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 take to reach it?

Explanation

Ships A and B are moving with same `20 kmh^(-1)` in the directions shown in figure. It is a two dimensional , two bosy problem with zero acceleration. Let us find `v_(BA)`.

`v_(BA)=v_(B)-v_(A)`
Here, `|v_(BA)|=sqrt((20)^(2)+(20)^(2))=20sqrt(2)kmh^(-1)`
i.e., `v_(BA)` is `20sqrt(2)kmh^(-1)` at an angle of `45^(@)` from East towards North. Thus, the given problem can be simplified as

A is at rest and B is moving with `v_(BA)` in the direction shoen in figure. Therefore, the minimum distance between the two is
`s_(min)=AC=AB sin 45^(@)=10((1)/(sqrt(2)))km = 5sqrt(2)km`

and the desired time is
`t=(BC)/(v_(BA)|)=(5sqrt(2))/(20sqrt(2))(BC=AC=5sqrt(2)km)`
`=(1)/(4)h = 15 min`