Question

Six particles situated at the corners of a regular hexagon of side ‘a’ move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particle will take to meet each other.
$
  (a){\text{ }}\dfrac{{2a}}{{3v}} \\
  (b){\text{ }}\dfrac{a}{v} \\
  (c){\text{ }}\dfrac{{2a}}{v} \\
  (d){\text{ }}\dfrac{a}{{3v}} \\
$

Answer 1

Hint: In this question consider particles at the ends of a hexagon moving towards each other in an anti-clockwise direction. Take into consideration any two particles say F and E. Use the relationship between distance, speed and time along with the concept of relative velocity with the fact that for the particle on E the direction of the velocity is rotated by (180 -120) that is 60 degree. This will help approaching the problem.

Complete step-by-step answer:

Six particles A, B, C, D, E and F are situated on a regular hexagon as shown in the above diagram.
Each particle maintains a direction towards the particle at the next corner as shown in the figure.
The speed of each particle is constant at v m/s.
The side of the regular hexagon is a.
As we know in a regular hexagon the interior angle of a regular hexagon is 120 degree as shown in the figure.
Now concentrate on two particles F and E.
The displacement between these particles is the side of the regular hexagon i.e. a.
Let it take time to meet each other.
Now as we know the relation between velocity, time and distance, as time taken to meet these two particles is the ratio of the distance travelled by these two particles to the relative velocity of these two particles.
As the speed of every particle is the same, but the direction is different so the velocities of particles are changed due to direction of the particles.
If we consider the direction of F particle as reference so the velocity of particle F is v m/s.
Now as we see the particle on E the direction of the velocity is rotated by (180 -120) = 60 degree.
So the velocity of E particle becomes $v\cos {60^o}$
So the relative velocity (${V_{rel}}$) of particle F and E is
$ \Rightarrow {V_{rel}} = v - v\cos {60^o}$..................... (1)
So the time taken by the particles F and E to travel (a) distance is
$ \Rightarrow t = \dfrac{a}{{{V_{rel}}}}$
Now from equation (1) we have,
$ \Rightarrow t = \dfrac{a}{{v - v\cos {{60}^o}}}$
Now as we know that $\cos {60^o} = \dfrac{1}{2}$ so we have,
$ \Rightarrow t = \dfrac{a}{{v - \dfrac{v}{2}}} = \dfrac{a}{{\dfrac{v}{2}}} = \dfrac{{2a}}{v}$
So this is the required time particles will take to meet each other.
Hence option (C) is the correct answer.

Note – As the particles are moving towards each other along the direction of the sides of the hexagon thus the velocity is resolved into its two components that is $v\cos \theta $ and $v\sin \theta $, taking into consideration that the interior angles of hexagon is equal to 60 degrees thus $\theta = {60^0}$. While calculating the relative velocity of one particle with respect to another particle one particle is taken at rest and its velocity is given in exactly the opposite direction to the other particle in motion.