The steam train travels from Zalakomár, which is 21 km away, to Kanizsa. It covers the distance in 16 minutes, starting with uniform acceleration, then traveling at a constant speed of $90 \frac{\mathrm{km}}{\mathrm{h}}$, and finally decelerating uniformly to a stop. How long does the steam train travel at $90 \frac{\mathrm{km}}{\mathrm{h}}$?
Let the time of acceleration be $t_{1}$, the time of deceleration be $t_{2}$, and the time during which the train traveled at $90 \frac{\mathrm{km}}{\mathrm{h}}$ be $t$. The total travel time is 16 minutes $=\frac{4}{15}$ hours. That is,

$$
t_{1}+t+t_{2}=\frac{4}{15}
$$

The average speed of the train during acceleration and deceleration, according to the known formula, is $\bar{v}=\frac{0+90}{2}=45 \frac{\mathrm{km}}{\mathrm{h}}$. We can write the following equation for the distance traveled:

$$
\left(t_{1}+t_{2}\right) 45+90 t=21
$$

From (1), $t_{1}+t_{2}=\frac{4}{15}-t$, which we substitute into (2):

$$
\left(\frac{4}{15}-t\right) 45+90 t=21
$$

From this, $t=\frac{9}{45}=\frac{1}{5}$ hours, or the train traveled at $90 \frac{\mathrm{km}}{\mathrm{h}}$ for 12 minutes.