question_answer

Four persons \[K,\,L,\,M\] and \[N\] are initially at the corners of a square of side of length \[d\]. If every person starts moving, such that \[K\] is always headed towards \[L,\,L\] towards \[M,\,M\] is headed directly towards \[N\] and \[N\] towards\[K\], then the four persons will meet after                            [IIT 1984]

A)             \[\frac{d}{v}\] sec

B)               \[\frac{\sqrt{2d}}{v}\] sec

C)             \[\frac{d}{\sqrt{2v}}\] sec

D)               \[\frac{d}{2v}\] sec

Correct Answer: A

Solution :

                It is obvious from considerations of symmetry that at any moment of time all of the persons will be at the corners of square whose side gradually decreases (see fig.) and so they will finally meet at the centre of the square O.             The speed of each person along the line joining his initial position and O will be \[v\ \cos 45=v/\sqrt{2}\].             As each person has displacement \[d\cos 45=d/\sqrt{2}\] to reach the centre, the four persons will meet at the centre of the square O after time.             \[\therefore \] \[t=\frac{d/\sqrt{2}}{v/\sqrt{2}}=\frac{d}{v}\]