13.113. A material particle entered the pipe through an opening, and 6.8 minutes later, a second particle entered the same opening. Upon entering the pipe, each particle immediately began linear motion along the pipe: the first particle moved uniformly at a speed of 5 m/min, while the second particle covered 3 m in the first minute and 0.5 m more in each subsequent minute than in the previous one. How many minutes will it take for the second particle to catch up with the first?
## Solution.

Let $t$ be the time (in minutes) it takes for the second particle to catch up with the first. The distance traveled by the second particle is equal to the sum of $t$ terms of an arithmetic progression with $a_{1}=3, d=0.5$; therefore, $s=\frac{2 a_{1}+d(t-1)}{2} t=\frac{6+0.5(t-1)}{2} t$. The same distance traveled by the first particle is $5(6.8+t)=34+5 t$. Thus, $\frac{6+0.5(t-1)}{2} t=34+5 t$, from which $t=17$ (min).

Answer: 17 min.