# Thread: Motorcycle and Walking Rates and Times

1. ## Re: Motorcycle and Walking Rates and Times

Originally Posted by Wilmer
After M drives A back, M then drops A off, then M goes forward to pick up C.
But A will also go forward : both leave together going forward, and M will never
be able to catch up, since speed slower....
Right! So m must be quicker than walkers.

Originally Posted by Wilmer
Remark on case where a walker is driven backward:
it is possible to get a solution where a walker is driven back BEHIND (West of) starting point X.
The wording of your problem does not prevent that.
Indeed...

2. ## Re: Motorcycle and Walking Rates and Times

Originally Posted by godelproof
Right! So m must be quicker than walkers.
NO...only in certain cases...

You can have m < both walkers; like m=10, a=30, b=20 ; with distance k=400:
M meets A after 10 hours; drives A back for 5 hours : A now has 150 to go.
A takes 5 hours to travel this 150: so total of 10 + 5 + 5 = 20 hours.

Meanwhile, B happily travels the 400 distance in 20 hours; so Time simply k/b.

The time that A is on that slow Motorcycle: k(a - b) / [b(a + m)]

3. ## Re: Motorcycle and Walking Rates and Times

Originally Posted by Wilmer
NO...only in certain cases...

You can have m < both walkers; like m=10, a=30, b=20 ; with distance k=400:
M meets A after 10 hours; drives A back for 5 hours : A now has 150 to go.
A takes 5 hours to travel this 150: so total of 10 + 5 + 5 = 20 hours.

Meanwhile, B happily travels the 400 distance in 20 hours; so Time simply k/b.

The time that A is on that slow Motorcycle: k(a - b) / [b(a + m)]
Of course!!! When you have only 2 walkers anything will go! That's what I meant by ${B}^{4}={\mathbb{R}}_{++}^{4}$!!!

But I don't quite understand the case for 3 walkers ( ${B}^{5}$). I'll think about it... See http://www.mathhelpforum.com/math-he...rd-183226.html for new posts

4. ## Re: Motorcycle and Walking Rates and Times

Originally Posted by bjhopper
Hello.
I have followed this post from the beginning.I made calculations to see if I could estimate the time for the walkers to reach the other end.My inital estimate was 4 hours I started by sending M to pick up the slowest walker and to drop him at a point where the time to do the same for the others would keep him from crossing the finish line.Rate of walkers .1 K per minute,.133, .166. Rate of M 2K per minute.It takes M to meet a 57 minutes.They approach each other at 2.1 K per minute. The trip back to the 90 mile mark takes 47 minute. The pick up point was at mile 5.7. Picking up the others took a total of 247 minutes. I can see that the drop offs should have been closer to the end.Final back and forth movements ofM to get walkers in correct positions would take little time.

bjh
least time takes about 4.5 hours

a question for you too at http://www.mathhelpforum.com/math-he...rd-183226.html

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