A ship A is moving West wards a speed of $10 k m h^{- 1}$ and a ship B 100 km South of A, is moving North wards with a speed of $10 k m h^{- 1}$ .The time after which the distance between them becomes shortest, is :

10√2 h

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0 h

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5√2 h

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Solution

The correct option is C
5h

$V_{B A} = 10 \sqrt{2} k m, \propto= 45^{\circ}$

AN = minimum distance between the two

= d cos $\propto$

Time taken to reach at N = $\frac{d c o s α}{V_{B A}} =$ $\frac{100 \times \frac{1}{\sqrt{2}}}{10 \sqrt{2}} = 5 h$

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