# Trains

From station 130 km away started passenger train and after 2.2 hours after the express train, which travels 37 km an hour more. Express train finish journey 7 minutes early.

Calculate the average speed of this two trains.

Correct result:

v1 =  185.4 km/h
v2 =  222.4 km/h

#### Solution:

v_1t_1 = v_2t_2 = 130 \ \\ v_2 = v_1 + 37 \ \\ t_2 = t_1 -7/60 \ \\ \ \\ v_1t_1 = 130 \ \\ v_1t_1 - 0.117 v_1 + 37 t_1 - 37 \cdot 0.117 = 130 \ \\ \ \\ t_1 = 0.003 v_1 + 0.117 \ \\ \ \\ v_1 ( 0.003 v_1 + 0.117) = 130 \ \\ \ \\ 0.0031531531531532v_1^2 +0.117v_1 -130 =0 \ \\ \ \\ a=0.0031531531531532; b=0.117; c=-130 \ \\ D = b^2 - 4ac = 0.117^2 - 4\cdot 0.0031531531531532 \cdot (-130) = 1.6532507508 \ \\ D>0 \ \\ \ \\ v_1_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ -0.12 \pm \sqrt{ 1.65 } }{ 0.0063063063063063 } \ \\ v_1_{1,2} = -18.5 \pm 203.88923813819 \ \\ v_1_{1} = 185.38923813819 \ \\ v_1_{2} = -222.38923813819 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 0.0031531531531532 (v_1 -185.38923813819) (v_1 +222.38923813819) = 0 \ \\ \ \\ v_1 = 185.4 \ \text{km/h}
$v_2 = v_1 + 37 = 222.4 \ \text{km/h}$

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