Question

A body of mass $2kg$ has an initial velocity of $3m/s$ along $x-$axis and it is subjected to a force of $1.25N$ 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. $13m$
• B. $28m$
• C. $20m$
• D. $48m$

Answer 1

Answer: (C)

$20m$

$V_x=3m/s$

Because applied force is perpendicular to $x$, so the

velocity $v_x$ will be constant.

So, displacement along $x-$ axis in 4 sec

$S_x=3\times 4=12m$

Along y-a x i s

$F=4N$

$m=2kg$

$a=\frac{F}{m}=2ms-2$

$\begin{array}{cc}{S}_y& =ut+\frac{1}{2}a{t}^2\\ & =0+\frac{1}{2}\times 2\times 4^2\\ & =16m\end{array}$

$\therefore d^2=S_x^2+S_y^2$

$\Rightarrow d^2={12}^2+{16}^2$

$\Rightarrow d^2=144+256$

$\Rightarrow d=\sqrt{400}$

$\therefore d=20m$ ans