Question 334273
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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 *[tex \Large l] be the length of the train and let *[tex \Large r_t] represent the speed of the train.


For the car going back-to-front using *[tex \Large 2\,min\ =\ \frac{1}{30}\,hr]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{65\ -\ r_t}{30}]


With a bit of manipulation becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 65\ -\ 30l\ =\ r_t]


The other car, using *[tex \Large 10\,sec\ =\ \frac{1}{360}\,hr]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{65\ +\ r_t}{360}]


With a bit of manipulation becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 360l\ -\ 65\ =\ r_t]


Now, since the speed of the train is a constant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 360l\ -\ 65\ \ =\ 65\ -\ 30l]


Just solve for *[tex \Large l], the length of the train in miles. My innate sense of neatness says that you should leave the answer in reduced fractional form.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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