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A ship A is moving Westwards with a speed of $10 km h^{-1}$ and a ship B 100 km South of A, is moving Northwards with a speed of $10 km h^{-1}$. The time after which the distance between them becomes shortest, is
Explanation
$\vec{V}_{A} = 10(-\hat{i})$ $\vec{V}_{B} = 10(\hat{j})$ $\vec{V}_{BA} = 10\hat{j} + 10\hat{i} = 10\sqrt{2} km/h$ Distance $OB = 100 \cos 45^{\circ} = 50\sqrt{2} km$ Time taken to each the shortest distance between A and $B = OB/\vec{V}_{BA} = 50\sqrt{2}/10\sqrt{2} = 5 h$