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

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) (Show Source): You can put this solution on YOUR website!
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.

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