Question 67706: A train leaves a station at 9:00 A.M. traveling at a constant rate of speed. Two hours later a second train, traveling at an average rate of 25 miles per hour more than the first train, leaves the same station, going in the same direction, on a parallel track. If the faster train overtakes the first train at 3:00 P.M. on the same day, what is the average rate of speed of the first train.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A train leaves a station at 9:00 A.M. traveling at a constant rate of speed. Two hours later a second train, traveling at an average rate of 25 miles per hour more than the first train, leaves the same station, going in the same direction, on a parallel track. If the faster train overtakes the first train at 3:00 P.M. on the same day, what is the average rate of speed of the first train.
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If train two overtakes train one at 3:00 train two must have traveled
from 11:00 to 3:00 or 4 hours. Train one left two hours earlier so
it must have traveled for 6 hours.
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Train one DATA:
Time = 6 hours ; Rate = x mph; Distance = r*t = 6x miles
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Train two DATA:
Time = 4 hurs ; Rate = x+25 mph ; Distance = r*t = 4(x+25)
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EQUATION:
distance = distance
6x=4(x+25)
6x=4x+100
2x=100
x=50 mph (speed of 1st train)
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Cheers,
Stan H.
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