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)

5 h

Answer 2

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