Q 11-03-028JEE MainTop questionHard
Ship A is $20$ km due north of ship B. A sails due west at $15$ km/h and B sails due north at $20$ km/h. The distance of closest approach between the ships is
Answer: (A) $12$ km
Concept. Work in A's frame: B moves in a straight line with $\vec v_{BA}$; closest approach is the perpendicular distance from A to that line.
Step 1. Relative velocity. $\vec v_{BA} = 20\hat N + 15\hat E$, magnitude $25$ km/h; it makes angle $\phi$ with north, $\sin\phi = \dfrac35$.
Step 2. Closest distance. B starts $20$ km south of A:
$$d_{\min} = 20\sin\phi = 12\ \text{km}$$
Answer. $12$ km (Option A).
Solution by Sreeraj P, M.Sc Physics