Question

Six particles situated at the corners of a regular hexagon with side ${}^{\prime }{a}^{\prime }$ move at a constant speed ${}^{\prime }{v}^{\prime }$. Each particle maintains a direction towards the particle at the next corner. The time taken by the particles to meet each other is

• A. $\frac{a}{v}$
• B. $\frac{a}{2v}$
• C. $\frac{2a}{v}$
• D. $\frac{a(1-\cos {72}^{\circ })}{v}$

Answer 1

The correct option is C $\frac{2a}{v}$

They all will meet at the center $O$ of the hexagon as shown in the figure.



$V_{AB}=V_A-V_B=v-v\cos \theta$
$\theta =\frac{2\pi }{(\text{no. of sides)}}$
$v_{AB}=v-v\cos \frac{2\pi }{6}$

The length of the side of the hexagon $a$. This means, displacement of particle $AO=a$
$\therefore \text{time taken}=\frac{\text{total displacement}}{\text{relative velocity}}$
$=\frac{a}{v-v\cos \frac{2\pi }{6}}=\frac{a}{v(1-1/2)}$
$=\frac{2a}{v}$