Acceleration of an instantaneous centre is always 0? I came across the following question recently:

My understanding of the question is that it wants us to show that end A is an instantaneous centre.
The mark scheme does so by showing that A has no linear acceleration (A also has no velocity - the instant shown is the instant the force P was applied, and prior to that, the rod was at rest, so at this instant, linear and angular velocity of the bar is assumed to be 0):

This is confusing to me - is it always true that an instantaneous centre at a particular instant has 0 acceleration? I know that an IC always has 0 velocity, but I didn't know a similar property applied for acceleration.
Or am I misunderstanding the question, and this isn't related to instantaneous centres?
 A: In this case, the instantaneous tangential acceleration of end, A, relative to the center of mass is equal and opposite to the linear acceleration of the center of mass. Since it starts with no velocity, that does make it an instantaneous center of rotation.
A: For an unconstrained body, the motion is typically described as translation of the center of mass (CM) with rotation about the CM.  The translational motion of the CM can be determined using $$(1)\vec F_{net} = M\vec a_{CM}$$ where $\vec F_{net}$ is the net external force, $M$ the total mass, and $\vec a_{CM}$ the acceleration of the CM.  The rotational motion about the CM can be determined using $$(2) {d\vec J_{CM} \over dt} = \vec N_{CM}$$ where $\vec J_{CM}$ is the angular momentum with respect to the CM and $\vec N_{CM}$ is the total external torque with respect to the CM.  (2) is valid even of the CM is accelerating.
For your problem, consider the motion for a short time $\Delta t$ after the force $P$ is applied.  Using (1) the velocity of the CM after $(3) \Delta t$ is $v_{CM} = {P \Delta t \over M}$ up.  Using (2), the angular velocity of the rod with respect to the CM after time $\Delta t$ is $\omega = {{2P \Delta t} \over {ML}}$, and the velocity of point A with respect to the CM is $ (3) \omega {L \over 2} = { {P\Delta t} \over M}$ down.  The total velocity of A after the short time $\Delta t$ is $(3) up \enspace – (4) down = 0$.  Since A has total velocity $0$, after a short time, a line into the page at A is an instantaneous fixed axis of rotation.  This is not a permanent fixed axis of rotation since A will move if P is continuously applied.
