Four persons are initially at the four corners of a square whose side is equal to $\mathrm{d}$. Each person now moves with a uniform speed $\mathrm{V}$ in such a way that the first moves directly towards the second, the second directly towards the third, the third directly towards the fourth and the fourth directly towards the first. The four persons will meet after a time equal to  $\mathrm{nd}\text{/}\mathrm{V}$ . Find  $\text{n}$

From considerations of symmetry, the four persons meet at the centre of the square. The displacement from the corner to the centre of the square for each person is given by

$S_r=\frac{\sqrt{d^2+d^2}}{2}=\frac{d}{\sqrt{2}}$

The speed of each person can be resolved into two components: the radial component and the perpendicular component. Throughout the journey, the radial component of velocity towards the centre is given by           

$V_r=V\cos 45°=\frac{V}{\sqrt{2}},\text{\hspace{0.17em}\hspace{0.17em}Time}=\frac{S_r}{V_r}=\frac{d/\sqrt{2}}{V/\sqrt{2}}=\frac{d}{V}$