[tex]dane:\\v_{o} = 0\\m = 2,5 \ kg\\F = 4 \ N\\t = 2 \ s\\szukane:\\s = ?\\\\Rozwiazanie\\\\s = \frac{1}{2}at^{2}}\\\\ale:\\\\a = \frac{F}{m} = \frac{4 \ N}{2,5 \ kg} = 1,6\frac{N}{kg} = 1,6\frac{m}{s^{2}}\\\\s = \frac{1}{2}\cdot1,6\frac{m}{s^{2}}\cdot(2 \ s)^{2}\\\\s = 3,2 \ m[/tex]
Odp. Przedmiot pokona drogę s = 3,2 m.
