Originally posted by PBE6
Question #2, for eldigglestick:
A conga line of length L is dancing forward with speed V1. The last partier in the conga line suddenly remembers that he let the first partier in the conga line hold his drink while he went to the loo, and he wants it back. The last partier runs to the front of the line with speed V2 (which is faster than V), grabs his drink a ...[text shortened]... irst partier move in the time it took for the last partier to get his drink and return to line?
The solution to this one is a formula, rather than a number.
It is, in essence, a general case scenario.
So the congo line is L ft long and moving forward at V1 ft per second.
The fellow at the end fetching his drink moves at V2 ft per second, a velocity we are told is faster than V1 (or else he'd never catch the front of the line.
So now, we can divide the calculations into 2 parts.
Fetching the Drink
Our hero gains on the front of the line based on his relative speed compared to the line. The velocity is equal to the difference in velocities given.
Time taken =
L / (V2 - V1)
Line travels =
V1 * Time Taken
Returning to the Back
Our hero, having succeeded in his quest returns to his position at a relative speed equal to the sum of the two velocities, since they are moving in opposite directions now.
Time taken =
L / (V2 + V1)
Line travels =
V1 * Time Taken
Now to add the two together and simplify..
L / (V2 - V1) + L / (V2 + V1)
Find a common denominator.
L * (V2 + V1) / (V2^2 - V1^2) + L * (V2 - V1) / (V2^2 - V1^2)
Combine terms and distribute..
L * (V2+V1 + V2-V1) / (V2^2 - V1^2)
Simplify the numerator..
2 * L * V2 / (V2^2 - V1^2)
And you have the formula for how long it takes for our hero to complete his mission and return with his drink.
Now multiply by V1 to find out how far forward the line has moved.
Answer =
2*L*V1*V2 / (V2^2 - V1^2)