Question

Four persons $K,L,M,N$ are initially at the four corners of a square of side $d$. Each person now moves with a uniform speed $v$ in such a way that $K$ always moves directly towards $L$, $L$ directly towards $M$, $M$ directly towards $N$, and $N$ directly towards $K$. The four persons will meet at a time

• A. $\frac{d}{v}$
• B. $\frac{2d}{v}$
• C. $\frac{3d}{v}$
• D. $\frac{4d}{v}$

Answer 1

The correct option is A $\frac{d}{v}$

Let us take a person and find his velocity of approach towards the center .

The velocity of $K$ throughout the motion towards the centre of the square is $v\cos {45}^{\circ }$ and the displacement covered by this velocity will be $KO$.

$\therefore t=\frac{KO}{v\cos {45}^{\circ }}=\frac{\left(\sqrt{2}d/2\right)}{v/\sqrt{2}}=\frac{d}{v}$


Hence, option $\left(\text{A}\right)$ is the right choice.

Alternatively: ${\overrightarrow{v}}_{KN}={\overrightarrow{v}}_K-{\overrightarrow{v}}_N$ $={\overrightarrow{v}}_K+(-{\overrightarrow{v}}_N)$ along the line $\text{KN}$ ${\overrightarrow{v}}_{KN}=0-\left[\right(-v\left)\right]=v$ Time taken for $\text{K}$ and $\text{N}$ to meet will be $=d/v$. Why this question? Concept: The rate at which the distance between any two points is decreasing is called velocity of approach.