Question

A ship $A$ is moving westward with a speed of $10km{h}^{-1}$ and a ship $B100km$ south of $A$ is moving northwards with a speed of $10km{h}^{-1}$. The time after which the distance between them becomes shortest is:

• A. $5\sqrt{2}h$
• B. $10\sqrt{2}h$
• C. $1h$
• D. $5h$

Answer 1

The correct option is D $5h$

It is clear from the figure that the shortest distance between the ship A and B is PQ

Here, $\sin 45=\frac{PQ}{OQ}$

$\Rightarrow PQ=100\times \frac{1}{\sqrt{2}}=50\sqrt{2}m$

Also,

$V_{AB}=\sqrt{V_A^2+V_B^2}=\sqrt{{10}^2+{10}^2}=10\sqrt{2}km/h$

So, time taken to reach shortest path is

$t=\frac{PQ}{V_{AB}}=\frac{50\sqrt{2}}{10\sqrt{2}}=5h$