Question

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

• A. $9\sqrt{2}\text{m}$
• B. $10\sqrt{2}\text{m}$
• C. $\frac{11}{\sqrt{2}}\text{m}$
• D. $12\sqrt{2}\text{m}$

Answer 1

The correct option is B $10\sqrt{2}\text{m}$

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



Using the figure, we get
$\tan \theta =\frac{20}{20}=1$
Again $\tan \theta =\frac{AD}{CD}=\frac{AD}{40}=1$
$\Rightarrow AD=40\text{m}$
$\Rightarrow DO=60-40=20\text{m}$
$\Rightarrow BC=20\text{m}$
$d_{\text{short}}=BC\cos \theta =20\cos \theta =\frac{20\times 1}{\sqrt{2}}=10\sqrt{2}\text{m}$