Imagine that you are stuck in the Newark Airport because of a silly security breach or thunderstorms and you are going out of your way keeping your kids awake and entertained for a few extra hours. Their favorite thing to do is go back and forth on the moving walkway. Your daughter is daring you to a race. She says: "Let's see who is faster: you going ALONGSIDE the walkway all the way to the end and then coming back, or me going ON the moving walkway first time in the direction of the motion and then back, against the direction of the motion. I will try to keep the same speed both ways." You are tired and frustrated because of the delay, but you are a parent. So, you agree. Who do you think is going to win?
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5 comments:
I think I will win. All things being equal, we should go the same speed. But I can walk faster than my daughter, therefore I will will.
The parent will win, because the parent is not being helped or hindered by the slidewalk. Although the kid is being helped by the slidewalk going one way, the kid is being hindered by the slidewalk going the other way. You might think that those would cancel out, but the kid is spending more time fighting the slidewalk and is affected by it for longer. So more of the trip is spent going at the slower speed.
If the slidewalk speed was the *same* as the base walking speed, then the kid wouldn't go *anywhere* on the upstream leg, and would be completely stopped.
Phil's answer is correct but his note that "All things being equal, we should go the same speed." is not. It looks like it should be the same time, but it is not. And Lynnet very nicely explained why not in her answer. So, I think she deserves a puzzle point.
For those of us, who need numbers to support Lynnet's explanation, here their are:
Let S be the legth of the walkway,
Let V be the speed of you and your child. Assume they are the same.
Let v be the speed of the walkway.
Then, your time going along the walkway would be:
t = 2S/V
Your child's time would be:
t = S/(V+v) + S/(V-v) = (SV-Sv+SV+Sv)/(VV-vv)=2S/(V-v/V)
So, we need to compare your time and your child's time:
2S/V and 2S/(V-v/V)
Walkway speed, v, is smaller than your child's speed, V. Otherwise she won't be able to walk against its motion. So denominator in the right fraction is smaller than demoninator in the left fraction. This means that right fraction will be larger than left fraction. Your child's time will be more than yours. What a surprise! I was shocked to find this out when my father told me this puzzle.
Distance formula: rate*time = distance
so time = distance/rate
rm = the mother's rate of walking
rw = the walkway's rate of moving
rd = the daughter's rate of walking
as Lynnet said, rd > rw or the daughter won't be able to go against the flow.
tm = the mother's time to walk out and back, not on the walkway
td = the daughter's time to walk out and back on the walkway
for the mother to beat the daughter, tm < td
D = length of the walkway
then
tm = 2*D/rm
and
td = D/(rd + rw) + D/(rd - rw)
(the time to go downstream plus the time to go upstream)
We want to find when tm < td.
2*D/rm < D/(rd + rw) + D/(rd - rw)
2*D/rm < [D*(rd - rw)+ D*(rd + rw)]/(rd^2 - rw^2)
2*D/rm < 2*D*rd/(rd^2 - rw^2)
1/rm < rd/(rd^2 - rw^2)
(rd^2 - rw^2)/rd < rm
So if rm > rd - rw^2/rd, mom wins the footrace.
Mom may walk at a slower pace than the daughter, yet still win!
It will depend on the speed of the walker, I think. If the kid walks slower than the track, he'll never get off it once heading against the grain, and would already have found this out, and so we must assume he walks a little (or a lot) faster than the track travels.
And I'll assume that both parent and child walk at the same speed?
The "contribution" of the track (let's imagine it goes 100 feet a minute, tho that's probably too fast?) is maybe not as easy as I thought at first. Hmm.
Ya know what? The parent is probably a winner in either case, getting some exercise, and keeping the kid happy. OR, the kid and the parent are both losers, because it's not really a good/safe idea to walk against the direction of the track, due to other users.
Tom
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