Question

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?

Answer 1

`\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)\`