What is the Average Speed Paradox?

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Discussion Overview

The discussion revolves around a problem regarding average speed, specifically the scenario where a journey is divided into two halves with differing speeds. Participants explore the implications of achieving a specific average speed given the constraints of the first half of the journey being completed at a low speed due to traffic. The conversation touches on mathematical reasoning and the paradox that arises from the conditions set by the problem.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes a journey where the first half is traveled at 10 mph and questions how fast the second half must be to achieve an average speed of 20 mph, suggesting that it leads to an impossible scenario of traveling faster than light.
  • Another participant calculates the average speed based on the distances and times involved, concluding that the average speed would actually be approximately 16.36 mph, which contradicts the book's answer of 45 mph.
  • A later reply acknowledges a mistake in the reference to the problem's chapter, indicating that the book states the problem is impossible, thus supporting the earlier calculations that suggest the average speed cannot be achieved.
  • Participants express personal reflections on their mathematical understanding and previous discussions on related topics, such as the Euler Product formula and techniques in number theory.

Areas of Agreement / Disagreement

There is disagreement regarding the correct average speed and the validity of the book's answer. Some participants support the idea that the problem is impossible, while others challenge the book's conclusion based on their calculations. The discussion remains unresolved with multiple competing views.

Contextual Notes

The discussion highlights limitations in the problem's assumptions and the definitions of average speed, as well as unresolved mathematical steps that lead to differing conclusions among participants.

DeaconJohn
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I take a certain journey and due to heavy traffic crawl along the first half of the complete distance of my journey at an average speed of 10 mph.

How fast would I have to travel over the second half of the journey to bring my average speed to 20 mph?

[This is Problem 10 on page 18 in Chapter 1 of Carter and Russell, "The Complete Book of Fun Maths," Wiley publisher, 2002, verbatim.] On page 95, Carter and Russell give the answer as ...
45 mph
However, suppose that the "complete distance" of my journey is 20 miles. Then, after one hour at 10 mph, I will have traveled 10 miles, and that is "the first half of the complete distance."

So, if I'm going to average 20 mph by the time I reach the end of my journey, then I must transverse the remaining 10 miles instantaneously, that is, faster than the speed of light, and that's not physically possible!

Paradox!

Can anybody help me figure out what is going on here?

What's even worse, if I set t2 = the time for the second half of the journey and write down the equations, I come out with t2 = 0:
t1 = time for first half of journey
t2 = time for second half of journey
ri = rate for the ith half of journey (r1 = 10 mph)
di = distance the ith half of journey
i = 1,2.

d1 = d2
di = ri ti
(d1 + d2)/(t1 + t2) = 2r1 (= 20 mph)

2r1t1 / (t1 + t2) = 2r1
t1 = t1 + t2
t2 = 0 (==> r2 = infinity, and that can't be right!)
 
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This 45 mph answer doesn't make sense. You did some distance d at 10 mph in t1=d/10. Then you did identical distance d in t2=d/45. Total time t1+t2, total distance 2d. Average speed = 2d/(d/10 + d/45) = 2/(1/20 + 1/45) = 16.36.
 
Borek said:
This 45 mph answer doesn't make sense. You did some distance d at 10 mph in t1=d/10. Then you did identical distance d in t2=d/45. Total time t1+t2, total distance 2d. Average speed = 2d/(d/10 + d/45) = 2/(1/20 + 1/45) = 16.36.

So, de book was wrong!

Thanks, Borek, I was hoping that was the case!

It seems to me that I've been missing a lot of simple things lately. That one really had me worried.

I'm just coming out of a three year period of MCI/MCD - meaning - my brain was not working right. It seems to be back to the way it was now.

By the way, I really appreciated the proof of the Euler Product formula that I got from your brillant musings -- I think it was on the number theory board. Very impressive.

I don't know if you noticed my posts to that effect. I was able to take your "musings" and use the Mobius inversion formula (also called the Krockner inversion formula) to pull out what I though was a really slick proof of the Euler product formula.

My favorite reference for the Mobius and Krockner stuff is Analytic Number theory by Tom Apostle.

At the same time, I learned to appreciate the power of the Mobius inversion formula and the Krockner products that it is related to.

I remember when i was a beginning graduate student. I felt intuitively that such a formula as you found existed, but, I couldn't find it.

BTW, your "prime probability" technique can also be applied to give a "heuristic" derivation of the prime number theorem. In fact, if one is really careful about it, and works with "densities" or "the probability that a number around size N is prime" and then takes an integral to get the number of primes less than N, one actually gets the more accurate formula that involves the integral from 0 to N of 1/log and is often called li(N).

The book that Manfred Schroeder "Number Theory in Scinece and Communication" fleshes out the details on that last one.

DJ
 
My Mistake!

DeaconJohn said:
I take a certain journey and due to heavy traffic crawl along the first half of the complete distance of my journey at an average speed of 10 mph.

How fast would I have to travel over the second half of the journey to bring my average speed to 20 mph?

[This is Problem 10 on page 18 in Chapter 1 of Carter and Russell, "The Complete Book of Fun Maths," Wiley publisher, 2002, verbatim.] On page 95, Carter and Russell give the answer as ... 45 mph

Ah Borek,

Maybe my mind's not working so well after all! The problem is in Chapter 2, not Chapter 1!

The "book" answer (in with the Chapter 2 answers) is that the problem is impossible!

Thanks for your help. At least, my math was right!

DJ
 
DeaconJohn said:
By the way, I really appreciated the proof of the Euler Product formula that I got from your brillant musings -- I think it was on the number theory board. Very impressive.

You mistake me for someone else. My most intelligent remarks on math forums would be "errr..." and "ummm...". However, if you want to discuss different techniques used for calculation of chemical equilibria - that's me :wink: And don't worry about your mental abilities - if you are able to deal with things like Euler product or Krockner product (geez, I have no idea what I am writing about) you are still light years ahead of me.
 

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