Calculation of an exact answer would be doable by a good A level maths or Physics student;
Method would be something like this:
Envisage both aircraft heading north at the waypoint. Set up an X Y coordinate system about the waypoint. Perhaps: X is aircraft distance east of the start point. Y is aircraft distance North of the start point.
You then some equations that tell you where each aircraft is in space at time T. This is not that tricky if you view it as circular motion and invoke basic trigonometry.
Proportion of turn completed at time T = Speed X Time / circumference of turn
(convert TAS in knots into feet per second)
(circumference = 2 x pi x Turn Radius (10 or 20K depending on the aircraft)
The multiply by 360 to find out how many degrees around the turn the aircraft has gone, which since we started on North is actually the heading. Call it HDGa for aircraft A etc
You can then work out X and Y positions in space, bearing in mind the centre of the turn is 10 or 20K to the east of the waypoint: The Y distances may be negative, don't worry about it just apply + - rules rigorously.
Aircraft A:
Xa = 10,000 - (10,000 * Cos HDGa)
Ya = 10,000 * Sin HDGa
Aircraft B:
Xb = 20,000 - (20,000 * Cos HDGb)
Yb = 20,000 * Sin HDGb
Then work out the difference between them, first in each dimension
Xseperation = Xa - Xb
Yseperation = Ya - Yb
Then two dimensionally by using pythagoras
Seperation = square root of ((Xseperation^2) + (Yseperation^2))
For two aircraft at 250 kts, one with a 10 K radius of tuen the other 20K, after 5 seconds the position differnce would be 111.2 feet.
(of course that's a silly level of accuracy).
(I think thats correct, haven't check it rigorously)
pb