Alice and Bob ran a marathon ($26.2$ miles) with Alice running at a uniform $8$ minutes per mile pace and Bob running erratically, but taking exactly $8$ minutes and $1$ second to complete each mile interval - that is to say all intervals of the form $[t, t + 1]$ with $t$ in $[0, 25.2]$. Can Bob have finished ahead of Alice?
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Yes, it is possible, given that there are no limits for Bob's speed. At 208 minutes, Alice would have covered 26 miles while Bob would have covered the same in 208 minutes 26 seconds (which is still not sufficient for Alice to finish the race). The moment Bob covers 26 miles, he can accelerate and finish the race.
ghosts_in_the_code
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But the speed at which Bob can run the final .2 miles is limited by the amount of time it took him to get from 25.2 to 26 -- if he ran that portion too quickly, he can't accelerate too much. – Barry Cipra Nov 26 '14 at 15:21
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@BarryCipra Bob will reach the 26 mile mark in 208 min. 8 sec. Alice will finish the race in 209 min. 36 sec. If Bob can cover the last 0.2 miles in less than 1 min. 26 sec. he will win. – Dan Christensen Nov 26 '14 at 15:53
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@DanChristensen, I agree. The point of the puzzle is to show that he can do that. You have to show that he ran in such a way that he took more than 6 min. 35 sec. to get from the 25.2 mile mark to the 26 mile mark. It has to take him exactly 8 min. 1 sec. to get from 25.2 to 26.2. Let me be clear about what I'm saying: This answer is not a solution. – Barry Cipra Nov 26 '14 at 16:59
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@DanChristensen, my previous reply used your numbers, but actually I think Bob reaches the 26 mile mark in 208 min. 26 sec., not 8 sec. Whatever the correct arithmetic is, the point is the same: Bob's speed on the last stretch is constrained by how he ran the latest stretch. In fact, his speed has to be a periodic function, with period 8 min. 1 sec. – Barry Cipra Nov 26 '14 at 17:13
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Let Bob repeatedly run two tenths of a mile in one second followed by eight tenths of a mile in eight minutes. You need to verify two things: First, any one-mile stretch consists of intervals totalling two tenths of a mile where Bob is running fast and eight tenths where he's running slowly, so that his running time on any one-mile stretch totals eight minutes and one second; and second, his total running time is less than Alice's.
Barry Cipra
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Yes. The time at which Bob uses position $x$ is given by a continuous, monotonic function $f$ and we are given that $f(x+1)-f(x)=8\frac1{60}$ for all $x\in[0,25.2]$ and of course $f(0)=0$. The question is: Can it be that $f(26.2)<8\cdot 26.2$?
Let $g\colon[0,1]\to[0,1]$ be any continuous monotonic function with $g(0)=0$, $g(1)=1$, e.g., $g(x)=x^2$. Then $$f(x)=8\tfrac1{60} \cdot \bigl(\lfloor x\rfloor +g(x-\lfloor x\rfloor)\bigr)$$ describes a possible run by Bob, and we note that he takes $$ f(26.2)=8\tfrac1{60}(26+g(0.2))=208\tfrac{13}{30}+8\tfrac1{60}g(0.2)$$ minutes to complete the run, wheras Alice takes $209\tfrac35$ minutes. So as long as $g(0.2)$ is small enough ($<\tfrac{70}{481}$, to be precise, which does hold for $g(t)=t^2$ for example), Bob can win. For the sake of physics, one may want to require the $g$ is smooth, with matching derivatives at $0$ and $1$ and with $g'$ not too small (bounded from below by the reciprocal of the speed of light) and so on, but all this doesn't forbid to find suzitable $g$ (and hence $f$).
Hagen von Eitzen
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