[tex]\it Fie\ v_1, \ respectiv\ v_2\ vitezele \ celor \ dou\breve a \ mobile,\ cu\ v_1 > v_2.\\ \\ \\ La \ mi\d sc\breve a rile\ \hat\imath n\ acela\d si\ sens,\ presupunem\ c\breve a\ al\ doilea \ mobil\ parcurge\\ \\ \hat\imath n\ 20s \ distan\c{\it t}a\ x.\ Primul\ mobil\ va\ parcurge\ 100+x.\\ \\ Vom\ avea:[/tex]
[tex]\it \left.\begin{aligned} \it v_1=\dfrac{100+x}{20}\\ \\ \\\it v_2=\dfrac{x}{20}\ \ \ \ \ \ \ \ \ \ \end{aligned}\right\} \Rightarrow v_1-v_2=\dfrac{100+x-x}{20} \Rightarrow v_1-v_2=5\ \ \ \ \ (1)[/tex]
[tex]\it Pentru \ mi\d sc\breve arile\ de\ sens\ contrar, presupunem\ c\breve a\ al\ doilea\ mobil\\ \\ parcurge\ \hat\imath n\ 4s\ \ distan\c{\it t}a\ \ y,\ iar\ primul\ mobil\ va\ parcurge\ 100-y.\\ \\ Vom\ avea:[/tex]
[tex]\it \left.\begin{aligned}\ \it v_1=\dfrac{100-y}{4}\\ \\ \\ \it v_2=\dfrac{y}{4}\end{aligned}\right\} \Rightarrow v_1+v_2=\dfrac{100-y+y}{4}\ \Rightarrow v_1+v_2=25\ \ \ (2)[/tex]
[tex]\it (1),\ (2) \Rightarrow\begin{cases}\ \it v_1-v_2=5\\ \\ \it v_1+v_2=25\end{cases} \Rightarrow v_1=15\ m/s;\ \ \ v_2=10\ m/s[/tex]
