Question

Two particles A and B are moving with uniform velocities as shown in the figure given below at $t=0$. Find out the shortest distance between the two particles

• A. $9\sqrt{2}m$
• B. $20\sqrt{2}m$
• C. $11\sqrt{2}m$
• D. $12\sqrt{2}m$

Answer 1

The correct option is B $20\sqrt{2}m$

First of all, let us find the velocity of A w.r.t B in the problem.
Let $\theta$ be the angle made by $\overrightarrow{V_{AB}}$ with $-\overrightarrow{V_B}$.


Using the figure, we get
$\tan \theta =\frac{|{\overrightarrow{V}}_A|}{|{\overrightarrow{V}}_B|}=\frac{20}{20}=1\Rightarrow \theta ={45}^{\circ }$
Again, $\tan \theta =\frac{AD}{CD}=\frac{AD}{40}=1$
$\Rightarrow AD=40m$
$\Rightarrow DO=80-40=40m$
$\Rightarrow BC=40m$
$d_{short}=BC\cos \theta =40\cos {45}^{\circ }=\frac{40\times 1}{\sqrt{2}}=20\sqrt{2}m$