6.1. (12 points) From point $A$ to point $B$, a bus and a cyclist departed simultaneously at 13:00. After arriving at point $B$, the bus, without stopping, headed back and met the cyclist at point $C$ at 13:10. Upon returning to point $A$, the bus again, without stopping, headed to point $B$ and caught up with the cyclist at point $D$, which is located $\frac{2}{3}$ km from point $C$. Find the speed of the bus (in km/h), if the distance between points $A$ and $B$ is 4 km, and the speeds of the bus and the cyclist are constant.
Answer: 40.

Solution. Let $v_{1}$ and $v_{2}$ be the speeds (in km/h) of the cyclist and the bus, respectively. By the time of the first meeting, they have collectively traveled $\frac{1}{6} v_{1}+\frac{1}{6} v_{2}=8$ km. Until the next meeting, the time that has passed is equal to $\frac{s_{0}}{v_{1}}=\frac{2 \cdot \frac{1}{6} v_{1}+s_{0}}{v_{2}}$ hours, where $s_{0}=\frac{1}{3}$ km. From this, we find $\frac{1}{4} v_{1}^{2}+v_{1}-24=0, v_{1}=8$ km/h (the second root is negative), $v_{2}=40 \text{km}/$ h.