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It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

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Answer:


\Delta d =(y-x)/(y)*d_(Hs)\

Explanation:

Fist the velocity of high-speed train will be given by:


V_(Hs) =(z)/(x)

And the velocity of the regular will be given by:


V_(R) =(z)/(y)

The position equations of each movement will be given by:


d_(Hs) =(z)/(x) t


d_(R) =z-(z)/(y) t

We can get the encounter point isolating t in the first equation and reepalcing it in the second one:

First isolating:


d_(Hs)(x)/(z) = t

Second reeplacing t:


d_(R) =z-(z)/(y)*d_(Hs)(x)/(z)


d_(R)=z-(x)/(y)*d_(Hs)\

In the moment of the encounter the high-speed train have gone round
d_(Hs)\ and the regular one
(x)/(y)*d_(Hs)\

Then the difference between the distance traveled the high-speed train and the regular one is:


\Delta d =d_(Hs)-(x)/(y)*d_(Hs)\


\Delta d =(y-x)/(y)*d_(Hs)\

User Gregy
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