Question

A and B are two ships at some distance apart moving at different velocities along different directions. The relative velocity of B relative to A is 20 km/hour along a direction as shown. If the two ships take $\frac{1}{4}$ hour to reach their distance for closest approach, find the distance of closest approach. (in km)

Answer 1

In $\frac{1}{4}$ hour distance parallel to ship Ais $x$ and perpendicular to A is y.

$x=20\sin 30°\times \frac{1}{4}=\frac{5}{2}kmy=20\cos 30°\times \frac{1}{4}=5\sqrt{3}km$

Actual closest distance D$=\sqrt{{\left(\frac{5}{2}\right)}^2}+{\left(\frac{5\sqrt{3}}{2}\right)}^2=\sqrt{\frac{25}{4}+\frac{75}{4}}=\frac{10}{2}=5km$