A terrorist ‘A’ is walking at a constant speed of 7.5 km/hr due West. At time t=0, he was exactly South an army camp at a distance 1 km. At this instant a large number of army men scattered in every possible direction from their camp in search of the terrorist. Each army person walked in straight line at a constant speed of 6 km/hr. At what time (in minutes) will the terrorist in this get nearest to an army person?

Let the ${\overrightarrow{V}}_a$ be velocity of an army person and ${\overrightarrow{V}}_T$ be velocity of the terrorist. With respect to the terrorist, velocity of ${\overrightarrow{V}}_{aT}={\overrightarrow{V}}_a+(-\overrightarrow{V_T})$ 
$\overrightarrow{AO}$ Represents $(-\overrightarrow{V_T})$  
$\overrightarrow{O{B}_1},\overrightarrow{O{B}_2}$ ….represent ${\overrightarrow{V}}_a$ in various possible directions.
$\overrightarrow{A{B}_1},\overrightarrow{A{B}_2}$ … are relative velocities for various possible direction ${\overrightarrow{V}}_a.$ When AB gets tangential to the circle shown. We get maximum value of angle $\alpha .$ This is the vase fro which an army person will get closest to the terrorist.
For this case $\sin \alpha =\frac{6}{7.5}=\frac{3}{5}$ 
Terrorist sees one army person walking along a direction $\alpha ={\sin }^{-1}\left(\frac{3}{5}\right)$ south of East (with respect to himself) who $\frac{d_{\min }}{1km}=\cos \alpha$
$d_{\min }=1\times \frac{4}{5}=\frac{4}{5}km$ 

(b) Time required $t=\frac{3/5km}{\sqrt{{7.5}^2-6^2}}=\frac{0.6}{4.5}hr=8 min$