Question

Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. What is their distance of closest approach and how long do they to reach it?

Answer 1

Ships A and B are moving with same speed 20 km/h in the directions shown in figure. It is two dimensional, two body problem with zero acceleration.
Let us find ${\overrightarrow{v}}_{BA}$
${\overrightarrow{v}}_{BA}={\overrightarrow{v}}_B-{\overrightarrow{v}}_A$
Here, $\left|{\overrightarrow{v}}_{BA}\right|=\sqrt{{\left(20\right)}^2+{\left(20\right)}^2}=20\sqrt{2}km/h$
i.e., ${\overrightarrow{v}}_{BA}is20\sqrt{2}$ km/h 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
$s_{min}=AC=AB\sin {45}^{\circ }$
$=10\left(\frac{1}{\sqrt{2}}\right)km$
$=5\sqrt{2}km$

And the desired time is:
$t=\frac{BC}{\left|{\overrightarrow{v}}_{BA}\right|}=\frac{5\sqrt{2}}{20\sqrt{2}}(BC=AC=5\sqrt{2}km)$

$=\frac{1}{4}h$
$=15min$