First slide

Motion in a plane

Question

A ship A is moving Westwards with a speed of $10 {kmh}^{- 1}$ 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

Difficult

A

$5 \sqrt{2} h$
B

$10 \sqrt{2} h$
C

0 h
D

5 h

Solution

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
$v_{A B} = \sqrt{v_{A}^{2} + v_{B}^{2}} = 10 \sqrt{2} km h^{- 1}$
$\cos 45 ° = \frac{O P}{O S}; \frac{1}{\sqrt{2}} = \frac{O P}{100}$
$O P = \frac{100}{\sqrt{2}} = \frac{100 \sqrt{2}}{2} = 50 \sqrt{2} km$
The time after which distance between them equals to OP is given by
$t = \frac{O P}{v_{A B}} = \frac{50 \sqrt{2}}{10 \sqrt{2}} \Rightarrow t = 5 h$

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