• # question_answer 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  [NEET 2015 ] A) 0 h        B) 5 h C) $5\sqrt{2}h$ D) $10\sqrt{2}h$

 It is clear from the diagram that the shortest distance between the ship A and B is PQ. Here       $\sin {{45}^{{}^\circ }}=\frac{PQ}{OQ}$ $\Rightarrow$   $PQ=100\times \frac{1}{\sqrt{2}}=50\sqrt{2}m$ Also, ${{v}_{AB}}=\sqrt{v_{A}^{2}+v_{B}^{2}}=\sqrt{{{10}^{2}}+{{10}^{2}}}=10\sqrt{2}km/h$ So, time taken for them to reach shortest path is $t=\frac{PQ}{{{v}_{AB}}}=\frac{50\sqrt{2}}{10\sqrt{2}}=5h$