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

• A. 5 $ \sqrt 2 $ h
• B. 10 $ \sqrt 2 $ h
• C. 0 h
• D. 5 h

Answer 1

The Correct Option is D

Given situation is shown in the figure.

Velocity of ship A
$ v_A = 10 \, km \, h^{ - 1} $ towards west
Velocity of ship B
$ v_B = 10 \, km \, h^{ - 1} $ towards north
OS= 100 km
OP = shortest distance
Relative velocity between A and B is
$ v_{ AB} = \sqrt{ v_A^2 + v_B^2 } = 10 \sqrt 2 \, km \, h^{ - 1} $
cos $ 45^\circ = \frac{ OP}{ OS} ; \frac{ 1}{ \sqrt 2} = \frac{ OP}{ 100} $
OP = $ \frac{ 100 }{ \sqrt 2} = \frac{ 100 \, \sqrt 2}{ 2} = 50 \sqrt 2 $ km
The time after which distance between them equals to OP is given by
$ t = \frac{ OP}{ v_{ AB}} = \frac{ 50 \sqrt 2}{ 10 \sqrt 2} \Rightarrow 1 = 5 $ h