Question

two ships 10 km apart on a line running from south to north,the one farther north is streaming towards west at 20 kmhr.the other is streaming towards north at 20 kmhr.then the distance of closest approach ?

what is the time taken to reach the position wid the closest approach

Answer 1

Please find below the answer.

Answer 2

Answer:

Required time is 0.25 hour.

Explanation:

Suppose ship A moves with a velocity 20 km/hr towards north snd Ship B moves with a velocity of 20 km/h towards west.

Then velocity of B relative to A= V₍A₎ towards east - V₍A₎ towards north

$= \sqrt{ {20}^{2} + {20}^{2} } = 20 \sqrt{2}$

Distance of closest approach =

$10 \sin( {45}^{o} ) = 10 \times \frac{1}{ \sqrt{2} } = \frac{10}{ \sqrt{2} } km$

Time taken =

$\frac{distance}{speed} = \frac{ \frac{10}{ \sqrt{2} } }{20 \times \sqrt{2} } = 0.25 \: hour$