3.16 A material particle entered the hole of a pipe, and 6.8 minutes later, a second particle entered the same hole. 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 did it take for the second particle to catch up with the first?
3.16 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, where $a_{1}=3, d=0.5$; 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).