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Who Will Win the Race?

Date: 01/14/98 at 23:13:48
From: casey smith
Subject: Problem solving

Tortisha and Harry were arguing about who could run the fastest.  
To end the dispute, they decied to have a race. Harry was so sure 
that he was faster than Tortisha that he said, "I'll give you a 
2-mile head start and still beat you in the race." If Harry runs 
3/4 mile every 6 minutes and Tortisha runs 1/2 mile every 5 minutes, 
who will win the race?

Date: 01/15/98 at 10:41:49
From: Doctor Jerry
Subject: Re: Problem solving

Hi Casey,

Imagine an x-axis and a clock. Assume that the runners head towards 
the positive side, starting from the origin. The clock will start 
at t = 0 at the time both start to run. Tortisha's velocity is 
(1/2)/5 = 0.1 mile/minute and Harry's velocity is (3/4)/6 = 1/8 = 
0.125 mile/minute. 

After t minutes, Tortisha will be at

   T = 2+0.1*t

while Harry will be at

   H = 0.125*t

To answer the question, we must know how long the race is.  But, 
meanwhile, we can ask for the time when H=T.  That would be

    2+0.1*t = 0.125*t 
    0.025*t = 2 or

t = 2/0.025 = 80.

So, with a 2 mile headstart, Harry will catch up with Tortisha after 
80 minutes. Tortisha will have run 2+0.1*80 = 2+8 = 10 miles by then.

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Date: 01/15/98 at 11:07:12
From: Doctor Pipe
Subject: Re: Problem solving


Since we don't know the distance of the race (4 miles, 6 miles, etc.) 
we can't say who will win the race.  

But, we can figure out the distance for a race that would end in a tie 
and then say that races shorter than this would be won by Tortisha and 
races longer than this would be won by Harry.

If we let t be the number of minutes and s be the distance in miles 
than we can use the information in the problem to calculate how far 
Tortisha is from the start after t minutes and how far Harry is from 
the start after t minutes.

First, how long does it take each of them to run a mile?  If Tortisha 
runs 1/2 of a mile every 5 minutes, then she runs a mile every 10 
minutes. If Harry runs 3/4 of a mile every 6 minutes, then he runs a 
mile every 8 minutes.

For Tortisha, the distance, s, from the start after t minutes is 
(don't forget, Tortisha gets a 2 mile head start):

     s = 2 + (t / 10)

For Harry,  the distance, s, from the start after t minutes is:

     s = t / 8

To calculate the amount of time for the two of them to arrive at the 
same distance from the start (or, how long will it take for Harry to 
make up the two mile head start that he is giving Tortisha) we create 
a new equation using the righthand sides of each of the two previous 
equations and then solve for t:

            2 + (t / 10) = t / 8

     40 x (2 + (t / 10)) = 40 x (t / 8)

         80 + (40t / 10) = 40t / 8

                 80 + 4t = 5t

            80 + 4t - 4t = 5t - 4t

                      80 = t

This tells us that after running for 80 minutes (whew!), both Harry 
and Tortisha will reach the same spot. Let's check this by plugging 
t = 80 into the first two equations and verify that they both give the 
same distance.

For Tortisha:

     s = 2 + (t / 10)

     s = 2 + (80 / 10)

     s = 2 + 8

     s = 10

For Harry:

     s = t / 8

     s = 80 / 8

     s = 10

So, if the race is a ten mile race, then it will end in a tie. 
Tortisha will have a two mile head start and require the 80 minutes 
to run the additional eight miles. Harry will require 80 minutes to 
run 10 miles.

If the race is shorter than 10 miles, Harry will not be able to catch 
up and Tortisha will win!  If the race is longer than 10 miles, Harry 
will pass Tortisha at the ten mile mark and go on to win!

See you at the finish line!

-Doctor Pipe,  The Math Forum
 Check out our web site!   

Date: 01/15/98 at 18:59:16
From: Drako1027
Subject: Thank you!

Thank you Dr. Math, you really helped me get that problem done.  
I thought that I was never going to understand how to figure it out.  
I thank you again and hope to hear from you soon the next time I need 
your help.
Associated Topics:
High School Basic Algebra

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