# Trolleybus

Trolleybus line No. 206 measured 24 km. If the trolleybus goes faster by 5 km/h, the way there and back would is shorter by 33 minutes. Calculate the trolleybus speed and how much time it takes a return trip.

Result

Trolleybus speed is:  18.5 km/h
The return trip took:  155.4 min

#### Solution:

$s = 24 \ km \ \\ 33 \ min = 0.55 \ h \ \\ 2s = vt \ \\ 2s = (v+5)(t-0.55) \ \\ vt = vt +5 t - 0.55 v - 2.75 \ \\ 0 = 5 t - 0.55 v - 2.75 \ \\ 0 = 5 t^2 - 0.55 \cdot 2 \cdot 24 - 2.75 t \ \\ 0 = 5 t^2 - 2.75 t - 26.4 \ \\ t_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 2.75 \pm \sqrt{ 535.56 } }{ 10 } \ \\ t_{1,2} = 0.275 \pm 2.31422233158 \ \\ t_{1} = 2.58922233158 \ \\ t_{2} = -2.03922233158 \ \\ t>0; t= t_1 = 155.4 \ \text { min } \ \\ \ \\ v = \dfrac{ 2 \cdot 24 }{ t_1 } = 18.54 \ km/h$

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