What is the Average Speed Paradox?

In summary, the conversation centers around problem 10 in Chapter 2 of Carter and Russell's book "The Complete Book of Fun Maths." The problem involves a journey and finding the average speed over the first and second halves of the journey. After discussing the answer given in the book, the conversation shifts to a different topic about prime numbers and the prime number theorem.
  • #1
DeaconJohn
122
0
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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  • #2
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.
 
  • #3
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
 
  • #4
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
 
  • #5
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.
 

What is the Average Speed Paradox?

The Average Speed Paradox is a phenomenon that occurs when calculating the average speed of a moving object. It states that the average speed of an object over a given distance and time may not accurately represent the true speed of the object, as it does not account for any changes in speed during the journey.

How does the Average Speed Paradox affect our understanding of speed?

The Average Speed Paradox challenges our traditional understanding of speed as a constant value. It highlights the importance of considering changes in speed and the overall pattern of motion in determining the true speed of an object.

What are some real-life examples of the Average Speed Paradox?

One example of the Average Speed Paradox is rush hour traffic. Even if a car travels at a high speed on the highway, it may have a lower average speed due to slower speeds during congested periods. Another example is a runner who runs at different speeds throughout a race, but their average speed may not accurately reflect their performance.

How can the Average Speed Paradox be resolved?

To accurately determine the speed of an object, the Average Speed Paradox can be resolved by considering the entire journey and calculating the total distance traveled divided by the total time taken. This takes into account any changes in speed and provides a more accurate representation of the object's speed.

How does the Average Speed Paradox impact scientific research and experiments?

In scientific research and experiments, the Average Speed Paradox can lead to inaccurate results if not properly accounted for. It is important for scientists to understand the concept and apply it in their calculations to ensure the validity of their findings.

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