Question

Un tren pleaca dintr-o statie avand o miscare uniforma accelerata.Dupa ce a parcurs 600 m , atinge viteza de 45 km\h si isi continua apoi drumul cu o miscare uniforma timp de 10 minute .Cu 50 de secunde inainte de a se opri in statia urnatoare , incepe franarea.Sa se determine :a)timpul si acceleratia in miscarea uniform accelerata ;b)acceleratia de franare in miscarea uniform incetinita;c)distanta dintrecele doua statii.

Answer 1

$v_{1}=45 \frac{km}{h}=12.5 \frac{m}{s}$
Timpul initial e 0
$\Delta t=(t-t_{0})=t= \frac{d_{1}}{v_{1}}= \frac{600}{12.5}=48s$
Acceleratia
$v_{1}^{2}=v_{0}^{2}+2*a*d_{1}$
$a_{1}= \frac{v_{1}^2}{2d_{1}}= \frac{156.25}{1200}=0.13m/s^{2}$
Distanta parcursa in miscarea uniforma
$d_{2}= v_{1}*t_{2}=12.5*600=7500m$
Accelerarea incetinita
$a_{2}= \frac{\Delta v}{\Delta t}= \frac{0-12.5}{50}=-0.25m/s^{2}$
Distanta parcursa intre cele doua statii
$d=d_{1}+d_{2}+v_{1}*t_{2}- \frac{at_{2}^{2}}{2}=600+7500+625-312.5=8412.5m$
$d=8.41km$