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=v1:v2  s=s1+s2  s1=v1 t s2=v2 t  s2=9v1 s1=4v2  4v2=v1 t 9v1=v2 t  9v14v2=v2v1 9v124v22=1 v12v22=49 v1v22=49 v1v2=49=23  r=v1v2 r=2:30.66670.6667=2:3r = 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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