### Neet Solved Paper 2015 Question 2

##### Question: 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

#### Options:

A) 0 h

B) 5 h

C) $ 5\sqrt{2}h $

D) $ 10\sqrt{2}h $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

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 $