$6 \cdot 93$ The bus route consists of a circular loop with a circumference of 10 miles and a straight line from the main station to point $Q$ on the loop, which is 1 mile long (as shown in the figure). Two buses serve on this route. Each loop takes 20 minutes. Bus 1 leaves the main station, first travels along the straight road, then goes clockwise around the loop, and finally returns to the main station along the straight road. Bus 2 follows the same route but travels counterclockwise around the loop, arriving at the main station 10 minutes later than Bus 1. Both buses run continuously without stopping at any point on the route, and the time for passengers to board and disembark can be ignored.

A person plans to wait for the bus at location $P$, which is $x$ miles from the main station along the route of Bus 1 $(0 \leqslant x<12)$. Knowing that he wants to take the bus to the main station and will board the bus that gets him there the fastest, let the longest time required for his journey be $w(x)$ minutes (waiting time plus travel time).
(1) Find $w(2), w(4)$.
(2) For what value of $x$ is $w(x)$ the greatest?
(3) For $0 \leqslant x<12$, sketch the graph of $y=w(x)$.
[Solution] From the given information, the car's speed is $\frac{3}{5}$ miles per minute, and $w(x) = w(12-x)$, so it is sufficient to consider the case where $0 < x \leq 6$.

If $0 < x \leq 1$, the person will obviously take the car heading to the main station, with a maximum waiting time of 10 minutes and a travel time of $\frac{5x}{3}$ minutes. Therefore, $w(x) = 10 + \frac{5x}{3} (0 < x \leq 1)$.

If $1 < x \leq 6$, the situation is more complicated. For this, we list the necessary data in the following table:
$A$ represents the range of $x$;
$B$ represents the number of the just-missed car;
$C$ represents the position of the other car (at the moment the person arrives at position $P$);
$D$ represents the travel time of car No. 1;
$E$ represents the travel time of car No. 2;
$F$ represents the number of the car he needs to wait for.
\begin{tabular}{|c|c|c|c|c|c|}
\hline A & $B$ & $C$ & $D$ & $E$ & $F$ \\
\hline $1 < x < 3 \mid$ & 1 & $6-x$ & \begin{tabular}{l}
$20 + \frac{5}{3}(12-x)$
\end{tabular} & \begin{tabular}{l}
$\frac{5}{3}(6-x)$
\end{tabular} & 2 \\
\hline & 2 & $6-x$ & \begin{tabular}{l}
$20 + \frac{5}{3}(6+x)$
\end{tabular} & $20 + \frac{5}{3}x$ & 2 \\
\hline $x = 3$ & 1,2 & 3 & 35 & 25 & 2 \\
\hline $3 < x < 6$ & 1 & $6-x$ & $20 + \frac{5}{3}(12-x)$ & \begin{tabular}{l}
$20 + \frac{5}{3}(6-x)$
\end{tabular} & 2 \\
\hline & 2 & $6-x$ & \begin{tabular}{l}
$\frac{5}{3}(6+x)$
\end{tabular} & \begin{tabular}{l}
$20 + \frac{5}{3}x$
\end{tabular} & 1 \\
\hline $x = 6$ & 1 & 0 & 30 & 20 & 2 \\
\hline & 2 & 0 & 20 & 30 & 1 \\
\hline
\end{tabular}
$$
\therefore w(x) = \left\{\begin{array}{l}
0, x = 0, \\
10 + \frac{5}{3}x, 0 < x \leq 1, \\
20 + \frac{5}{3}x, 1 < x < 3, \\
25, x = 3, \\
20 + \frac{5}{3}(6-x), 3 < x < 6, \\
20, x = 6 .
\end{array}\right.
$$
(1) $w(2) = \frac{70}{3}, w(4) = \frac{70}{3}$.
(2) When $x = 3$ and $x = 9$, $w(x)$ reaches its maximum value of 25.
(3) The rough sketch of $w(x)$ is