[tex]dane:\\s_1 = 4 \ km\\v_1 = 20\frac{km}{h}\\s_2 = 15 \ km\\v_2 = 60\frac{km}{h}\\szukane:\\v_{sr} = ?\\\\Rozwiazanie\\\\v_{sr} = \frac{s_{c}}{t_{c}}\\\\s_{c} = s_1 + s_2 = (4+15) \ km = 19 \ km\\\\t = \frac{s}{v}\\\\t_{c} = t_1+t_2 = (\frac{4}{20} + \frac{15}{60}) \ h = (0,2 + 0,25) \ h = 0,45 \ h\\\\v_{sr} = \frac{19 \ km}{0,45 \ h} = 42,(2)\frac{km}{h} \approx42,2\frac{km}{h}[/tex]
