Average acceleration versus instantaneous acceleration Does instantaneous acceleration also changes when direction of velocity changes. I know average acceleration does change.
 A: A nice way to compare both is to invoke the definitions:
$${\vec a}_{\rm avg} = \frac{\Delta {\vec v}}{\Delta t}$$
and 
$${\vec a}_{\rm inst} = \lim_{\Delta t \rightarrow 0}\frac{\Delta {\vec v}}{\Delta t} = \frac{d{\vec{v}}}{dt}$$
Graphically,

and if you consider change over an infinitesimal time period $\Delta t \rightarrow 0$, the same definition gives you the instantaneous accelaration.  
Thus, in instantaneous acceleration, you are only shrinking the considered time interval to an infinitesimal. While that means that your instantaneous acceleration could have changes many times in the larger time duration $\Delta t$, on route to producing the net average acceleration, it doesn't change anything as far as the general conclusion of acceleration changing with a change in the velocity direction only, even if the magnitude is preserved. This originates from the fact that in both these definitions, acceleration is defined as the change in a vector, and  a vector can change even when its magnitude is held constant, via a change in its direction. For example, $v = 4$ units along x-axis, and $v = 4$ units along y-axis, are two different vectors.   
Hope that helps.

{Image source = Wikipedia}
A: The instantaneous acceleration is the time derivative of the velocity vector:
$$ \vec{a} = \frac{d\vec{v}}{dt} $$
If the velocity is changing then the acceleration will be non-zero.
