Question

Two ships $A$ and $B$ are $10\text{km}$ apart on a line running south to north. Ship $A$ farther north is streaming west at $20{\text{kmhr}}^{-1}$ and ship $B$ is streaming north at $20{\text{kmhr}}^{-1}$. How long do they take to reach distance of closest approach?

• A. 10 minute
• B. 15 minute
• C. 20 minute
• D. 25 minute

Answer 1

The correct option is B 15 minute

Ships $A$ and $B$ are moving with same speed $20{\text{kmhr}}^{-1}$ in the directions shown in figure. It is a two dimensional, two body problem with zero acceleration. Let us find ${\overrightarrow{v}}_{BA}$.


Here, $\left|{\overrightarrow{v}}_{BA}\right|=\sqrt{(20{)}^2+(20{)}^2}=20\sqrt{2}{\text{kmhr}}^{-1}$
i.e., ${\overrightarrow{v}}_{BA}$ is $20\sqrt{2}{\text{kmhr}}^{-1}$ at an angle of ${45}^{\circ }$ from east towards north. Thus, the given problem can be simplified as:
$A$ is at rest and $B$ is moving with ${\overrightarrow{v}}_{BA}$ in the direction shown in figure.


therefore, the minimum distance between the two is
$s_{min}=AC=AB\sin {45}^{\circ }=10\left(\frac{1}{\sqrt{2}}\right)\text{km}=5\sqrt{2}\text{km}$
and the desired time is
$t=\frac{BC}{\left|{\overrightarrow{v}}_{BA}\right|}=\frac{5\sqrt{2}}{20\sqrt{2}}\{BC=AC=5\sqrt{2}\text{km}\}\Rightarrow t=\frac{1}{4}\text{hr}=15\text{minutes}$