s = 180 km v - srednia predkosc t - wyjściowy czas mamy układ równań: [latex]left { {{v = frac{180 km}{t}} atop {v + 20 frac{km}{h} = frac{180 km}{t - 0,75 h}}} ight[/latex] rozwiązujemy go: [latex]frac{180 km}{t} + 20 frac{km}{h} = frac{180 km}{t - 0,75 h} |:20km\ frac{9}{t} + frac{1}{h} = frac{9}{t - 0,75 h}\ 9h(t -0,75h) + t(t - 0,75h) = 9th\ 9th - 6,75h^2 + t^2 - 0,75th = 9th\ t^2 - 0,75th - 6,75h^2 = 0 \ Delta = 0,5625h^2 + 27h^2 = (5,25)^2 \ t > 0 Rightarrow t = frac{0,75h + 5,25h}{2} = 3h[/latex] wyliczamy prękość: [latex]v = frac{180 km}{t} = frac{180 km}{3 h} = 60frac{km}{h}[/latex] jak masz pytania to pisz na pw
V=s/t V=180/t V+20=(180)/(t-3/4) (180/t) + 20=(180)/(t-3/4) 180-135/t+20t-15=180 15 = -135/t+20t ..../*t 15t=-135+20t² 20t²-15t-135=0 4t²-3t-27=0 Δ=9+432=441 √Δ=21 t₁=(3+21)/8=24/8=3 t₂=(3-21)/8=-18/8 <0, więc odrzucamy t=3h V=180km/3h=60km/h
