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 !!!
But I don't quite understand the case for 3 walkers ( ). I'll think about it... See http://www.mathhelpforum.com/math-he...rd-183226.html for new posts
least time takes about 4.5 hours
a question for you too at http://www.mathhelpforum.com/math-he...rd-183226.html