[tex]Dane:\\t_1 = 6:55 - 6:20 = 35 \ min\\s_1 = 25 \ km\\t_2 = 20 \ min=20\cdot60 \ s = 1200 \ s\\v_2 = 15\frac{m}{s}\\szukane:\\v_{sr} = ?\\\\Rozwiazanie\\\\v_{sr} = \frac{s_{c}}{t_{c}}\\\\s_2 = v_2\cdot t_2 = 15\frac{m}{s}\cdot1200 \ s = 18 \ 000 \ m = 18 \ km\\\\s_{c} = s_1+s_2 = 25 \ km + 18 \ km = 43 \ km[/tex]
[tex]t_{c} = t_1 + t_2 = 35 \ min + 20 \ min = 55 \ min = \frac{55}{60} \ h = \frac{11}{12} \ h[/tex]
[tex]v_{sr} = \frac{43 \ km}{\frac{11}{12} \ h} =\frac{43\cdot12}{11} \ \frac{km}{h} = 46,(90) \ \frac{km}{h}\\\\\boxed{v_{sr}\approx46,91\frac{km}{h}}[/tex]
Odp. Prędkość srednia na całej trasie wynosiła ok. 46,91 km/h.
