[tex]dane:\\s_1 = 2 \ km\\v_1 = 15\frac{km}{h}\\s_2 = 4 \ km\\v_2 = 20\frac{km}{h}\\szukane:\\v_{sr} = ?\\\\Rozwiazanie\\\\v_{sr} = \frac{s_{c}}{t_{c}}\\\\s_{c} = s_1 + s_2 = 2 \ km + 4 \ km = 6 \ km\\\\t = \frac{s}{v}\\\\t_{c} = t_1 + t_2 = \frac{s_1}{v_1} + \frac{s_2}{v_2} = \frac{2 \ km}{15\frac{km}{h}}+\frac{4 \ km}{20\frac{km}{h}} = \frac{2}{15} \ h + \frac{1}{5} \ h = (\frac{2}{15}+\frac{3}{15}) \ h = \frac{5}{15} \ h = \frac{1}{3} \ h\\\\v_{sr} = \frac{6 \ km}{\frac{1}{3}h}=18\frac{km}{h}[/tex]
Odp. Średnia prędkość Izy wynosiła 18 km/h.
