It is $v_{d}=90 \frac{\mathrm{km}}{\mathrm{h}}=25 \frac{\mathrm{m}}{\mathrm{s}}, v_{s}=144 \frac{\mathrm{km}}{\mathrm{h}}=40 \frac{\mathrm{m}}{\mathrm{s}}$.

The solution can be found through the following reasoning: If the D-train were not moving, but the locomotive had the speed $v_{r}=v_{d}+v_{s}$ (or vice versa), the relative speed between the D-train and the fast locomotive would be the same. Therefore, if $t_{d}$ and $t_{s}$ denote the respective times, it must hold that

$$
\begin{gathered}
\left(v_{d}+v_{s}\right) t_{s}=l_{s} \quad \text { and } \quad\left(v_{d}+v_{s}\right) t_{d}=l_{d} \quad \text { thus } \\
t_{s}=\frac{l_{s}}{v_{d}+v_{s}}=\frac{6}{13} \approx 0.5 \mathrm{~s} ; \quad t_{d}=\frac{l_{d}}{v_{d}+v_{s}}=\frac{260}{65}=4 \mathrm{~s}
\end{gathered}
$$

The time $t_{s}$ for the fast locomotive to pass the traveler in the D-train is thus $t_{s} \approx 0.5 \mathrm{~s}$, and the time $t_{d}$ for the D-train to pass the traveler in the locomotive is $t_{d}=4 \mathrm{~s}$.
## Task $7 / 62$

On a railway line, a D-train and a fast railcar meet. The D-train has a length of $l_{d}=260 \mathrm{~m}$ and a speed of $v_{d}=90 \frac{\mathrm{km}}{\mathrm{h}}$, the fast railcar is $l_{s}=30$ $\mathrm{m}$ long and has a speed of $v_{s}=144 \frac{\mathrm{km}}{\mathrm{h}}$.

How long does it take for a passenger in the D-train for the railcar to pass by, and for a passenger in the railcar for the D-train to pass by?