[tex]Dane:\\m = 100 \ t = 100 \ 000 \ kg = 10^{5} \ kg\\F_{w} = 120 \ kN = 120 \ 000 \ N = 1,2\cdot10^{5} \ N\\v_{o}=0\\v = 72\frac{m}{s}\\Szukane:\\s = ?\\\\Rozwiazanie\\\\a = \frac{F_{w}}{m}\\\\a = \frac{1,2\cdot10^{5} \ N}{10^{5} \ kg} = 1,2\frac{N}{kg} = 1,2\frac{m}{s^{2}}\\\\a = \frac{v}{t} \ \ \rightarrow \ \ t = \frac{v}{a}\\\\t = \frac{72\frac{m}{s}}{1,2\frac{m}{s^{2}}} = 60 \ s[/tex]
[tex]s = \frac{1}{2}at^{2}\\\\s = \frac{1}{2}\cdot1,2\frac{m}{s^{2}}\cdot(60 \ s)^{2}\\\\s = 0,6\frac{m}{s^{2}}\cdot3600 \ s\\\\\boxed{s = 2 \ 160 \ m}[/tex]
Odp. Pas startowy musi mieć długość s = 2160 m.
