3. Locations A and B are respectively upstream and downstream on a lake. Each day, there is a boat that departs on time from both locations and travels at a constant speed towards each other, usually meeting at 11:00 AM on the way. ... Due to a delay, the boat from location A left 40 minutes late, and as a result, the two boats met at 11:15 AM on the way. It is known that the boat departing from location A travels at a speed of 44 kilometers per hour in still water, and the boat departing from location B travels at a speed of $v$ kilometers per hour in still water. Then $v=$ $\qquad$ kilometers per hour.
3. 6 .

As shown in the figure, the two ships usually meet at point $A$. On the day when ship B is late, they meet at point $B$. According to the problem, ship A takes $15^{\prime}$ to navigate segment $AB$, and ship C takes
$$
\begin{array}{l}
(40-15)^{\prime}=25^{\prime} . \\
\therefore \frac{15}{60}(44+v)=\frac{25}{60}\left(v^{2}-v\right),
\end{array}
$$

which simplifies to $5 v^{2}-8 v-132=0$.
$$
(v-6)(5 v+22)=0 \text {. }
$$

Solving this, we get $v=6$ (the negative root is discarded).