3.46 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 $B$.
3.46 In Fig. 3.9, the graphs of train movements before the speed changes are shown.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-110.jpg?height=506&width=546&top_left_y=1035&top_left_x=350)

Fig. 3.9

We find $\operatorname{tg} \angle M C N=v^{(0)}{ }_{\text {пас }}=288: 3.2=90$ (km/h). Let $v^{(0)}{ }_{\text {тов }}=x$ km/h; then $A B=N L=x t$ km; $C D=90(t-3.2)=x t$, from which

$$
t=\frac{288}{90-x}
$$

After the speed changes, we have $v^{(1)}{ }_{\text {пас }}=100 \text{ km/h}$, $v_{\text {тов }}^{(1)}=x+10$ km/h. According to the condition,

$\frac{x t}{x+10}-\frac{x t}{100}=2.4$.

Solving the system of equations (1) and (2), we get $x=50 \text{ km/h}$, $t=7.2$ h. Thus, $A B=x t=50 \cdot 7.2=360$ (km).

Answer: $360 \text{ km}$.