Question

biegacz pokonal dystans 20 km z predkoscia 12 kmh a nastepnie 5km z predkoscia 8 kmh jaka byla jego predkosc srednia calej drogi

Answer 1

Odpowiedź:

[tex]\huge\boxed{v_{sr}\approx10,91\frac{km}{h}}[/tex]

Wyjaśnienie:

Prędkość średnia - iloraz przemieszczenia do czasu trwania ruchu:

[tex]v_{sr} = \frac{s_{c}}{t_{c}}[/tex]

[tex]Dane:\\s_1 = 20 \ km\\v_1 = 12\frac{km}{h}\\s_2 = 5 \ km\\v_2 = 8\frac{km}{h}\\Szukane:\\v_{sr} = ?\\\\Rozwiazanie\\\\v_{sr} = \frac{s_{c}}{t_{c}}\\\\s_{c} = s_1+s_2 = 20 \ km + 5 \ km = 25 \ km\\\\t = \frac{s}{v}\\\\t_{c} = t_1+ t_2 = \frac{s_1}{v_1}+\frac{s_2}{t_2} = \frac{20 \ km}{12\frac{km}{h}}+\frac{5 \ km}{8\frac{km}{h}}=\frac{5}{3} \ h + \frac{5}{8} \ h = (\frac{40}{24}+\frac{15}{24}) \ h = \frac{55}{24} \ h[/tex]

[tex]v_{sr} = \frac{25 \ km}{\frac{55}{24} \ h} = \frac{25\cdot24}{55} \ \frac{km}{h}\\\\\boxed{v_{sr} = 10,(90)\frac{km}{h}\approx10,91\frac{km}{h}}[/tex]