Two particle A and B are moving with uniform velocity as shown in the figure given below at t =0.
I. Will the two particle collide
II. Find out shortest distance between two particles

The correct option is C

3

Observing A from the frame of B
Let us draw $\overrightarrow{V_{AB}},\overrightarrow{V_{AB}}=\overrightarrow{V_A}=\overrightarrow{V_B}=\overrightarrow{V_A}+(-\overrightarrow{V_B})$
We get $tan\theta =\frac{10}{20}=\frac{1}{2};againtan\theta =\frac{AD}{CD}=\frac{AD}{40}=\frac{1}{2}$
$\Rightarrow AD=20\Rightarrow DO=10$ and BC = 10
Hence shortest distance between the particle $d_{short}=BCcos\theta =10cos\theta =\frac{10\times 2}{\sqrt{5}}=4\sqrt{5}m$
Since closest distance is non zero therefore they will not collide