- #1

Monsterboy

- 303

- 96

- Homework Statement
- 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 particles will take to meet each other.

- Relevant Equations
- Speed = distance/time

When I first read the question, it didn't occur to me that these particles would ever meet or catch up with their neighbors. They are all traveling from one vertex to another with a velocity ##v## and a distance ##a##, all either clockwise or anticlockwise right ?

The question says "Each particle maintains a direction towards the particle at the next corner." it's not clear if all the particles are moving around the center of the hexagon in the same direction (clockwise or anticlockwise) or whether some particles are moving towards each other.

If we consider two particles moving towards each other, we get ##time = \frac{distance}{speed}## so ##t = \frac{\frac{a}{2}}{v}##

But the answer is ## \frac{2a}{v} ##

The question also says "Calculate the time the particles will take to meet each other." does this mean the

.

I saw one answer online which said for the particle at point C.

##velocity = v cos \theta = v cos 60 = \frac{v}{2}##

## t = \frac{a}{v/2} = \frac{2a}{v} ## but why would the particle at B stay there ? It would have moved to point A.

The question says "Each particle maintains a direction towards the particle at the next corner." it's not clear if all the particles are moving around the center of the hexagon in the same direction (clockwise or anticlockwise) or whether some particles are moving towards each other.

If we consider two particles moving towards each other, we get ##time = \frac{distance}{speed}## so ##t = \frac{\frac{a}{2}}{v}##

But the answer is ## \frac{2a}{v} ##

The question also says "Calculate the time the particles will take to meet each other." does this mean the

*all*particles are going to meet each other ?.

I saw one answer online which said for the particle at point C.

##velocity = v cos \theta = v cos 60 = \frac{v}{2}##

## t = \frac{a}{v/2} = \frac{2a}{v} ## but why would the particle at B stay there ? It would have moved to point A.