A ship A is moving Westwards with a speed of l0 km ${\mathrm{h}}^{-\mathrm{l}}$ and a ship B 100 km South of A is moving northwards with a speed of 10 km${\mathrm{h}}^{-\mathrm{l}}$ . The time after which the distance between them becomes shortest is

$\mathrm{Velocity} \mathrm{of} \text{'}\mathrm{A}\text{'} \overrightarrow{{\mathrm{V}}_{\mathrm{A}}} = -10 \overbrace{\mathrm{i}} \mathrm{km}/\mathrm{h}$

$\mathrm{Velocity} \mathrm{of} \text{'}\mathrm{B}\text{'} {\overbrace{\mathrm{V}}}_{\mathrm{B}} = 10 \overbrace{\mathrm{j}} \mathrm{km}/\mathrm{h}$

Velocity of A w.r.t B

$\stackrel{\to }{{\mathrm{V}}_{\mathrm{AB}}} = {\overrightarrow{\mathrm{V}}}_{\mathrm{A}}-{\overrightarrow{\mathrm{V}}}_{\mathrm{B}}= {\overrightarrow{\mathrm{V}}}_{\mathrm{A}}+\left(\overrightarrow{-{\mathrm{V}}_{\mathrm{B}}}\right) = 10 \overbrace{\mathrm{i}}+10\overbrace{\mathrm{j}} \mathrm{km}/\mathrm{h}$

$\left|{\overrightarrow{\mathrm{V}}}_{\mathrm{AB}}\right| = \sqrt{{\left(10\right)}^2+{\left(10\right)}^2}$

$\left|{\overrightarrow{\mathrm{V}}}_{\mathrm{AB}}\right| = 10\sqrt{2} \mathrm{km}/\mathrm{h} \mathrm{directed} \mathrm{along} \mathrm{AC}$

The shortest distance between the ships will be at position ‘C’. Hence relative displacement of A w.r.t B when distance between A and B is shortest will be equal to AC.

From figure AC = $\frac{100}{\sqrt{2}} \mathrm{km}$

so time t = $\frac{\mathrm{AC}}{\left|{\overrightarrow{\mathrm{V}}}_{\mathrm{AB}}\right|} = \frac{{\displaystyle \frac{100}{\sqrt{2}}}}{10\sqrt{2}}$

t =  5 h