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

Given situation is shown in the figure. 

Velocity of ship A

$v_A=10 \mathrm{km}{ \mathrm{h}}^{-1}\text{towards west }$

Velocity of ship B

$v_B=10 \mathrm{km}{ \mathrm{h}}^{-1}\text{towards north }$

OS = 100 km

OP = shortest distance

Relative velocity between A and B is

$v_{AB}=\sqrt{v_A^2+v_B^2}=10\sqrt{2} \mathrm{km}{ \mathrm{h}}^{-1}$

$\cos 45°=\frac{OP}{OS};\frac{1}{\sqrt{2}}=\frac{OP}{100}$

$OP=\frac{100}{\sqrt{2}}=\frac{100\sqrt{2}}{2}=50\sqrt{2} \mathrm{km}$

The time after which distance between them equals to OP is given by

$t=\frac{OP}{v_{AB}}=\frac{50\sqrt{2}}{10\sqrt{2}}\Rightarrow t=5 \mathrm{h}$