13.263. The path from $A$ to $B$ is traveled by a passenger train 3 hours and 12 minutes faster than by a freight train. In the time it takes the freight train to travel from $A$ to $B$, the passenger train travels 288 km more. If the speed of each train is increased by $10 \mathrm{~km} / \mathrm{h}$, the passenger train will travel from $A$ to $B$ 2 hours and 24 minutes faster than the freight train. Determine the distance from $A$ to $B$.
Solution.

In Fig. 13.7, the graphs of train movements before the speed changes are considered.

We have $\operatorname{tg} \angle M C N=v_{\text {pas. }}^{(0)}=288: 3.2=90$ km/h. If $v_{\text {tov. }}^{(0)}=x$ km/h, then $A B=N L=x t \mathrm{km}, C D=90(t-3.2)=x t \Rightarrow t=\frac{288}{90-x}$. After the speed changes, we get $v_{\text {pas. }}^{(1)}=100 \mathrm{km} / \mathrm{h}, v_{\text {tov. }}^{(1)}=x+10$ km/h. According to the condition $\frac{x t}{x+10}-\frac{x t}{100}=2.4$. Substituting $t=\frac{288}{90-x}$ here and solving the resulting equation, we find $x=50 \mathrm{km} / \mathrm{h}, t=7.2$ h. Therefore, $A B=x t=50 \cdot 7.2=360 \mathrm{km}$.

Answer: 360 km.