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

Answer: (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$