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

Answer 1

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

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


Using the figure, we get
$\tan \theta =\frac{\overrightarrow{V_A}}{\overrightarrow{V_B}}=\frac{10}{20}=\frac{1}{2}$
Again $\tan \theta =\frac{AD}{CD}=\frac{AD}{40}=\frac{1}{2}$
$\Rightarrow AD=20\text{m}$
$\Rightarrow DO=30-20=10\text{m}$
$\Rightarrow BC=10$
$|\overrightarrow{V_{AB}}|=\sqrt{V_A^2+V_B^2}=\sqrt{{10}^2+{20}^2}=10\sqrt{5}$
$\cos \theta =\frac{|\overrightarrow{V_B}|}{|\overrightarrow{V_{AB}}|}=\frac{20}{10\sqrt{5}}=\frac{2}{\sqrt{5}}$
$d_{short}=BC\cos \theta =10\cos \theta =\frac{10\times 2}{\sqrt{5}}=4\sqrt{5}\text{m}$
Since closest distance is non zero therefore they will not collide.