Question

un tren avind viteza egata cu 16m/s incepe sa frineze cu acceleratie egala cu (-0,6))m/s.dupa cit timp tenul se va opri,determina disatanta parcursa?

Answer 1

$\hbox{Distanta de oprire} \longrightarrow d_{op}=-\frac{v^2}{2a} \ \ \ ; \ \ \ a\ \textless \ 0 \\\\\\\ d_{op}=- \frac{16^2 \ \frac{m^2}{s^2}}{2* (-0,6)\ \frac{m}{s^2}}= \\\\=\frac{256}{2* \frac{\not 6}{\not 10}} \frac{m^2}{\not s^2}* \frac{\not s^2}{m}= \\\\= \frac{\not 256}{\not 2* \frac{3}{5}}\frac{\not m^2}{\not m}= \\\\ =128m* \frac{5}{3} = \\\\=\frac{640}{3}m \\\\\\ \boxed{d_{op}=213,3 m}$