Question

A body of mass 2 kg has an initial velocity of 3 m/s along x-axis and it is subjected to a force of 4 N in y-direction. The distance of the body from origin after 4 seconds will be: (the body was subjected to force at the origin at t= 0 )

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

Answer 1

Answer: (C)

$u_x=3m{s}^{-1},as=0,u_y=2m/se{c}^2$ and $t=4$ sec

If $s_x$ an $s_y$ be the displacement along x-axis and y-axis respectively then

$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$=3\times 4+\frac{1}{2}\times 0\times {\left(4\right)}^2=12m$

and $s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$=0\times 4+\frac{1}{2}\times 2\times {\left(4\right)}^2=12$ m

$s=\sqrt{s_x^2+s_y^2}=\sqrt{{\left(12\right)}^2+{\left(16\right)}^2}$

$=\sqrt{144+256}=\sqrt{400}=20m$