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
Work in A's frame. Velocity of B relative to A:
$$\vec{v}_{BA} = 20\,\hat{N} - 15\,\hat{W} = 20\,\hat{N} + 15\,\hat{E}, \qquad |\vec{v}_{BA}| = 25\ \text{km/h}$$
In this frame A is fixed and B starts $20$ km south of it, moving along a straight line that makes angle $\phi$ with north, where $\sin\phi = \dfrac{15}{25} = \dfrac{3}{5}$.
The closest distance is the perpendicular distance from A to this line:
$$d_{min} = 20\sin\phi = 20 \times \frac{3}{5} = 12\ \text{km}$$
(It happens after $\dfrac{20\cos\phi}{25} = 0.64$ h.)
Solution by Sreeraj P, M.Sc Physics