Five particles situated at the corners of a pentagon 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 ?
Five particles situated at the corners of a pentagon 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 ?
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The particles will meet at the point o the center of hexagon.At any instance the particles will form a
hexagon with the same center O.
Let's focus on motion of A
At any instance velocity of A is making an angle 60∘ from the line AO
The component of velocity of A along AO will be vcos60∘=v2
This component is the rate of decrease of the distance AO
AO = a
⇒ time taken for AO to become 0 t = av2=2av
Alternate solution:
Velocity of A along AB is v
Velocity of B along AB is -v cos60∘
= −v2
So the rate at which the distance AB decrease is v−v2=v2
⇒t=av2
=2av
The particles will meet at the point o the center of hexagon.At any instance the particles will form a
hexagon with the same center O.
Let's focus on motion of A
At any instance velocity of A is making an angle 60∘ from the line AO
The component of velocity of A along AO will be vcos60∘=v2
This component is the rate of decrease of the distance AO
AO = a
⇒ time taken for AO to become 0 t = av2=2av
Alternate solution:
Velocity of A along AB is v
Velocity of B along AB is -v cos60∘
= −v2
So the rate at which the distance AB decrease is v−v2=v2
⇒t=av2
=2av
