11. 8 people are traveling to the railway station in two cars with the same speed, with 4 people in each car (excluding the driver). One of the cars breaks down 15 km away from the railway station, and there are still 42 minutes left before the ticket check stops. At this point, the only available means of transportation is the other car, which can accommodate 5 people including the driver, and the average speed of this car is 60 km/h, while the average walking speed of a person is 5 km/h. Design two plans and, through calculations, explain how these 8 people can reach the railway station before the ticket check stops.
Three, 11. [Plan One] When the car breaks down, the 4 people in this car get out and walk, while the other car takes the 4 people inside to the train station, immediately returns to pick up the 4 people walking, and takes them to the train station.

Let the distance walked by the 4 people in the broken-down car be $x \mathrm{~km}$. According to the problem, we have
$\frac{x}{5}=\frac{15+15-x}{60}$. Solving for $x$ gives $x=\frac{30}{13}$.
Therefore, the total time required for all 8 people to reach the train station is
$$
\begin{array}{l}
\frac{30}{13} \div 5+\left(15-\frac{30}{13}\right) \div 60=\frac{35}{52}(\mathrm{~h}) \\
=40 \frac{5}{13}(\mathrm{~min})<42(\mathrm{~min}) .
\end{array}
$$

Thus, this plan is feasible.
[Plan Two] When the car breaks down, the 4 people in this car get out and walk, while the other car takes the 4 people inside to a certain place, lets them get out and walk, and then immediately returns to pick up the 4 people who got out due to the car breaking down, ensuring that both groups arrive at the station at the same time.

Analyzing this plan, we know that the walking distances of the two groups are the same. As shown in Figure 7, $D$ is the location where the people in the non-faulty car get out, and $C$ is the location where the people in the faulty car get back in. 
Therefore, let $A C=D B=$ $y$, according to the problem, we have
$\frac{y}{5}=\frac{15-y+15-2 y}{60}$. Solving for $y$ gives $y=2$.
Therefore, the time required for all 8 people to arrive at the train station simultaneously is
$$
\frac{2}{5}+\frac{15-2}{60}=\frac{37}{60}(\mathrm{~h})=37(\mathrm{~min})<42(\mathrm{~min}) \text {. }
$$

Thus, this plan is also feasible.