MHB Calculating rates of two points moving along a circle

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The discussion focuses on calculating the speeds of two particles moving along a circular path. When moving in opposite directions, their rate of closure is the sum of their speeds, while moving in the same direction, the rate of opening is the difference between their speeds. A system of equations is set up to determine the speeds based on the distances traveled in a given time. The solution yields speeds of 18 ft/sec for the faster particle and 12 ft/sec for the slower particle. The distinction between speed and velocity is emphasized, clarifying that the problem specifically addresses speeds.
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Let $v_1 > v_2$ be the respective speeds (rates) in feet per second of the two particles.

Moving in opposite directions, their rate of closure is $(v_1+v_2)$

Moving in the same direction, their rate of opening is $(v_1-v_2)$

Set up a system of two equations and solve for both speeds.
 
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You have added them like they are moving on straight line...In circle velocity changes as direction is changing.
 
DaalChawal said:
You have added them like they are moving on straight line...In circle velocity changes as direction is changing.

If the two particles move in opposite directions, the sum of their respective distances traveled in 5 seconds will be one full circumference length when they meet again.

If the two particles move in the same direction, the faster particle will move ahead of the slower particle, hence the difference between their respective distances traveled in 25 seconds will be one full circumference length when they meet again.

Using this method, I get $v_1 = 18 \, ft/sec$ and $v_2 = 12 \, ft/sec$. You can check the results yourself.
 
This problem has nothing to do "velocity". The problem asks for their speeds, not their velocities.
 

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