A 1911 fires from a locked breach it isn't a blow back action.
Yes, it is. It's a "delayed blowback," - but it certainly is a blowback action.
The action DOES begin to move as soon as the bullet begins to move (simple physics). But the action moves less than a millimeter before the bullet leaves the barrel.
No, the physics are more complicated than that. From video evidence, it certainly does not move even a millimeter before the bullet is free from the barrel. Consider:
1) The interaction of the slide/cartridge base is, for the period the cartridge is trapped in the chamber, an "inelastic collision" in classical physics.
2) It's not a "pure" inelastic collision - there is tension besides the mass of the slide in the form of pressure from the recoil spring.
3) once the cartridge is allowed to ricochet off the ejector, you now have an elastic collision, and a different set of equations to apply.
3) peak pressure is not reached with the cartridge still in the case, it comes later, when the bullet is travelling already.
There is no single "simple physics" equation to deal with this complicated interaction, you have to use several and then integrate them. It is entirely within the physics of the situation that the slide in a given gun does not reach a point where its inertia and that of the spring are sufficiently overcome to start the motion with the bullet free of the barrel.