Question

A body of mass $2kg$ has an initial velocity of $5m{s}^{-1}$ along $OE$ and it is subjected to a force of $4N$ in a direction perpendicular to $OE$. The distance of the body from $O$ after $4s$ will be

• A. 20 m
• B. 48 m
• C. 26 m
• D. 12 m

Answer 1

Answer: (C)

From Newton's law
$F=ma$
where $m$ is mass and $a$ the acceleration.
Given, $F=4N,m=2kg$
$\therefore a=\frac{F}{m}=\frac{4}{2}=2m{s}^{-2}$
Also, distance covered by body is
$d=4\times 5=20m$
From equation of motion, displacement $(y)$ is given by
$y=ut+\frac{1}{2}a{t}^2$
Given, $u=0;a=2m{s}^{-2},t=4s$
$\therefore y=0+\frac{1}{2}\times 2\times (4{)}^2=16m$
In vector form, displacement is given by
$s=x\hat{i}+y\hat{j}$
$s=20\hat{i}+16\hat{j}$
Hence, distance from $O$ is
$|s|=\sqrt{20^2+16^2}$
$|s|=\sqrt{400+256}$
$|s|=25.6m$
$\Rightarrow s\approx 26m$