Question

Two particles $A$ and $B$ are moving with uniform velocity as shown in the figure given below at $t=0$.
Find out the shortest distance between the particles, if it occurs at $t=2.2\text{s}$

• A. $2\sqrt{3}\text{m}$
• B. $4\sqrt{5}\text{m}$
• C. $5\sqrt{5}\text{m}$
• D. $6\sqrt{3}\text{m}$

Answer 1

The correct option is B $4\sqrt{5}\text{m}$

${\overrightarrow{v}}_A=10\hat{i}\text{m/s}$
${\overrightarrow{v}}_B=20\hat{j}\text{m/s}$
velocity of $A$ w.r.t $B$ is,
${\overrightarrow{v}}_{AB}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$
${\overrightarrow{v}}_{AB}=(10\hat{i}-20\hat{j})\text{m/s}$

$|{\overrightarrow{v}}_{AB}|=\sqrt{{10}^2+{20}^2}=10\sqrt{5}\text{m/s}$

$\because$ velocity of $A$ w.r.t $B$ is directed along direction $AP$, so the minimum distance $\left(d_{min}\right)$ occurs where relative velocity of $A$ w.r.t $B$ i.e $v_{AB}$ is $\perp$ to $B$.
$\Rightarrow$ From $\Delta AOB$,
$AB=\sqrt{{30}^2+{40}^2}=50\text{m}$

$\Rightarrow$ Minimum distance has occured at $t=2.2\text{s}$
$AP=|{\overrightarrow{v}}_{AB}|\times t=10\sqrt{5}\times 2.2$
$\therefore AP=22\sqrt{5}\text{m}$

Now, from $\Delta ABP$,
$A{B}^2=A{P}^2+P{B}^2$
$\Rightarrow 2500=2420+d_{min}^2$
$\therefore d_{min}=\sqrt{80}=4\sqrt{5}\text{m}$