Question

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

Answer 1

Given, mass of the body $m=2kg$
Initial velocity of the particle, $\overrightarrow{u}=3\hat{i}$
Force on the particle at $t=0,\hat{F}=4j$
So, acceleration of the particle, $\overrightarrow{a}=\frac{\overrightarrow{F}}{m}$
$\overrightarrow{a}=\frac{4j}{2}=2j$
Now, position of the body from origin after $t=4s$.
$\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$
$\overrightarrow{s}=\left(3\hat{i}\right)\left(4\right)+\frac{1}{2}\left(2\hat{j}\right)(4{)}^2$
$\overrightarrow{s}=(12\hat{i}+16\hat{j})$
S0, distance (in $m$) of the body from origin after $t=4s$,
$|\overrightarrow{s}|=\sqrt{(12{)}^2+(16{)}^2}$
$|\overrightarrow{s}|=20m$