SOLUTION: A TRAIN LEAVES D.C. AT NOON TRAVELING NORTH AT 44 MPH. ONE HOUR LATER, ANOTHER TRAIN GOING 52 MPH TRAVELS NORTH ON A PARALLEL TRACK. HOW MANY HOURS WILL THE SECOND TRAIN TRAVEL B

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Question 312688: A TRAIN LEAVES D.C. AT NOON TRAVELING NORTH AT 44 MPH. ONE HOUR LATER, ANOTHER TRAIN GOING 52 MPH TRAVELS NORTH ON A PARALLEL TRACK. HOW MANY HOURS WILL THE SECOND TRAIN TRAVEL BEFORE IT OVERTAKES THE FIRST TRAIN?
Found 2 solutions by mananth, nyc_function:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
train A leaves 12 noon at 44 mph.
Train B leaves at 1 pm at 52 mph
..
Before train B leaves DC train A has traveled 44*1 = 44 miles.
From this point let train A travel x miles before they meet.
..
So train B travels x+44 miles to catch up with train.
..
time taken by train A = x/44
time taken by train B = x+44 / 52
..
x/44 = x+44 / 52
52x =44*(x+44)
52x=44x +1936
8x=1936
x= 242 miles.
Train B has to travel 242+44=286 miles to catch up.
its speed is 52 mph
time taken = 286/52
=5.5 hours

Answer by nyc_function(2741)   (Show Source): You can put this solution on YOUR website!
Here is your equation:
44(x + 1) = 52x
Can you take it from here?

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