SOLUTION: A train left a railroad station in N.Y. at 9:00 AM bound for Washington, D.C. traveling at a uniform speed. At 10:30 AM another train left the same station bound for Washington D.C

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Question 371335: A train left a railroad station in N.Y. at 9:00 AM bound for Washington, D.C. traveling at a uniform speed. At 10:30 AM another train left the same station bound for Washington D.C. (on a different track) averaging 20 mph more than the first train and passed the first train 180 miles from N.Y. Find the speed of each train.
Answer by stanbon(75887) About Me  (Show Source):
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A train left a railroad station in N.Y. at 9:00 AM bound for Washington, D.C. traveling at a uniform speed. At 10:30 AM another train left the same station bound for Washington D.C. (on a different track) averaging 20 mph more than the first train and passed the first train 180 miles from N.Y. Find the speed of each train.
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1st train DATA:
distance = 180 mi ; time = x hrs ; rate = 180/x mph
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2nd train DATA:
distance = 180 mi ; time = (x-(3/2)) hrs ; rate = 180/(x-(3/2)) mph
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Equation:
2nd train rate - 1st train rate = 20 mph
180/(x-(3/2)) - 180/x = 20
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2/(2x-3) - 1/x = 1/9
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Multiply thru by 9x(2x-3) to get:
18x - 9(2x-3) = x(2x-3)
27 = 2x^2-3x
2x^2 - 3x - 27 = 0
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x = 4.5 hrs
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1st train rate = 180/4.5 = 40 mph
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2nd train rate = 180/(4.5-1.5) = 180/3 = 60 mph
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Cheers,
Stan H.