Thread: 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

You need to use the equation distance = rate * time.

We know that
and

Solve for r. that will be the rate of runner 2.

your working equation is wrong. . t = 790 which is too large.

i guess this is by far the most complex problem ever

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.

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.

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?

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

Given

Multiply by the common denominator



Multiply the expressions out





Combine like terms



It should look familiar from here.

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.