Closest Approach In Two-dimensional Relative Motion

Question

Two ships start from the same instant: ship A is 100 m due east of ship B and sails north at 5 m/s, while ship B sails east at 5 m/s. What is the minimum separation distance between the two ships during their subsequent motion?

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Minimum separation problems are most cleanly solved in the relative frame, where one ship is treated as stationary and the other moves with the relative velocity along a straight line; the closest approach is the perpendicular distance from the stationary ship to that line. Place ship B at the origin and ship A initially at $(100, 0)$. Ship A's velocity is $(0, 5)$ and ship B's is $(5, 0)$, so A's velocity relative to B is $(0 - 5, 5 - 0) = (-5, 5)	ext{ m/s}$. In B's frame, A starts at $(100, 0)$ and travels along the direction $(-5, 5)$, i.e. $(-1, 1)$ normalized. The closest approach is the perpendicular distance from the origin to the line through $(100, 0)$ with direction $(-1, 1)$. That line is $x + y = 100$, and the distance from the origin is $\frac{|0 + 0 - 100|}{\sqrt{1^2 + 1^2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2}	ext{ m}$. The option 100 m is the initial separation, not the minimum. The option 50 m and 100√2 m arise from arithmetic mis-scaling of the perpendicular distance. This is the JEE Advanced relative-frame technique for closest approach. As a check, $50\sqrt{2} \approx 70.7	ext{ m}$ is sensibly less than the starting 100 m, confirming the ships first draw nearer before separating.

Closest Approach In Two-dimensional Relative Motion

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