Question 115995This question is from textbook
: I've been working on this one for two days. Any help is appreciated.
Girl is hiking and standing on the middle of a train bridge over a river. She is 300 ft from point P which is also the direction the train is approaching from. She is 500 ft from point Q in the opposite direction of the train's approach. Heather can reach point P at the same time as the train will reach point P while she is running 10mph. She can reach point Q at the same time as the train will reach point Q while she is running 10mph. How fast in mph is the train traveling?
I've tried various d=rt combinations, using the table format, but nothing is correct. I know that the girl can reach point P in 30 units of time or reach point Q in 50 units of time. The train can reach point Q in 50 units of time + the unknown amount of time it takes to reach the bridge. There is an unknown in the distance the train is from the bridge. I think I've looked at this too long.
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The train travels x ft while the girl travels 300 ft.
The Train travels x + 300 ft + 500 ft while the girls travels 500 ft.
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Info: 10mph = 10(mi/hr)(5280 ft/1 mi)(1 hr/60 min) = 880 ft/min
Time for the girl to arrive at P.
Distance = 300 ft ; Rate = 880 ft/min ; time = 300/880 = 0.341 minute
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Time for the girl to arrive at Q.
Distance = 500 ft ; rate = 880 ft/min ; time = 500/880 = 0.568 minute
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Train to P DATA:
Distance = x ft ; time = 0.341 min;
Rate = d/t = x/0.341 ft/min
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Train to Q DATA:
Distance = (800+x) ft ; time = 0.568; rate = (800+x)/0.568 ft.min
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EQUATION:
Train rate to P = train rate
x/0.341 = (800+x)/0.568
0.568x = 0.341*800 + 0.341x
0.227x = 272.8
x = 1201.762 ft
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Train Rate = x/0.341 = 1201.762/0.341 = 3524.229 ft/min
Info: 3524.229(ft/min)(1mi/5280 ft)(60 min/1 hr) = 40.048 mph
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Checking:
(800+1201.762)/0.568 = 3524.229 ft/min = 40.048 mph
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Whew!!! That was some problem.
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Cheers,
Stan H.
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