SOLUTION: "It took a faster runner 10 seconds longer to run a distance of 1500 feet than it took a slower runner to run a distance of 1000 feet. if the rate of the faster runner was 5 ft/s m

Question 244764: "It took a faster runner 10 seconds longer to run a distance of 1500 feet than it took a slower runner to run a distance of 1000 feet. if the rate of the faster runner was 5 ft/s more than the rate of the slower runner, what was the rate of each?"
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
"It took a faster runner 10 seconds longer to run a distance of 1500 feet than it took a slower runner to run a distance of 1000 feet.
Let t = time in sec for the faster runner to run 1500 ft
then
(t-10) = time of the slower runner to run 1000 ft
:
if the rate of the faster runner was 5 ft/s more than the rate of the slower runner, what was the rate of each?"
:
Write a rate equation: Rate = dist/time:
:
Fast rate - slow rate = 5 ft/sec
- = 5
:
Multiply equation by t(t-10), results
1500(t-10) - 1000t = 5t(t-10)
:
1500t - 15000 - 1000t = 5t^2 - 50t
:
500t - 15000 = 5t^2 - 50t
:
0 = 5t^2 - 50t - 500t + 15000
A quadratic equation
5t^2 - 550t + 15000 = 0
Simplify, divide by 5
t^2 - 110t + 3000 = 0
Factor to
(t - 50)(t - 60) = 0
two solutions
t = 50 sec
t = 60 sec
:
Assume t = 50
1500/50 = 30 ft/sec for the fast guy
and
1000/40 = 25 ft/sec for the slow guy (5 ft/sec slower)
:
You can check the other solution
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