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.


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


s=24 km 33 min=0.55 h 2s=vt 2s=(v+5)(t0.55) vt=vt+5t0.55v2.75 0=5t0.55v2.75 0=5t20.552242.75t 0=5t22.75t26.4 t1,2=b±D2a=2.75±535.5610 t1,2=0.275±2.31422233158 t1=2.58922233158 t2=2.03922233158 t>0;t=t1=155.4  min   v=224t1=18.54 km/hs = 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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