It can be solved using calculus.
The integral of acceleration is velocity, and the integral of velocity is distance.
Asg (acceleration of Sir George) = .377 m/s^2 * t^0
Vsg = 3.77t + 0t^0 (there is no initial velocity)
Dsg = (3.77/2) t^2 + 0t^1 +0t^0 (there is no initial velocity nor is there an initial distance)
Asa (acceleration of Sir Alfred) = -0.364 m/s^2 * t^0 (it's negative because he's moving in the opposite direction of Sir George)
Vsa = -0.364t
Dsa = (-0.364/2) t^2 + 80.5 (the initial distance is 80.5 meters away from Sir George)
Now, just set Dsg and Dsa equal to each other and solve for t, and you should come up with your answer in terms of time. You can then go back and plug the value for t into either distance equation and it should give you your answer in meters, and distance from Sir George's starting point. I'll let you work it out from here.
Chris