3. Two people simultaneously step onto an escalator from opposite ends, which is moving downward at a speed of $u=1.5 \mathrm{~m} / \mathrm{s}$. The person moving downward has a speed of $v=3 \mathrm{~m} / \mathrm{s}$ relative to the escalator, while the person moving upward has a speed of $2 v / 3$ relative to the escalator. At what distance from the bottom of the escalator will they meet? The length of the escalator is $l=100 \mathrm{~m}$.
3. The speeds of the people relative to the ground are

$$
\text { descending } v_{c n}=v+u \text {, ascending } v_{\text {nоо }}=\frac{2}{3} v-u \text {. }
$$

Therefore, the time it takes for the people to meet is

$$
t=\frac{l}{v_{c n}+v_{n o d}}=\frac{3 l}{5 v}
$$

And since the ascending person will travel a distance until the meeting

$$
l_{\text {noo }}=v_{\text {noo }} t=\frac{3 l}{5 v}\left(\frac{2}{3} v-u\right)=10 \mathrm{M} .
$$