Question

Two ships are 10 km apart on a line joining south to north. The one farther north is steaming west at $20{\text{km h}}^{-1}$. The other is steaming north at $20{\text{km h}}^{-1}$. What is their distance of closest approach? How long do they take to reach it?

• A. $5\sqrt{2}\text{km},15\text{min}$
• B. $10\text{km},12\text{min}$
• C. $5\sqrt{2}\text{km},12\text{min}$
• D. $5\sqrt{2}\text{km},10\sqrt{2}\text{min}$

Answer 1

The correct option is A $5\sqrt{2}\text{km},15\text{min}$


The velocity component of both ships 1 and 2 are equal. Hence, the angle between the two vectors is 45 degrees.
Solving from the frame of particle - 1
We get $d_{short}=10\sin {45}^{\circ }=\frac{10}{\sqrt{2}}=5\sqrt{2}\text{km}$
$t=\frac{10\sin 45}{|{\overrightarrow{v}}_{21}|}=\frac{10\times \frac{1}{\sqrt{2}}}{20\sqrt{2}}=\frac{1}{4}=15\text{min}$