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Alice and Bob's marathon


bonanova
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Alice and Bob run a standard 26.2-mile marathon race.

Bob's pace is a constant 8 minutes/mile, start to finish.

Alice's pace is such that for every 1-mile segment of the race, she takes 8:01 minutes.

That is, between mile markers 10 and 11, Alice consumes 8:01 minutes.

And between miles 6.374 and 7.374, Alice consumes 8:01 minutes.

 

Is Bob the sure winner of the marathon?

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Alice and Bob run a standard 26.2-mile marathon race.

Bob's pace is a constant 8 minutes/mile, start to finish.

Alice's pace is such that for every 1-mile segment of the race, she takes 8:01 minutes.

That is, between mile markers 10 and 11, Alice consumes 8:01 minutes.

And between miles 6.374 and 7.374, Alice consumes 8:01 minutes.

 

Is Bob the sure winner of the marathon?

I think, yes.

 

The pace can be 8:01 for every 1-mile segment if and only if it is a constant pace of 8:01 min/mile.. I think :)

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Alice could run fast for 0.2 mile then slow for the rest of the first mile,


and repeat this pattern for the whole marathon.
This way her pace is constant on every 1-mile segment.
Now, after 26 miles of the race she is 26 seconds behind Bob,
and she is going to run fast on remaining 0.2 mile segment of the race.
If she runs fast enough the fast segments, she wins.
The question remains: is it possible to run like this?

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Alice could run fast for 0.2 mile then slow for the rest of the first mile,

and repeat this pattern for the whole marathon.

This way her pace is constant on every 1-mile segment.

Now, after 26 miles of the race she is 26 seconds behind Bob,

and she is going to run fast on remaining 0.2 mile segment of the race.

If she runs fast enough the fast segments, she wins.

The question remains: is it possible to run like this?

A graphical approach..

 

 

post-55015-0-15601000-1415802145_thumb.p

 

The dashed slanted line represents a constant pace of 8:01 min/mile. The red-dashed line represents a possible pace Alice ran at.

 

Now, select any two points on the "distance" axis, separated by equivalent of 1 mile unit. Draw verticals as shown to intersect both the pace lines. Draw a line that connects the point where the verticals meet the red-dashed lines. This line represents the average pace for Alice for that one mile.

 

Now, the requirement is that the average pace should be equal to 8:01 min/mile. Which means, the average pace line is parallel to the 8:01 min/mile pace line.

 

Further, if we slide the 1-mile window (left or right), the average pace line must still be parallel to the 8:01 min/mile pace line.

 

Without proving formally, I assert that is possible iff the red-dashed line is exactly same as the 8:01 min/mile pace line.

 

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From the problem statement, we get that Alice takes 8:01 for any complete
mile in the course [0.00 to 1.00, 0.10 to 1,10, ..., 25.2 to 26.2]. Therefore at
the end of

 

  • 1 mile her time should be 8.017 min [8:01].
     
  • 26 miles her time should be 208.43 min [208:26]
     
  • 26.2 miles her time should be 210.04 min [210:02]

 

While Bob's time should be 209.60 [209:36]

 

 

The only way for her to win is if she can speed up the last fraction of
a mile [where there are no more race markers] such that she can make up the
difference  -- is this possible?? [this is a math problem, not a
realistic athletic question].

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To beat Bob in this marathon by .01 of a second, Alice would have to run at constant speed of 27.7 mph for the first 0.2 miles, then slow down to about 6.329 mph for the next 0.8 miles. Then repeat this pattern for the rest of the marathon. This way Alice covers the first 0.2 miles in just under 26 seconds and takes 455 seconds for the remaining 0.8 miles of each mile. At the mile marker 26 Alice is 26 seconds behind, so she catches up and beats Bob by less than .01 seconds while beating the current world record of sprinting speed of 27.44 mph (according to Wikipedia 

http://en.wikipedia.org/wiki/Footspeed)
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Alice could run fast for 0.2 mile then slow for the rest of the first mile,

and repeat this pattern for the whole marathon.

This way her pace is constant on every 1-mile segment.

Now, after 26 miles of the race she is 26 seconds behind Bob,

and she is going to run fast on remaining 0.2 mile segment of the race.

If she runs fast enough the fast segments, she wins.

The question remains: is it possible to run like this?

A graphical approach..

 

 

attachicon.gifMarathon_Alice_Bob.png

 

The dashed slanted line represents a constant pace of 8:01 min/mile. The red-dashed line represents a possible pace Alice ran at.

 

Now, select any two points on the "distance" axis, separated by equivalent of 1 mile unit. Draw verticals as shown to intersect both the pace lines. Draw a line that connects the point where the verticals meet the red-dashed lines. This line represents the average pace for Alice for that one mile.

 

Now, the requirement is that the average pace should be equal to 8:01 min/mile. Which means, the average pace line is parallel to the 8:01 min/mile pace line.

 

Further, if we slide the 1-mile window (left or right), the average pace line must still be parallel to the 8:01 min/mile pace line.

 

Without proving formally, I assert that is possible iff the red-dashed line is exactly same as the 8:01 min/mile pace line.

 

 

 

 

Let s(t) be the distance Alice covers in time 't'.

We assume s(t) is a continuous and differentiable function of 't' - a reasonable assumption: Since Alice can't disappear and appear in a different spot in zero time (implying infinite speed), and can't instantaneously change her speed to a completely different value (implying infinite acceleration). It is ok for her to cover a large distance (say 20 miles) in a small fraction of a second (say, 10-100).

We further assume that s(t) is increasing function of t. In other words, if x > y, then s(x) >= s(y). Basically, Alice doesn't start running backwards.

Obviously, we stick to a universe governed by classical/Newtonian mechanics and where time flows continuously

Let,

N = 1 mile

M = 26.2 miles

T = 8:02 minutes

Ta = Time Alice takes to complete the race

s(t) = M for t > Ta, but we don't bother about what happens after the race.

As per the OP, the following assertion is true:

If,

  s(x) - s(y) = N,

then,

  x - y = T, for any s(x), s(y) < M (or equivalently, for x, y < Ta).

We first note that the converse is also true: In every 8:02 minute interval during the race, she covers 1 mile. Because: 1. if there is a 8:02 minute interval in which she covers more than 1 mile, then there is a 1 mile distance that she does in less than 8:02 minutes 2. if there is a 8:02 minute interval in which she covers less than 1 mile, continue ticking the clock until she covers one mile - and note that she has taken more than 8:02 minutes for covering that one mile.

So,

  s(t + T) - s(t) = N, if (t+T) < Ta and t > 0.

and/or,

  s(t) - s(t - T) = N, if t < Ta and t > T

For any time 't' during the race, at least one of the conditions above will hold true.

Increasing time by 'dt', we get

  s(t + T + dt) - s(t+dt) = N

and/or,

  s(t + dt) - s(t - T + dt) = N

Subtracting from earlier equation, we get:

  s(t+T+dt) - s(t+T) = s(d+dt) - s(t)

and/or,

  s(t-T+dt) - s(t-T) = s(d+dt) - s(t)

If,

  ds = Incremental change in 's' at time t,

then from equation above,

  ds = s(t+T+dt) - s(t+T)

and/or

  ds = s(t-T+dt) - s(t-T)

So the rate of change of speed (= velocity here) at any arbitrary point 't' must equal

a. The velocity after time T (if race is still in progress after time T)

b. The velocity that was before T (if race had started by then)

Since 't' is any arbitrary point of time during the race, the conditions above is only possible if the velocity is constant through out the race.

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Let s(t) be the distance Alice covers in time 't'.

We assume s(t) is a continuous and differentiable function of 't' - a reasonable assumption: Since Alice can't disappear and appear in a different spot in zero time (implying infinite speed), and can't instantaneously change her speed to a completely different value (implying infinite acceleration).

This is exactly what meant by the last sentence of my answer:

The question remains: is it possible to run like this?

I'd also add one more condition: s(0)=0, since the runners start moving when the the race starts.

Same restrictions should also apply to Bob,

so another question is: how can he run with a constant positive speed?

Obviously he can't without braking his speed function at 0.

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Let s(t) be the distance Alice covers in time 't'.

We assume s(t) is a continuous and differentiable function of 't' - a reasonable assumption: Since Alice can't disappear and appear in a different spot in zero time (implying infinite speed), and can't instantaneously change her speed to a completely different value (implying infinite acceleration).

This is exactly what meant by the last sentence of my answer:

The question remains: is it possible to run like this?

Ok - I misunderstood then.

 

I'd also add one more condition: s(0)=0, since the runners start moving when the the race starts.

Same restrictions should also apply to Bob,

Agree!

so another question is: how can he run with a constant positive speed?

Obviously he can't without braking his speed function at 0.

True :) I didn't bring that up.. but one possibility is to do a "rolling start", starting from behind the race line.

 

 

Annotated my comments above..

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This is only an issue for Bob; he needs Karthikgururaj's suggestion of a rolling start.

For Alice, it's not a problem; she only needs to slow to a momentary stop at the integer mile markers.

She can still win. Her peak speed for the first .2 miles of every lap would just need to be a little greater.

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This is only an issue for Bob; he needs Karthikgururaj's suggestion of a rolling start.

For Alice, it's not a problem; she only needs to slow to a momentary stop at the integer mile markers.

She can still win. Her peak speed for the first .2 miles of every lap would just need to be a little greater.

I agree - I was wrong in my earlier post. The diagram in particular has a big mistake.

It looks like Alice has to change her speed instantaneously, at 0.2 mile mark

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