How many cars traveling in the same direction will overtake (and leave behind) the pedestrian, when and where? (Keep the previous data in mind!)





Translate the above text into English, please retain the source text's line breaks and format, and output the translation result directly.
The pedestrian's path intersects with the route of 8 cars starting from $A$. These cars can only depart from $A$ after $8^{h} 15^{\prime}$ and arrive in $B$ before $10^{h} 15^{\prime}$; among them, the first departs at $8^{h} 20^{\prime}$, and the last at $9^{h} 30^{\prime}$. (Indeed, 8 cars depart from $A$ during this period.

The meeting time. The pedestrian starts $[5+10(k-1)]$ minutes before each of these cars, where $k=1,2, \ldots 8$, and thus gains an advantage of $[5+10(k-1)] \frac{200}{3}$ meters;

indeed, the pedestrian covers $\frac{4000}{60}=\frac{200}{3} \mathrm{~m}$ per minute.

Each car covers $\frac{600}{3} \mathrm{~m}$ per minute, which means they gain $\frac{400}{3} \mathrm{~m}$ per minute over the pedestrian. The pedestrian's advantage is

$$
[5+10(k-1)] \frac{200}{3}: \frac{400}{3}=\frac{1}{2}[5+10(k-1)] \text { minutes }
$$

to catch up. Therefore, from the time of departure,

$$
\begin{array}{rccccc}
\text { the first car } & \frac{5}{2} \text { minutes }=2^{\prime} 30^{\prime \prime} & \text { later, } & 8^{h} 22^{\prime} 30^{\prime \prime} \text {-at } \\
\text { the second } & \quad \frac{15}{2} \quad " & =7^{\prime} 30^{\prime \prime} & , & 8^{h} 37^{\prime} 30^{\prime \prime} \text {-at } \\
\text { the third } " & \frac{25}{2}, & =12^{\prime} 30^{\prime \prime} & , & 8^{h} 52^{\prime} 30^{\prime \prime} \text {-at }
\end{array}
$$

and so on, each car from $A$ catches up with the pedestrian 15 minutes after the previous one. The last one

$$
8^{h} 22^{\prime} 30^{\prime \prime}+7 \cdot 15^{\prime}=10^{h} 7^{\prime} 30^{\prime \prime} \text {-at }
$$

catches and leaves the pedestrian behind.

The meeting place. The pedestrian covers 1 km in 15 minutes. After starting, the first car catches up with the pedestrian at $5^{\prime}+2^{\prime} 30^{\prime \prime}=7^{\prime} 30^{\prime \prime}$; during this time, the pedestrian covers 500 meters.

Each subsequent car catches up 15 minutes later, thus 1 km apart.

Therefore, the cars starting from $A$ catch up with the pedestrian at a distance of $500+k \cdot 1000 \mathrm{~m}$ from $A$. $k=1,2, \ldots 7$.