# runner

• Aug 1st 2010, 07:22 PM
aeroflix
runner
it took a faster runner 10 seconds longer to run a distance of 1500 ft. Then it took a slower runner to run 1000 ft. If the rate of faster runner was 5 feet per second more than the slower runner, what was the rate of each runner
• Aug 1st 2010, 07:52 PM
Math Major
You need to use the equation distance = rate * time.

We know that $\displaystyle 1500 = (r + 5)(t + 10)$
and $\displaystyle 1000 = r * t$

Solve for r. that will be the rate of runner 2.
• Aug 1st 2010, 10:08 PM
aeroflix
your working equation is wrong. . t = 790 which is too large.
• Aug 1st 2010, 10:11 PM
aeroflix
i guess this is by far the most complex problem ever
• Aug 1st 2010, 10:29 PM
Math Major
Show me your work. You should get the rate of runner 1 as 25 feet per second and the rate of runner 2 as 20 feet per second. t solves out to be 50, with runner 1 taking 60 seconds.
• Aug 1st 2010, 10:37 PM
Math Major
I'll even set it up for you.

$\displaystyle 1000 = r * t$

Solve for t

$\displaystyle t = \frac{1000}{r}$

Plug this value in for t in equation 1

$\displaystyle 1500 = (r + 5)(\frac{1000}{r} + 10)$

Foil it

$\displaystyle 1500 = 1000 + 10r + \frac{5000}{r} + 50$

Combine like terms

$\displaystyle 450 = 10r + \frac{5000}{r}$

Multiply through by an r

$\displaystyle 0 = 10r^2 - 450r + 5000$

Divide by 10

$\displaystyle r^2 - 45r + 500 = 0$

Solve the quadratic. You will get two rates. One will be the rate for runner 1. The other will be the rater for runner 2.
• Aug 1st 2010, 10:59 PM
aeroflix
25 and 20??? is that how easy it is???? but why does my formula doesnt work the way it should be???
distace rate time
faster 1500 ft x+5 1500/x+5
slower 1000 ft X 1000 / x

equation: distance of faster is equal to distance travveled by slower:

1500/ x +5 = (1000 / x) + 10

plus 10 to make the slower runner distance 1500 feet also... why my equation doesnt work?? but its correct right?
• Aug 1st 2010, 11:01 PM
aeroflix
also why is 10 needed to add to faster runner? isnt it is supposed to be added to the slower runnner?? i need good analysis. pls help
• Aug 1st 2010, 11:03 PM
Math Major
Given $\displaystyle \frac{1500}{x + 5} = \frac{1000}{x} + 10$

Multiply by the common denominator $\displaystyle (x)(x+5)$

$\displaystyle 1500x = 1000(x+5) + 10(x)(x+5)$

Multiply the expressions out

$\displaystyle 1500x = 1000x + 5000 + 10(x^2 + 5x)$

$\displaystyle 1500x = 1000x + 5000 + 10x^2 + 50x$

Combine like terms

$\displaystyle 450x = 10x^2 + 5000$

It should look familiar from here.
• Aug 1st 2010, 11:05 PM
Math Major
Quote:

Originally Posted by aeroflix
also why is 10 needed to add to faster runner? isnt it is supposed to be added to the slower runnner?? i need good analysis. pls help

Reread your problem statement. You said that it took the faster runner 10 seconds longer to run 1500 feet than it took the slower runner to run 1000 feet. The faster runner's time is 10 seconds longer.