A fast train consisting of 8 carriages, each $10 \mathrm{~m}$ long, and its locomotive and tender together $24 \mathrm{~m}$ long, travels at a speed of $60 \mathrm{~km}$ per hour past a passenger train. How long is the fast train visible from the window of the passenger train if the passenger train is stationary, if the fast train is coming towards it, and if it is traveling in the same direction at a speed of $42 \mathrm{~km}$ per hour?
Let $A$ be the express train, $B$ the person, and $C$ the appropriate window of the passenger train. $A$ is $80+24=104 \mathrm{~m}$ long, if we do not take into account the distance between the cars.

1. If $C$ is stationary, we need to calculate how long it takes for $A$ to completely pass in front of $C$, that is, how long it takes for $A$ to travel $104 \mathrm{~m}$ at a speed of $60 \mathrm{~km} /$ hour. Let this time be $t_{1}$ :

$$
t_{1}=\frac{104 \mathrm{~m}}{\frac{60 \cdot 1000 \mathrm{~m}}{60 \cdot 60 \mathrm{sec}}}=\frac{3 \cdot 52}{25} \mathrm{sec}=6.24 \mathrm{sec}
$$

2. If $C$ is also moving, at a speed of $42 \mathrm{~km} /$ hour relative to $A$, then $C$ will see $A$ moving much faster than $A$'s own speed, as if $A$ were moving at $60+42 \mathrm{~km} /$ hour relative to $C$. Therefore, in this case, $A$ will be visible for

$$
t_{2}=\frac{104}{\frac{60 \cdot 1000}{60 \cdot 60}+\frac{42 \cdot 1000 \mathrm{~m}}{60 \cdot 60 \mathrm{sec}}}=\frac{104}{\frac{85}{3}}=\frac{312}{85}=3.67 \mathrm{sec}
$$

3. If $C$ is moving in the same direction as $A$, then $C$ will see $A$ moving at a lower speed relative to $C$ than it does relative to the stationary tracks, because $C$ is moving with $A$. The relative speed of $A$ to $C$ is now $60-42 \mathrm{~km} /$ hour, so $A$ will be visible for

$$
t_{3}=\frac{104}{\frac{60 \cdot 1000}{60 \cdot 60}-\frac{42 \cdot 1000}{60 \cdot 60}} \frac{\mathrm{m}}{\mathrm{m}}=\frac{104}{\frac{50}{3}-\frac{35}{3}}=\frac{3 \cdot 104}{15}=20.8 \mathrm{sec}
$$