1.
[tex]Dane:\\s = 2,5 \ km = 2 \ 500 \ m\\v = \frac{1}{2}\frac{m}{s}\\Szukane:\\t = ?\\\\v = \frac{s}{t} \ \ \rightarrow \ \ t = \frac{s}{v}\\\\t = \frac{2500 \ m}{\frac{1}{2}\frac{m}{s}}=2500\cdot2 \ s\\\\\boxed{t = 500 \ s = 8 \ min \ i \ 20 \ s}[/tex]
2.
[tex]Dane:\\s_1 = 200 \ km\\t_1 = 2 \ h\\s_2 = 140 \ km\\v = 30\frac{m}{s} = 30\cdot\frac{1000 \ m}{3600 \ s} = 8,(3)\frac{km}{h} = 8,33\frac{km}{h}\\Szukane:\\v_{sr} = ?[/tex]
Rozwiązanie
Prędkość średnia jest ilorazem przemieszczenia do czasu trwania ruchu.
[tex]v_{sr} = \frac{s_{c}}{t_{c}}\\\\\\s_{c} = s_1+s_2 = 200 \ km + 140 \ km = 340 \ km\\\\t_2 = \frac{s_2}{v_2} = \frac{140 \ km}{8,33\frac{km}{h}} =16,87 \ h\\\\t_{c} = t_1+t_2 = 2 \ h +16,87 \ h = 18,87 \ h\\\\\\v_{sr} = \frac{340 \ km}{18,87 \ h}\\\\\boxed{v_{sr} = 18\frac{km}{h}}[/tex]