1. Jul 1, 2008

### DeaconJohn

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 travelled 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!

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!)

Last edited: Jul 1, 2008
2. Jul 1, 2008

### Staff: Mentor

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. Jul 1, 2008

### DeaconJohn

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. Jul 1, 2008

### DeaconJohn

My Mistake!

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. Jul 2, 2008

### Staff: Mentor

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 - thats me 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.