The time it would take the cyclist, travelling at speed v, to reach the end of the tunnel, length x, as the truck is approaching is 2x/(5v). In that time the truck travels 80*2x/(5v)=160x/(5v), so this is how far the truck is initially from the tunnel exit. The truck has to travel the length of the tunnel plus this starting distance in the same time that the cyclist has to reverse direction and head out of the tunnel in the same direction as the truck. The time for the cyclist to cover 3x/5km is 3x/(5v) and this is equal to the time the truck takes to reach the entrance=(x+160x/(5v))/80. Therefore, 3x/(5v)=(x+160x/(5v))/80=(5vx+160x)/(400v). The x's cancel out: 3/(5v)=(5v+160)/400v. Multiply through by 400v: 240=5v+160; 5v=80; v=16km/h.
Check: let the length of the tunnel be 80m (it doesn't matter what length we choose because the length is arbitrary). The cyclist reaches 3/5 of the length=48m. He has 32m further before the exit. It will take him 0.002hr=0.002*3600=7.2 secs to reach the exit. In that time the truck would travel 0.002*80km=160m to meet the cyclist at the exit. When the cyclist turns round and heads back, the truck has to cover 160+80=240m to catch up. It takes the cyclist 0.048/16=0.003hr=10.8 secs to reach the entrance. It takes the truck 0.240/80=0.003hr=10.8 secs to catch up. So the cyclist reaches the entrance at the same time as the truck.
