2-p and LB, thorns and barbs aside, you two have done a commendable job of debating the issue from the Thomas-Burrard points of view. Thanks.

2-p, the acceleration will follow the wad base force, as generated by the pressure, minus friction drag. The amount of friction drag will be a function of sliding coefficient of friction between the plastic wad's obturation section and the hull wall followed by the steel barrel (think pretty low) and the wad's velocity. 10,000 psi is going to exert about 4200 force pounds on the wad base (0.42 square inches). Suppose that about 0.2" of the wad skirt is in contact with the hull wall/barrel and has a cirumference of 2.3" for a contact area of 0.46 square inches. So, the coefficient of friction would have to be 0.91 (0.42/0.46) to stop acceleration. Another way to look at this issue would be the possibility of the wad holding the force on its base with friction. Suppose the thickness of the wad skirt is 0.050" and has a circumference of 2.3". The area available to resist the 4200 lbf load is 0.115 square inches. 4200 lbf over 0.115 sq in requires a tensile strength of 36,500 psi. Both the case of a friction coefficient of 0.91 and a tensil strength of 36,500 psi are implausible. Somebody please check these numbers. Unless I calculate wrong, the acceleration is going to follow the pressure curve pretty close for the first part of the shot.