Four persons K, L, M and N are initially at t | vedclass

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(A) Due to symmetry,at any instant,the four persons will be at the corners of a square whose side length gradually decreases. They will eventually mee

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$\frac{d}{v} 	ext{ s}$

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(A) Due to symmetry,at any instant,the four persons will be at the corners of a square whose side length gradually decreases. They will eventually meet at the center $O$ of the square.The velocity of each person is $v$. At any instant,the velocity component of a person directed towards the center $O$ is $v \cos(45^{\circ}) = \frac{v}{\sqrt{2}}$.The initial distance of each person from the center $O$ is half the diagonal of the square,which is $\frac{d\sqrt{2}}{2} = \frac{d}{\sqrt{2}}$.The time taken to reach the center is the distance divided by the velocity component along the radial direction:$t = \frac{	ext{distance}}{	ext{velocity component}} = \frac{d/\sqrt{2}}{v/\sqrt{2}} = \frac{d}{v}$.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

Four persons $K, L, M$ and $N$ are initially at the corners of a square of side length $d$. If every person starts moving with speed $v$ such that $K$ is always headed towards $L$,$L$ towards $M$,$M$ towards $N$,and $N$ towards $K$,then the four persons will meet after

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