Tricky train problem involving ratios

**Intsecxtanx** · April 22nd 2010, 11:25 AM

The ratio of the speeds of two trains is equal to the ratio of the time they take to pass each other going in the same direction to the time they take to pass each other in the opposite directions. how can you find the ratio of the speeds of the two trains?

**Soroban** · April 22nd 2010, 05:32 PM

Hello, Intsecxtanx!

The ratio of the speeds of two trains is equal to
the ratio of the time they take to pass each other going in the same direction
to the time they take to pass each other in the opposite directions.
Find the ratio of the speeds of the two trains.

Let: . $\begin{Bmatrix} F &=& \text{speed of faster train} \\<br /> L_1 &=& \text{length of faster train} \end{Bmatrix} \qquad \begin{Bmatrix} F &=& \text{speed of slower train} \\ L_2 &=& \text{length of slower train} \end{Bmatrix}$

Going in the same direction, the faster train has a relative speed of $F \!-\! S$ ft/sec
. . and it travels a distance of $L_1\!+\!L_2$ feet.

This will take: . $T_{\text{same}} \;=\;\frac{L_1+L_2}{F-S}$ seconds.

Going in opposite directions, the faster train has a relative speed of $F\!+\!S$ ft/sec
. . and it travels a distance of $L_1\!+\!L_2$ feet.

This will take: . $T_{\text{opp}} \;=\;\frac{L_1+L_2}{F+S}$ seconds.

The ratio of their speeds equals the ratio of their two times:

. . $\frac{F}{S} \;=\;\frac{\dfrac{L_1+L+2}{F-S}}{\dfrac{L_1+L_2}{F+S}} \quad\Rightarrow\quad \frac{F}{S} \:=\:\frac{F+S}{F-S}$

. . . . $F^2 - FS \:=\:FS + S^2 \quad\Rightarrow\quad F^2 - 2SF - S^2 \:=\:0$

Divide by $S^2\!:\;\;\frac{F^2}{S^2} - 2\,\frac{F}{S} - 1 \:=\:0 \quad\Rightarrow\quad \left(\frac{F}{S}\right)^2 - 2\left(\frac{F}{S}\right) - 1 \:=\:0$

Quadratic Formula: . $\frac{F}{S} \;=\;\frac{2\pm\sqrt{4+4}}{2} \;=\;1 \pm \sqrt{2}$

The ratio of their speeds is: . $F:S \;=\;(1\!+\!\sqrt{2}) : 1$

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