Originally posted by geepamoogleWrong geep, you're a liar. There are others who have agreed with me, even on this thread, that take exception to the wording of the problem.
Why then, does it seem to be relatively common that a number of others seem to understand the problem as the poser intended it, but you usually seem to misunderstand the problem and think it poorly worded?
Originally posted by geepamoogleWell then you could add a float, but whatever just as equally contrived i guess. But this idea of the flower on the pond sticking up a foot with it's stem cut exactly at water level that just stretches out exactly to the edge of the pond as long a you use a 5 foot stick just looked so implausible and all that just looked so bogus, PBE6 could only add childish ad hominems to his problem description instead of taking a real interest in it. He could have just thrown some measurements at us without all this silly hokum attached.
I would also suggest to you that a beach ball on a string would not float a foot above the water like a flower on a stiff stem would, but that is tangential to the matter at hand.
Originally posted by eldragonflyWhy don't you offer up a "well worded" problem? By the way, bud, string does not stay straight. It's not rigid, but it bends and twists, so your idea is just as bad.
Well then you could add a float, but whatever just as equally contrived i guess. But this idea of the flower on the pond sticking up a foot with it's stem cut exactly at water level that just stretches out exactly to the edge of the pond as long a you use a 5 foot stick just looked so implausible and all that just looked so bogus, PBE6 could only add chi ...[text shortened]... in it. He could have just thrown some measurements at us without all this silly hokum attached.
Originally posted by eldragonflyClearly you think the solution offered is going to be wrong on the order of feet, not just inches, since the number given is in feet. Right?
Well then you could add a float, but whatever just as equally contrived i guess. But this idea of the flower on the pond sticking up a foot with it's stem cut exactly at water level that just stretches out exactly to the edge of the pond as long a you use a 5 foot stick just looked so implausible and all that just looked so bogus, PBE6 could only add chi ...[text shortened]... in it. He could have just thrown some measurements at us without all this silly hokum attached.
For someone as critical as you about wording, you need to work on your English too.
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eldragonfly is the worst kind of troll. The greatest contribution he makes to this forum is the excessive amount of horse manure that drops directly from his brain to his fingertips - it keeps plants stems like the one in the original question nice and taught.
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 and turns around in a negligible amount of time, and runs back with the same speed V2 to join the line in his original position.
How far did the first partier move in the time it took for the last partier to get his drink and return to line?
PS This is a very easy question. If you want an undergraduate engineering level question, I can provide one of those too, although I'm afraid eldickinfly won't be able to solve it because they all REQUIRE THAT YOU STATE YOUR ASSUMPTIONS!!! 😲
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Originally posted by PBE6my first guess is to redefine the problem as two objects travelling towards each other at different velocities; one from one of the end points at velocity V2 of Line A which is 2L in length, and one from the other endpoint of Line A, at velocity V1. Since the actual rate/velocity of both objects is unknown, it is impossible to determine where these two objects would meet, or in the case of the congo line, how far it moved.
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 and turns around in a negligible ...[text shortened]... first partier move in the time it took for the last partier to get his drink and return to line?
Originally posted by eldragonflyActually since you are not given V1 there is no solution.
my first guess is to redefine the problem as two objects travelling towards each other at different velocities; one from one of the end points at velocity V2 of Line A which is 2L in length, and one from the other endpoint of Line A, at velocity V1. Since the actual rate/velocity of both objects is unknown, it is impossible to determine where these two objects would meet, or in the case of the congo line, how far it moved.
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Originally posted by PBE6The solution to this one is a formula, rather than a number.
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?
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)