Question

Two particles A and B are moving with uniform velocity as shown in the figure given below at t =0.

Find out shortest distance between two particles.

• A. $2\sqrt{5}m$
• B. $4\sqrt{5}m$
• C. $\frac{8}{\sqrt{5}}m$
• D. $8\sqrt{5}m$

Answer 1

The correct option is B $4\sqrt{5}m$



Solving from the frame of B
we get $tan\theta =\frac{10}{20}=\frac{1}{2}$
again $tan\theta =\frac{AD}{CD}=\frac{AD}{40}=\frac{1}{2}$
$\Rightarrow AD=20\Rightarrow DO=10\Rightarrow BC=10$
$d_{short}=BCcos\theta =10cos\theta =\frac{10\times 2}{\sqrt{5}}=4\sqrt{5}m$