Error in a project Eureka problem?

1. May 17, 2012

Bagbo

On project Eureka the following problem was posted:

You and a friend like running, but your friend is much faster than you. So you can run together, you two have arranged the following system. You both start out at the beginning of a path and run at constant speeds. Your friend runs to the end of the path, turns around, and then meets you along the path. Your friend then repeats this process two more times (running to the end of the path, turning around, and then meeting you). Your first meeting happens after one hour; the third meeting happens after two hours.
What portion of the path do you cover in the first hour?

The accepted solution 38.2 and a description was available from another site

The 'solution' is described as follows:

""Alas, no one got the solution.... Here's how I did the problem. Use units of length so that the length of the path is 1. Let s1 be your speed (in units per hour), while s2 is your friend's speed. During the first hour, both of you together run the path twice. So we know that s1 + s2 = 2 and that you are at position s1, leaving you a distance of 1 - s1 to the end of the path.

Between the first and second meeting, you and your friend will cover that distance twice. If that takes time t, we know that (s1 + s2) * t = 2t = 2(1 - s1). So the time to the second meeting is t = 1 - s1. You are now at position s1 + s1*(1 - s1), leaving a distance of 1 - 2s1 + (s1)2 to run.

Between the second and third meeting, you and your friend will cover that distance twice. Again, if that takes time t, we know that 2t = 2(1 - 2s1 + (s1)2). So the time to the third meeting is t = 1 - 2s1 + (s1)2. Thus the total time run so far is 1 (for the first meeting) plus 1 - s1 (for the second meeting) plus 1 - 2s1 + (s1)2 (for the third meeting). Setting this equal to 2 hours and solving gives us s1 = (3 - sqrt(5))/2 or about 38.2%.""

I don't think this is correct. There is no justification for claiming the (1 - s1) distance will be covered twice by the joggers between the first and second meeting. What will be covered by the joggers is 2s1 + 2s1*t hence t = 2s1(1 + t)/2 = s1(1 + t). Hence t = s1/(1 - s1). Since the second meeting must be between 1 and 2 hours and the joggers must meet a third time after 2 - 1 - t , I believe ANY s1 such that 0 < s1 < 0.5 is a valid answer including 38.2.

Is there something I am not seeing?

2. May 17, 2012

Diffy

It took me a while to figure out too. You being the slow runner neve make it to the end of the path.

Do you Accept in the first hour s1 + s2 = 2? The same concept applies to justify the 2*(1-s1) .

Try visualizing this with actual numbers. Say you run .25 units per hour. Then at the end of 1 hour you will have run .25 units, and your friend will have had to run 1.75 units. 1 unit to the end of the path and .75 units to get back to you.

Between the second and third meeting now your friend will have to only run .75 units to the end of the path, right? And then some amount of units to get back to you. Now he has to turn around to get back to you. When he meets you, he will have ran .75 minus the amount of distance you ran since the first meeting. Thus making it 2*.75 units that you both ran between the first and second meeting.

Do you see it?

I wish I could draw you a picture.

|____________________________|

3. May 17, 2012

Bagbo

With the numbers you give the runners will pass at 1 hour, 1.333 hours, and 2 hours, just as I said. ANY s1 between 0 and 0.5 will yield 3 meetings at 1 hour, 1/(1 - s1), and 2 hours respectively on a path of unit distance. With your numbers plugged in only twenty minutes pass between the first and second meetings, in that time the fast runner covers 1.75*.333 while you cover 0.25 * .333 which totals to 0.666 which is NOT equal to 2*(1 -0.25) = 1.5.

4. May 17, 2012

Bagbo

I see what I was doing wrong. I was misinterpreting the problem. I assumed he was passing each time saying "hello" and heading to opposite end of the path. It does not say he turns around when he meets you just that he meets you. and that he turns around when he reaches the end of the path.