A ship A is moving towards west with a speed of 10 km/h and another ship B, which is 100 km south of A, is moving northwards with a speed of 10 km/h. Find the time after which the distance between them becomes shortest.

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Solution

Suppose, after the ships move $x$ distance, the separation between them is the least.
From the Pythagoros theorem, $A B^{2} = (100 - x)^{2} + x^{2} = 2 x^{2} - 200 x + 10000$
When $A B$ will be minimum, $A B^{2}$ will also be minimum. Taking the first derivative and equating it to zero, we get
$4 x - 200 = 0 \Rightarrow x = 50 k m$
Speed of the ships $= 10 k m / h$
So, time taken to travel $50 k m = \frac{50}{10} h = 5 h$

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