\frac{F}{m}=\frac{4}{2}=2 m / s ^{2}$
Displacement along $O E ,$
$s_{1}=v t=3 \times 4=12 m$
Displacement perpendicular to $O E$
$ s_{2}=\frac{1}{2} a t^{2}$
$=\frac{1}{2} \times 2 \times(4)^{2}=16\, m$
The resultant displacement
$s=\sqrt{s_{1}^{2}+s_{2}^{2}}$
$=\sqrt{144+256}$
$=\sqrt{400} $
$=20 \,m$
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