Example 2 A pedestrian and a cyclist are traveling south simultaneously on a road parallel to a railway. The pedestrian's speed is $3.6 \mathrm{~km} / \mathrm{h}$, and the cyclist's speed is $10.8 \mathrm{~km} / \mathrm{h}$. If a train comes from behind and takes $22 \mathrm{~s}$ to pass the pedestrian and $26 \mathrm{~s}$ to pass the cyclist, then the length of the train is $\qquad$ $\mathrm{m}$.
(2002, Hubei Province Junior High School Mathematics Competition)
Explanation: First, convert the two speed units to: pedestrian $1 \mathrm{~m} / \mathrm{s}$, cyclist $3 \mathrm{~m} / \mathrm{s}$. Let the length of the train be $l \mathrm{~m}$, and the speed of the train be $v \mathrm{~m} / \mathrm{s}$. According to the problem, we have
$$
\left\{\begin{array}{l}
l=(v-1) \times 22, \\
l=(v-3) \times 26 .
\end{array}\right.
$$

Solving these equations gives $l=286, v=14$.
Therefore, the length of the train is $286 \mathrm{~m}$.