13.260. Two material particles, being 295 m apart from each other, started moving towards each other simultaneously. The first particle moves uniformly at a speed of $15 \mathrm{~m} / \mathrm{c}$, while the second particle moved 1 m in the first second and 3 m more in each subsequent second than in the previous one. Through what angle will the second hand of a clock move during the time that has passed from the start of the particles' movement until their meeting?
## Solution.

Let $t$ seconds have passed before the particles meet. In this time, the first particle has traveled $15 t \text{m}$, and the second particle has traveled $\frac{2+3(t-1)}{2} \cdot t$ (the formula for the sum of the first $n$ terms of an arithmetic progression was used).

According to the condition, $\frac{2+3(t-1)}{2} \cdot t + 15 t = 295$, from which $t = 10 \text{s}$.

The angle of the pointer's deflection $\alpha = \frac{360^{\circ}}{\frac{60}{10}} = 60^{\circ}$.
Answer: $60^{\circ}$.