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 $20\text{km/h}$. Then find the shortest distance between the ships.

• A. $20\sqrt{5}\text{km}$
• B. $25\sqrt{2}\text{km}$
• C. $10\sqrt{3}\text{km}$
• D. $5\sqrt{2}\text{km}$

Answer 1

The correct option is B $25\sqrt{2}\text{km}$

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

$\because$ 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 $BP$,
$\sin \alpha =\frac{BP}{AB}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{BP}{50}$
$\therefore BP=\frac{50}{\sqrt{2}}=25\sqrt{2}\text{km}$