SOLUTION: While traveling at 65 mph, a car passes a train from back to front in 2 minutes. At the same time, another car, also traveling at 65 mph, moves from the front to the back of the t
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Question 334273: While traveling at 65 mph, a car passes a train from back to front in 2 minutes. At the same time, another car, also traveling at 65 mph, moves from the front to the back of the train in 10 seconds. If the train travels at a constant speed, how long is the train?
I found the distance it took each car to pass the train in the given times using d=rt. The car moving from back to front takes 2 minutes, and therefore 2.1667 mi. The car moving from front to back takes 10 seconds, and therefore 0.1806 mi.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Well, not exactly...
You have forgotten to take into consideration that the train is moving as well as the cars. Think about it for a minute. If the train were stationary, then it wouldn't make any difference whether the car was going front to back or back to front -- it would take the same amount of time to go the length of the train either way.
What is happening is that since the train is moving at some unknown constant speed, the car that is going in the same direction as the train is covering the distance represented by the length of the train at a speed, not of 65 mph, but 65 mph MINUS the speed of the train. The other car is covering the length of the train in 65 mph PLUS the speed of the train.
Let 
be the length of the train and let 
represent the speed of the train.
For the car going back-to-front using 
:
With a bit of manipulation becomes:
The other car, using 
:
With a bit of manipulation becomes:
Now, since the speed of the train is a constant:
Just solve for , the length of the train in miles. My innate sense of neatness says that you should leave the answer in reduced fractional form.
John
My calculator said it, I believe it, that settles it
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