SOLUTION: 1. At 5:00 a.m., a train leaves a station and travels at a rate of 35 mph. At 7:00 a.m., a second train leaves the same station on the same track and travels in the direction of t

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Question 174165: 1. At 5:00 a.m., a train leaves a station and travels at a rate of 35 mph. At 7:00 a.m., a second train leaves the same station on the same track and travels in the direction of the first train at a speed of 45 mph. At 8:00 a.m., how many miles seperate the two trains?
solve:
2. 1/3x-2=5
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For both trains, the same formula applies, but I'll write it
differently for each train so I keep track of each train's data
d+=+r%2At
d=distance in miles
r= speed in mi/hr
t= elapsed time in hrs
For train a:
d%5Ba%5D+=+r%5Ba%5D%2At%5Ba%5D
For train b:
d%5Bb%5D+=+r%5Bb%5D%2At%5Bb%5D
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Given:
r%5Ba%5D+=+35 mi/hr
r%5Bb%5D+=+45 mi/hr
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I want to have
the start time for both trains the same, at 7 AM,
so I want to know how far train a has gone between
5 AM and 7 AM
d+=+35%2A2
d+=+70 mi
Now the trains will both travel for 1 hr, 7 AM to 8 AM
d%5Ba%5D+=+r%5Ba%5D%2At%5Ba%5D
d%5Ba%5D+=+35%2A1
d%5Ba%5D+=+35mi
------------
d%5Bb%5D+=+r%5Bb%5D%2At%5Bb%5D
s%5Bb%5D+=+45%2A1
d%5Bb%5D+=+45mi
------------
d%5Bb%5D+-+d%5Ba%5D+=+45+-+35
d%5Bb%5D+-+d%5Ba%5D+=+10mi
So, train b has gained 10 mi on train a, but train a
already had a 70 mi head start on train b
70+-+10+=+60
Train a is 60 mi ahead of train b
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check answer:
For train a:
d%5Ba%5D+=+r%5Ba%5D%2At%5Ba%5D
d%5Ba%5D+=+35%2A3
d%5Ba%5D+=+105mi
For train b:
d%5Bb%5D+=+r%5Bb%5D%2At%5Bb%5D
d%5Bb%5D+=+45%2A1
d%5Bb%5D+=+45mi
The difference is 60 mi