Cycling Motion Problem

Date: 6/26/96 at 19:3:13
From: Anna Booth
Subject: Motion Problem

Dear Dr. Math,

Please can you help me?

A cyclist travels from A to B over a high pass, C. His average speed 
going up the hill is 9 km/h; he goes down the hill at 24km/h. It takes 
him one and a quarter hours to go from A to B but one and a half hours 
to returnfrom B to A.  What is the distance from A to B?

If x is the time he takes to go from A to C and y is the time from C 
to B, he travels a distance of 9x + 24y and  x + y = 5/4 hours. 

What do I do now? I don't know how to take into account the return 
trip taking longer.

Julia Booth

Date: 6/28/96 at 8:52:30
From: Doctor Mike
Subject: Re: Motion Problem

Hello,

You set this up exactly right and you are almost there.
It seems you need to assume (reasonable?) that the cyclist's
up-hill and down-hill speeds are independent of the direction
of travel.

The cyclist travels the 24y km from B back up to C in
(24y)/9 hours, and the 9x km from C back down to A in (9x)/24
hours.  So, (24y)/9 + (9x)/24  equals the return trip time
of 3/2 hours.  You can do the rest!  The distance is 18 km.
  
Hope this is what you need.  
 
-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/