13.264. To rise in an ordinary elevator to the top floor of an eight-story building (height 33 m) with two 6-second intermediate stops, it takes as much time as it would to rise in an elevator of a high-rise building with one 7-second intermediate stop at the 20th floor (height 81 m). Determine the lifting speed of the elevator in the high-rise building, knowing that it exceeds the speed of the ordinary elevator by $1.5 \mathrm{~m} / \mathrm{c}$, but does not reach $5 \mathrm{~m} / \mathrm{c}$.
## Solution.

Let $v_{1}$ m/s be the speed of a regular elevator ($v_{1} \leq 3.5 \text{ m/s}$), and $v_{2}$ m/s be the speed of the elevator in a high-rise building, $v_{2}=v_{1}+1.5 \text{ m/s}$.

According to the condition,

$$
\frac{33}{v_{1}}+2 \cdot 6=\frac{81}{v_{1}+1.5}+7 \text{, from which } v_{1}=1.5 \text{ m/s}, v_{2}=v_{1}+1.5=3 \text{ m/s}
$$

Answer: $3 \text{ m/s}$.