Math Is Fun Forum

Asakura

Member

Registered: 2007-12-07

Posts: 10

Help needed

This was in my maths exam today, I just counldn't get it...

A cyclist and a runner start off simultaneously around a racetrack each going at a constant speed in the same direction. The cyclist completes one lap and then catches up with the runner. Instantly the cyclist turns around and heads back at the same speed to the starting point where he meets the runner who has just finished his first lap. Find the ratio of their speeds.

Please help...

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help needed

It's the amazing golden ratio, my friend.
1.618034 or so.
Woops, I'm wrong, but I'll keep trying...

How about this:
B = (B+1)(1-B)/B  can't explain it easily.
B is the small speed.  B+1 is the fast speed.

So the answer is 2.4142136 or so.
Fast one is 1.707; Slow one is 0.707 in speed.

Last edited by John E. Franklin (2007-12-07 12:32:24)

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igloo myrtilles fourmis

Asakura

Member

Registered: 2007-12-07

Posts: 10

Re: Help needed

I don't quite understand how u got that.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help needed

How to explain this.
First I tried 2, then I tried 3 for the ratios.
This gives me a feeling for the problem.
I usually try answers before I tackle algebra.
Then from my scribbles, I look for
relationships.
One is that 2 revolutions minus where they meet equals the slow distance left to go after meeting.
So say they meet at .6 and 1.6 revs, then 2 - 1.6 = .4
Then jogger would have .4 left to go and 100 - 40 = 60, so
fast guy would go back .6  (these numbers are made up)
Now the (1-B) is like the 2 - 1.6, which is 1 - .6 or 1 - B.
Here are some ideas that morphed into the final equation.
2 < n < 3, where n is the ratio after trying 2 and 3.
n[1-B] = B
n[2 - (B + 1)] = B
You gotta just try numbers and think about
the small speed as being just the same as the small distance.
And think about the fast speed just as the number for the fast distance.
I got the answer, but explaining it isn't always my fort/e.

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help needed

Oh, and the equation I wrote had a ratio in it:
(B+1)/B
Let's take the answer as about 0.7 for where they meet.
Suppose 70% around the circle they meet, or 1.70710678 for the
fast guy. (1 over sqroot of 2).
B is distance in revolutions, also can think of as speeds if units are made to be right.  (units like revolutions per 10 minute period or something).
Now 100 - 70.7 = 29.3, so here's the #'s in equation.
B = 0.707  (distance slow guy went until met)
(B + 1) = 1.707 (distance fast guy went before met)
(1 - B) = 0.293 (distance slow guy has left to go after meet)
So here's the relationship:
B = ((B+1)/B) (1-B)  or with #'s:
.707 = (1.707/.707) (.293)
I thought, "the slow speed of .293 times the upgrade ratio to get to the fast speed equals the fast speed".
The upgrade ratio is 1.707/.707
Hope that helps.  Yesterday, my equation had the divide by B more to the right, so the ratio was not as evident.

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igloo myrtilles fourmis