Who Will Win the Race?

Date: 11/26/2002 at 23:19:43
From: Sharonda Randall
Subject: Who will win the race and by how many meters?

Mike and Tim race for 100 meters and Tim wins by 10 meters. When they 
race again, Tim begins 10 meters behind the orginal starting line.  

If they both run the race at the same constant speed as the first 
time, who will win the race and by how many meters?

Date: 11/27/2002 at 21:28:02
From: Doctor Paul
Subject: Re: Who will win the race and by how many meters?

Suppose it takes Tim t seconds to run 100 meters.  Then it takes 
Mike t seconds to run 90 meters.

So Tim runs 100/t meters per second while Mike runs 90/t meters per 
second.

We need to answer the following question:

Which is longer - the time it takes Tim to run 110 meters or the time 
it takes Mike to run 100 meters?

We know that distance = rate * time, so time = distance / rate

Thus the time it takes Tim to run 110 meters is given by:

   110 / (100/t) = 11/10 * t = 1.1 * t

and

the time it takes Mike to run 100 meters is given by:

   100 / (90/t) = 10/9 * t = 1.11111111111... * t

Inasmuch as 1.111111... (repeating) is greater than 1.1, it takes Mike 
longer to run 100 meters than it takes Tim to run 110 meters. Thus Tim 
wins the race.

In order to determine the number of meters by which Tim wins, we need 
to know how far Mike ran in 1.1 * t seconds.

We know that Tim ran 110 meters in 1.1 * t seconds. If we know how far 
Mike has run when Tim crosses the finish, we can subtract Mike's 
distance from 110 to figure out how far ahead Tim is when he wins.

Again we use distance = rate * time.

Mike's distance = (90/t) * (1.1 * t) = 90 * 1.1 = 99 meters

So if they had started at the same point, Tim would be ahead by 
110 - 99 = 11 meters after 1.1 * t seconds. But remember that Tim 
started ten meters behind Mike. So Tim won by one meter.

This is a neat problem. Notice that we don't need to know how long it 
took Tim to run the 100-meter race to solve the problem.

I hope this helps.  Please write back if you'd like to talk about 
this some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/