[tex]Dane:\\m = 1400 \ kg\\v_{o} = 72\frac{m}{s} = 20\frac{m}{s}\\v = 0\\F = 7000 \ N\\Szukane:\\s = ?\\\\Rozwiazanie\\\\\underline{s = v_{o}t - \frac{1}{2}at^{2}}\\\\a = \frac{F}{m} = \frac{7000 \ N}{1400 \ kg} = \frac{7000 \ kg\cdot\frac{m}{s^{2}}}{1400 \ kg} = 5\frac{m}{s^{2}}\\\\a = \frac{\Delta v}{t} \ \ \rightarrow \ \ t = \frac{\Delta v}{a} = \frac{v_{o}-v}{a} = \frac{v_{o}}{a} = \frac{20\frac{m}{s}}{5 \ s}=4 \ s[/tex]
[tex]s = 20\frac{m}{s}\cdot 4 \ s - \frac{1}{2}\cdot5\frac{m}{s^{2}}\cdot(4 \ s)^{2}\\\\s = 80 \ m - 40 \ m\\\\\boxed{s = 40 \ m}[/tex]
Odp. Droga hamowania s = 40 m.
