SOLUTION: A sprinter who averaged 30ft/sec completed a race one second before another sprinter who averaged 28 ft/sec. what was the distance of the race?

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Question 93075: A sprinter who averaged 30ft/sec completed a race one second before another sprinter who averaged 28 ft/sec. what was the distance of the race?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A sprinter who averaged 30ft/sec completed a race one second before another sprinter who averaged 28 ft/sec. what was the distance of the race?
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1st sprinter DATA:
Rate = 30 ft/sec ; Time = x seconds ; distance = rt = 30x ft.
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2nd sprinter DATA:
Rate = 28 ft/sec ; Time = x+1 seconds ; diatance = rt = 28(x+1) = 28x + 28 ft.
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EQUATION:
distance = distance
30x = 28x + 28
2x = 28
x = 14 seconds
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Distance of the race = 30x = 30*14 = 420 ft.
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Cheers,
Stan H.