[tex]dane:\\m = 1200 \ kg\\v_{o} = 0\\t = 0,25 \ min = 0,25\cdot60 \ s = 15 \ s\\v = 72\frac{km}{h} = 72\cdot\frac{1000 \ m}{3600 \ s} = 20\frac{m}{s}\\szukane:\\F_{w} = ?\\\\Rozwiazanie\\\\a = \frac{\Delta v}{t} = \frac{v-v_{o}}{t} = \frac{v}{t} = \frac{20\frac{m}{s}}{15 \ s} = \frac{4}{3}\frac{m}{s^{2}}\\\\Z \ II \ zasady \ dynamiki:\\\\a = \frac{F_{w}}{m} \ \ /\cdot m\\\\F_{w} = m\cdot a\\\\F_{w} = 1200 \ kg\cdot\frac{4}{3}\frac{m}{s^{2}}\\\\F_{w} = 1\ 600 \ N = 1,6 \ kN[/tex]
Odp. Wartość siły wypadkowej Fw = 1,6 kN.