[tex]Dane:\\s = 800 \ m = 0,8 \ km\\v_{o} = 60\frac{km}{h}\\v=20\frac{km}{h}\\Szukane:\\a) \ t = ?\\b) \ a = ?[/tex]
Rozwiązanie
a) Czas hamowania:
[tex]s = v_{sr}\cdot t \ \ /:v_{sr}\\\\t = \frac{s}{v_{sr}}\\\\v_{sr} = \frac{60\frac{km}{h}-\frac{20 \ km}h}{2} = \frac{40\frac{km}{h}}{2} = 20\frac{km}{h}\\\\t = \frac{0,8 \ km}{20\frac{km}{h}} = 0,04 \ h =0,04\cdot3600 \ s = 144 \ s\\\\\boxed{t = 144 \ s}[/tex]
b) Wartość przyspieszenia tego pociągu:
[tex]s = \frac{at^{2}}{2} \ \ /\cdot\frac{2}{t^{2}}\\\\a = \frac{2s}{t^{2}}\\\\a = \frac{2\cdot800 \ m}{(144 \ s)^{2}} = \frac{1600 \ m}{20736 \ s^{2}}\\\\\boxed{a \approx0,08\frac{m}{s^{2}}}[/tex]
