Question

A ship $A$ is moving eastwards with a speed of $20\text{km/h}$ and a ship $B$ ,$50\text{km}$ south of $A$ is moving northwards with a speed of $20\text{km/h}$. The time after which the distance between ships become shortest is

• A. $1\text{h}$
• B. $3.5\text{h}$
• C. $1.25\text{h}$
• D. $2.5\text{h}$

Answer 1

The correct option is C $1.25\text{h}$

$\overrightarrow{v_A}=20\hat{i}\text{km/h}$
$\overrightarrow{v_B}=20\hat{j}\text{km/h}$
velocity of $A$ w.r.t $B$ is,
$\overrightarrow{v_{AB}}=\overrightarrow{v_A}-\overrightarrow{v_B}$
$\overrightarrow{v_{AB}}=(20\hat{i}-20\hat{j})\text{km/h}$
Hence $|\overrightarrow{v_{AB}}|=\sqrt{400+400}=20\sqrt{2}\text{km/h}$
$\therefore \tan \theta =\frac{|-20|}{20}=1$
$\Rightarrow \theta ={45}^{\circ }$

$\because$ 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 $\overrightarrow{v_{AB}}$ is $\perp$ to $B$.
$\Rightarrow$ From $\Delta APB$,
$\alpha ={90}^{\circ }-\theta ={45}^{\circ }$
Hence shortest distance between ships is $BP$,
$\sin \alpha =\frac{BP}{AB}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{BP}{50}$
$\therefore BP=\frac{50}{\sqrt{2}}=25\sqrt{2}\text{km}$
For ship $B$, the ship $A$ appears to be coming towards it along line $AP$.
From $\Delta BAP$,
$AP=\sqrt{A{B}^2-B{P}^2}=\sqrt{2500-1250}$
$AP=25\sqrt{2}\text{km}$
$\Rightarrow |\overrightarrow{v_{AB}}|\times t=AP$
$\therefore t=\frac{AP}{|\overrightarrow{v_{AB}}|}=\frac{25\sqrt{2}}{20\sqrt{2}}=1.25\text{h}$
$\Rightarrow t$ is the time at which shortest distance occured.