4. A car left point A for point B, which are 10 km apart, at 7:00. After traveling $2 / 3$ of the way, the car passed point C, from which a cyclist immediately set off for point A. As soon as the car arrived in B, a bus immediately set off from there in the opposite direction and arrived in point A at 9:00. At what distance from B did the bus catch up with the cyclist, if the cyclist arrived in point A at 10:00 and the speed of each participant in the movement was constant?
Solution. Let $v_{a}$ be the speed of the car, $v_{\varepsilon}$ be the speed of the cyclist, and $v_{a \varepsilon}$ be the speed of the bus. From the problem statement, we derive the following system of equations:

$$
\left\{\begin{array}{l}
\frac{20 / 3}{v_{a}}+\frac{20 / 3}{v_{s}}=3 \\
\frac{10}{v_{a}}+\frac{10}{v_{a \varepsilon}}=2
\end{array}\right.
$$

We need to find $\frac{\frac{10}{3}+\frac{10}{3 v_{a}} v_{g}}{v_{a \varepsilon}-v_{s}} v_{a \varepsilon}=\frac{\frac{10}{3 v_{s}}+\frac{10}{3 v_{a}}}{1 / v_{b}-1 / v_{a \varepsilon}}$.

From the system of equations, we get

$$
\left\{\begin{array} { l } 
{ \frac { 1 0 } { 3 v _ { a } } + \frac { 1 0 } { 3 v _ { s } } = \frac { 3 } { 2 } , } \\
{ \frac { 1 0 } { 3 v _ { a } } + \frac { 1 0 } { 3 v _ { a \varepsilon } } = \frac { 2 } { 3 } , \quad \frac { 1 0 } { 3 v _ { a } } + \frac { 1 0 } { 3 v _ { s } } = \frac { 3 } { 2 } , \quad \frac { 1 0 } { 3 v _ { s } } - \frac { 1 0 } { 3 v _ { a \varepsilon } } = \frac { 5 } { 6 } }
\end{array} \left\{\begin{array}{l}
\frac{10}{3 v_{a}}+\frac{10}{3 v_{s}}=\frac{3}{2} \\
\frac{1}{v_{s}}+\frac{1}{v_{a \varepsilon}}=\frac{1}{4}
\end{array}\right.\right.
$$

Substituting the obtained values into expression (*), we get $\frac{3 / 2}{1 / 4}=\frac{3}{2} \cdot 4=6$.

Answer: 6.