56. (8th grade) Two cyclists set out towards each other. If the first starts 1 hour earlier, and the second starts half an hour later, they will meet 18 minutes earlier. If the first starts half an hour later, and the second starts 1 hour earlier, the meeting point will move by 11200 meters.

Calculate the speed of each cyclist.
56. Let's introduce the following notations: $x$ and $y$ are the speeds of the cyclists in kilometers per hour; $AB$ is the distance between the points; $\frac{AB}{x+y}$ is the normal time of travel until they meet. $\left(\frac{AB}{x+y}+1-\frac{18}{60}\right)$ hours is the time the first cyclist travels until the meeting (first condition). $\left(\frac{AB}{x+y}-\frac{1}{2}-\frac{18}{60}\right)$ hours is the time the second cyclist travels. Together they cover the entire distance from $A$ to $B$, so,

$$
\left(\frac{AB}{x+y}+1-\frac{18}{60}\right) x+\left(\frac{AB}{x+y}-\frac{1}{2}-\frac{18}{60}\right) y=AB
$$

From the second part of the condition, it follows that the second cyclist will be on the road 1.5 hours longer than the first; therefore, before the first cyclist starts, the second cyclist manages to travel a distance of $1.5 y$ km. This condition of the problem can be expressed by the equation:

$$
\frac{AB-1.5 y}{x+y} x+11.2=\frac{AB}{x+y} x
$$

Considering these two equations together, we get $x=16, y=14$. The speed of the first cyclist is 16 km/h, and the speed of the second cyclist is 14 km/h.