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

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- Aug 1st 2010, 07:22 PMaeroflixrunner
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 PMMath 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 PMaeroflix
your working equation is wrong. . t = 790 which is too large.

- Aug 1st 2010, 10:11 PMaeroflix
i guess this is by far the most complex problem ever

- Aug 1st 2010, 10:29 PMMath 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 PMMath 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 PMaeroflix
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 PMaeroflix
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 PMMath 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 PMMath Major