[tex]dne:\\m = 1,6 \ t = 1 \ 600 \ kg\\v_{o} = 0\\t = 0,5 \ min = 30 \ s\\v = 108\frac{km}{h} = 108\cdot\frac{1000 \ m}{3600 \ s} = 30\frac{m}{s}\\szukane:\\F_{w} = ?\\\\Rozwiazanie\\\\F_{w} = m\cdot a\\\\ale:\\a = \frac{\Delta v}{t} = \frac{v-v_{o}}{t} = \frac{v}{t} = \frac{30\frac{m}{s}}{30 \ s} = 1\frac{m}{s^{2}}\\\\F_{w} = 1600 \ kg\cdot1\frac{m}{s^{2}}\\\\F_{w} = 1 \ 600 \ N = 1,6 \ kN\\\\(1 \ kN = 1000 \ N)[/tex]
Odp. Wartość siły wypadkowej wynosi 1,6 kN.
