7. Car A and Car B start from locations A and B simultaneously and travel towards each other. Car A travels at 40 kilometers per hour, and Car B travels at 60 kilometers per hour. After reaching B and A respectively, they immediately return. On the return trip, Car A's speed increases by half, while Car B's speed remains unchanged. It is known that the distance between the two points where the two cars meet for the second time is 50 kilometers, then the distance between A and B is $\qquad$ kilometers.
Analysis: $V_{\text {Jia: }} V_{\mathrm{Z}}=40: 60=2: 3$, when the two cars meet, the time is equal, $S_{\text {Jia: }} S_{\mathrm{Z}}=2: 3$, assume the total distance is “1”, the first meeting point is $\frac{2}{5}$ of the total distance from point A.
When car Jia reaches point B, car Z has already reached point A and traveled another $\frac{3}{2}-1=\frac{1}{2}$ of the total distance. At this point, car Jia's speed is $40+40 \div 2=60 \mathrm{~km} / \mathrm{h}$, the two cars have the same speed, and they complete the remaining $\frac{1}{2}$ together, each traveling $\frac{1}{4}$. Therefore, the second meeting point is $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$ of the total distance from point A. Thus, the total distance is: $50 \div\left(\frac{3}{4}-\frac{2}{5}\right)=\frac{1000}{7} \mathrm{~km}$.