Question

A ship $A$ is moving towards west with a speed of $10\text{km/h}$, and a ship $B$, $100\text{km}$ south of $A$, is moving towards north with the same speed of $10\text{km/h}$. The time after which the distance between them becomes shortest is

• A. $5\sqrt{2}\text{h}$
• B. $10\sqrt{2}\text{h}$
• C. $0$
• D. $5\text{h}$

Answer 1

The correct option is D $5\text{h}$


We know the time is given by,
$t=\frac{d}{V_{rel}}=\frac{d}{V_{AB}}$

Here, $\overrightarrow{V_{AB}}=\overrightarrow{V_A}-\overrightarrow{V_B}$


$|\overrightarrow{V_{AB}}|=\sqrt{{10}^2+{10}^2}=10\sqrt{2}km/h$
From figure, $d=AP$
$AP=AB\cos {45}^{\circ }=100\times \frac{1}{\sqrt{2}}km$

$t=\frac{AP}{V_{AB}}=\frac{100}{\sqrt{2}\times 10\sqrt{2}}=5h$