SOLUTION: A freight train heads north at 9a.m. at 40mph.2 hr. later a express train heads north at 60mph. What time is it when express is 20mi further from town than the freight?
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Rsub
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-> SOLUTION: A freight train heads north at 9a.m. at 40mph.2 hr. later a express train heads north at 60mph. What time is it when express is 20mi further from town than the freight?
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Question 139437: A freight train heads north at 9a.m. at 40mph.2 hr. later a express train heads north at 60mph. What time is it when express is 20mi further from town than the freight?
Work
RsubF(TsubF)+20=RsubE(TsubE) RsubF=40
RsubE=60
TsubF=TsubE-2
I got wrong answer and is this equation set up right? Answer by frostusna(7) (Show Source):
You can put this solution on YOUR website! Here's the easiest way for me to do the problem.
You know that by 11am the freight train has gone 80 miles.
With this information I would set up two equations, one for each train, using the distance = rate x time equations, keeping in mind that at a certain time t, which is the same time for each train, the express train will be 20 miles further from the town than the freight train.
Let the distance the express train travels starting at 11am equal De, then you get:
Let the distance the freight train travels starting at 11am be Df, then you get:
However, you also know that the express train at time t will be 20 miles further from the town than the freight train and the freight train got an 80 mile head start, so during time t the express train has to travel an additional
100 miles. In equation form that would be .
Substitute in for De and Df their respective rate-time values and you get the equation:
or or .
Starting at 11am and moving ahead for 5 hours makes it 4pm. At that time the express train will be or .
At the same time the freight train will be or .
The express train is 20 miles further from town.
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