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

Answer:

$\Delta d =\frac{y-x}{y}*d_{Hs}$

Step-by-step explanation:

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

$V_{Hs} =\frac{z}{x}$

And the velocity of the regular will be given by:

$V_{R} =\frac{z}{y}$

The position equations of each movement will be given by:

$d_{Hs} =\frac{z}{x} t$

$d_{R} =z-\frac{z}{y} t$

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

First isolating:

$d_{Hs}\frac{x}{z} = t$

Second reeplacing t:

$d_{R} =z-\frac{z}{y}*d_{Hs}\frac{x}{z}$

$d_{R}=z-\frac{x}{y}*d_{Hs}$

In the  moment of the encounter the high-speed train have gone round  $d_{Hs}$ and the regular one $\frac{x}{y}*d_{Hs}$

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

$\Delta d =d_{Hs}-\frac{x}{y}*d_{Hs}$

$\Delta d =\frac{y-x}{y}*d_{Hs}$