# Two trains

Two trains departed from City A and City B against each other. They met after some time. The first train then took 9 hours to reach city B, and the second train took 4 hours to reach city A. In what proportion were the train speeds?

Result

r =  2:3

#### Solution:

$r = v_{ 1 }:v_{ 2 } \ \\ \ \\ s = s_{ 1 }+s_{ 2 } \ \\ \ \\ s_{ 1 } = v_{ 1 } \cdot \ t \ \\ s_{ 2 } = v_{ 2 } \cdot \ t \ \\ \ \\ s_{ 2 } = 9v_{ 1 } \ \\ s_{ 1 } = 4v_{ 2 } \ \\ \ \\ 4v_{ 2 } = v_{ 1 } \ t \ \\ 9v_{ 1 } = v_{ 2 } \ t \ \\ \ \\ \dfrac{ 9v_{ 1 } }{ 4v_{ 2 } } = \dfrac{ v_{ 2 } }{ v_{ 1 } } \ \\ \dfrac{ 9v_{ 1 }^2 }{ 4v_{ 2 }^2 } = 1 \ \\ \dfrac{ v_{ 1 }^2 }{ v_{ 2 }^2 } = \dfrac{ 4 }{ 9 } \ \\ \dfrac{ v_{ 1 } }{ v_{ 2 } } ^2 = \dfrac{ 4 }{ 9 } \ \\ \dfrac{ v_{ 1 } }{ v_{ 2 } } = \sqrt{ \dfrac{ 4 }{ 9 } } = \dfrac{ 2 }{ 3 } \ \\ \ \\ r = \dfrac{ v_{ 1 } }{ v_{ 2 } } \ \\ r = 2:3 \doteq 0.6667≈ 0.6667 = 2:3$

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