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, 08:22 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, 08:52 PMMath Major
You need to use the equation distance = rate * time.

We know that

and

Solve for r. that will be the rate of runner 2. - Aug 1st 2010, 11:08 PMaeroflix
your working equation is wrong. . t = 790 which is too large.

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

- Aug 1st 2010, 11:29 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, 11:37 PMMath Major
I'll even set it up for you.

Solve for t

Plug this value in for t in equation 1

Foil it

Combine like terms

Multiply through by an r

Divide by 10

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, 11:59 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 2nd 2010, 12:01 AMaeroflix
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 2nd 2010, 12:03 AMMath Major
Given

Multiply by the common denominator

Multiply the expressions out

Combine like terms

It should look familiar from here. - Aug 2nd 2010, 12:05 AMMath Major