imc

First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds. $$4.8x-4.4x=500\Rightarrow x=1250$$ Because $4.4(1250)=5500$ is a multiple of 500, it turns out they just meet back at the start line. Now we must find a time that is a multiple of $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\boxed{2,500}$ seconds.