Question 179362
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Use the basic formula that relates distance, rate, and time.  *[tex \Large d = rt]


Both of the trains travelled the same distance, namely 60 miles, so *[tex \Large d = 60]


If we let the speed of the local train be <i>r</i>, and the express train travels twice as fast, the speed of the express train must be 2<i>r</i>.


If we let the elapsed time for the local train to make the trip be <i>t</i>, and the express train arrived two hours earlier, the elapsed time for the express train must be <i>t</i> - 2.


Now we can say, with respect to the local train's journey:  *[tex \Large 60 = rt] and with respect to the express train's journey: *[tex \Large 60 = 2r(t - 2)].


Since both of the right-hand expressions are equal to the same thing, namely 60, we can set them equal to each other:


*[tex \Large rt = 2r(t - 2) \rightarrow rt = 2rt - 4r \rightarrow -rt = -4r \rightarrow t = 4]


Telling us that the total time for the local train to make the 60 mile journey was 4 hours.


If *[tex \Large d = rt] then *[tex \Large r = \frac {d}{t} = \frac {60}{4} = 15]


Therefore the speed of the local train is 15 mph.  The speed of the express train is given as twice that, or 30 mph.


Hmm...I think I'll drive.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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