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

Velocity of `B` relative to `A` is `vecV_(BA)=vecV_(B)-vecV_(A)`
`|vecV_(BA)|=sqrt((20)^(2)+(20)^(2))=20sqrt2km//h`
i.e., `vecV_(AB)` is `20sqrt2km//h` at an angle of `45^(@)` from east towards north.
`A` is at rest and `B` is moving with `vecV_(BA)` in the direction shown in Fig.
Therefore the minimum distance between ships
`S_(min)=AC=AB sin 45^(@)=(1/sqrt2)km=5sqrt2km`
and time taken is
`t=BC/(|vecV_(BA)|)=(5sqrt2)/(20sqrt2)=1/4h=15min`