Example 2: The distance between location A and location B is 70 kilometers. Two cars start from the two locations simultaneously and continuously travel back and forth between A and B. The first car starts from A, traveling at 30 kilometers per hour, and the second car starts from B, traveling at 40 kilometers per hour. When the first car, which started from A, meets the second car after its second departure from A, the two cars have traveled $\qquad$ kilometers and $\qquad$ kilometers respectively.
(1999, Wuhan Selection Contest for the National Junior High School Mathematics League)
Solution: According to the problem, draw the graph, as shown in Figure 3.
The broken lines $O A B M$ and $D E F G$ represent the motion graphs of the first and second cars, respectively. Clearly, point $M$ indicates the second meeting of the first car, which departs from location A for the second time, with the second car.

Draw $M N \perp t$-axis, then $\frac{M N}{B N}$ and $\frac{M N}{G N}$ are the speeds of the first and second cars, respectively. Therefore,
$$
\begin{array}{l}
\frac{M N}{B N}=30, \\
\frac{M N}{G N}=40 .
\end{array}
$$
(1) $\div$ (2) gives $\quad \frac{G N}{B N}=\frac{3}{4}$.

Similarly, $\frac{O D}{O E}=40$.
From $O D=70$ we get $O E=\frac{7}{4}$;
From $\frac{O D}{A I}=30$ we get $A I=\frac{7}{3}$.
Since $O G=3 O E=\frac{21}{4}, O B=2 A D=\frac{14}{3}$,
then $B G=O\left(B-O B=\frac{21}{4}-\frac{14}{3}=\frac{7}{12}\right.$.
From (3) we know $B N=\frac{4}{7} B C=\frac{4}{7} \times \frac{7}{12}=\frac{1}{3}$;
From (1) we get $M N=30 B N=30 \times \frac{1}{3}=10$.
Therefore, the first car has traveled $70 \times 2+10=150$ (km), and the second car has traveled $70 \times 3-10=200$ (km).

The key here is equation (3). Its practical meaning is: the time ratio for two people to walk the same distance is inversely proportional to their speeds.