Train Speed Ratio 4Ad374

1. **Problem Statement:** We are given two trains with speeds $v_1$ and $v_2$ where $v_1 > v_2$. The ratio of their speeds is equal to the ratio of the time taken to pass each other when moving in the same direction to the time taken when moving in opposite directions. 2. **Define variables:** Let the time taken to pass each other when moving in the same direction be $t_s$ and when moving in opposite directions be $t_o$. 3. **Formula for passing time:** - When trains move in the same direction, relative speed = $v_1 - v_2$. - When trains move in opposite directions, relative speed = $v_1 + v_2$. If the length of the trains combined is $L$, then: $$t_s = \frac{L}{v_1 - v_2}$$ $$t_o = \frac{L}{v_1 + v_2}$$ 4. **Given ratio:** $$\frac{v_1}{v_2} = \frac{t_s}{t_o}$$ Substitute $t_s$ and $t_o$: $$\frac{v_1}{v_2} = \frac{\frac{L}{v_1 - v_2}}{\frac{L}{v_1 + v_2}} = \frac{v_1 + v_2}{v_1 - v_2}$$ 5. **Let** $x = \frac{v_1}{v_2}$, then: $$x = \frac{x + 1}{x - 1}$$ 6. **Solve for** $x$: Multiply both sides by $(x - 1)$: $$x(x - 1) = x + 1$$ $$x^2 - x = x + 1$$ Bring all terms to one side: $$x^2 - x - x - 1 = 0$$ $$x^2 - 2x - 1 = 0$$ 7. **Use quadratic formula:** $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2}$$ Simplify: $$x = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$ 8. **Since** $x = \frac{v_1}{v_2} > 1$, take the positive root: $$x = 1 + \sqrt{2}$$ **Final answer:** The ratio of the speed of the faster train to the slower train is: $$\boxed{1 + \sqrt{2}}$$