Question

A body of mass $2kg$ has an initial velocity of $3m/s$ along $OE$ and it is subjected to a force of $4N$ in a direction perpendicular to $OE$. The distance of body from $O$ after $4s$ will be

• A. 12 m
• B. 20 m
• C. 8 m
• D. 48 m

Answer 1

Answer: (B)

The acceleration of the body perpendicular to $OE$ is
$a=\frac{F}{m}=\frac{4}{2}=2m/s^2$
Displacement along OE,
$s_1=vt=3\times 4=12m$
Displacement perpendicular to OE
$s_2=\frac{1}{2}a{t}^2$
$=\frac{1}{2}\times 2\times {(4)}^2=16m$
The resultant displacement
$s=\sqrt{s_1^2+s_2^2}$
$=\sqrt{144+256}$
$=\sqrt{400}$
$=20m$