13.356. The distance between points $A$ and $B$ is 308 m. A point moves from point $A$ towards $B$, covering 15 m in the first second and 1 m less in each subsequent second. Another point moves from point $B$ in the opposite direction, covering 20 m in the first second and 3 m more in each subsequent second. At what distance from point $A$ will the meeting occur if the point that started from point $B$ began moving 3 seconds after the point that started from point $A$?
Solution.

Let's find the laws of motion of the points.

For the point from $A:\left\{\begin{array}{l}15=1 \cdot V_{1}-\frac{a_{1} \cdot 1^{2}}{2}, \\ 15+14=2 V_{1}-\frac{a_{1} \cdot 2^{2}}{2},\end{array}\right.$ from which $V_{1}=15.5(\mathrm{~m} / \mathrm{c})$, $a_{1}=1\left(\mathrm{m} / \mathrm{c}^{2}\right)$, and the law of its motion is $-S_{A}=15.5 t-\frac{t^{2}}{2}$.

For the point from $B$ :

$$
\left\{\begin{array}{l}
20=1 \cdot V_{2}+\frac{a_{1} \cdot 1^{2}}{2} \\
20+23=2 V_{2}+\frac{a_{1} \cdot 2^{2}}{2},
\end{array} \text { from which } V_{2}=18.5(\mathrm{m} / \mathrm{c})\right.
$$

$a_{2}=3\left(\mathrm{m} / \mathrm{c}^{2}\right)$, and the law of its motion is $-S_{B}=18.5 t+\frac{3 t^{2}}{2}$.

The point from $A$ will travel $15.5 \cdot 3-\frac{3^{2}}{2}=42$ (m) in 3 s, and its speed will become $V=V_{1}-3 a_{1}=12.5$ (m/s). According to the condition, $42+12.5 t-\frac{t^{2}}{2}+18.5 t+\frac{3 t^{2}}{2}=308$, from which $t=7(\mathrm{c})$.

Then the points will meet at a distance of $42+12.5 \cdot 7-\frac{7^{2}}{2}=105$ (m) from point $A$.

Answer: 105 m.