Beanyboy said:
The definition you've provided is one I'm vaguely familiar with. If I had to guess, I'd say that it's Instantaneous Acceleration. So, what's that? I'm guessing the rate of acceleration for a time interval that's infinitesimally small.
But,back to the question: Why will a body released from rest always move in the direction of acceleration? I noticed that in defining the acceleration we now have acceleration clearly notated as a vector; the direction is preordained. Consequently, on the right hand side of the equation, we must also have a vector quantity, which here, is the velocity. So, the change in position - the move in a particular direction - is defined, is determined by the acceleration vector. Motion must be in the direction of the acceleration.
Your answer is a good example of what pointed out earlier, namely that novices tend to rely on preconceptions and primitive but not well-formed notions about the physical world that put up a barrier to understanding it. It may look like I am criticizing you in what I will say below, but I that's not my intention. I just want to bring you face to face with the problem so that you can do something about it. Fixing it has to come from within, I cannot do it for you.
I gave you a definition for the acceleration and I asked you to interpret it. You responded that you are vaguely familiar with it, which probably means that you have seen it but ignored it. That equation is one of the first equations one sees in an intro physics text and is mostly ignored by most students because they deem that it's not useful because "it does not allow one to calculate anything" like the SUVAT equations. That's a normal reaction as most students believe that they are the sole arbiters of what they should know. Now look at your response when I asked you to interpret it and then used it. You did neither probably because you couldn't do it either. Instead you invoked "preordainment" and begged the question by asserting what you had to show. Aristotle asserted that masses fall because it's in their nature to fall. So what could you have done instead? I will spell it out.
(a) Identify the symbols in the equation.
Left side is the acceleration vector. Right side is the ratio of the change in velocity (final minus initial) over an infinitesimally small time interval.
(b) What is is the equation saying in plain English?
The acceleration is the same as the change in velocity over a time interval in the limit that this time interval is made very small. In other words, at any instant in time you can get the velocity at the next instant, by adding to the existing velocity vector a vector equal to the acceleration vector multiplied by the time difference between instants. That comes from ##\Delta \vec v =\vec {v}_f-\vec {v}_i= \vec a \Delta t## which gives ##\vec {v}_f=\vec {v}_i+\vec{a}\Delta t.##
(c) Special case, mass instantaneously at rest and it is understood that ##\Delta t## is very small.
When a mass is instantaneously at rest, ##\vec {v}_i=0.## Then the equation becomes ##\vec {v}_f=\vec{a}\Delta t.## The vector on the left is the same as the vector on the right. Two vectors are the same when they point in the same direction. Thus, this equation says that the instant after a mass is instantaneously at rest its velocity will be in the same direction as the acceleration vector. Furthermore, if the acceleration itself is zero, the mass will remain at rest.
There is quite a bit in that simple definition for the acceleration. I would strongly recommend that you learn how to interpret equations because they are shorthand notation of how we think the physical world is put together.