accelerationkinematicsvelocity
Can you have a zero velocity and nonzero average acceleration?
I am confused with the word "average" here. If the question would be, "Can you have a zero velocity and nonzero acceleration?" my answer would be yes. An example would be a ball thrown upward. At the highest point, the velocity is zero and instantaneous acceleration is -9.8 m/s$^2$. Since the question states that average acceleration, I can't think of an example that would satisfy the question.
The acceleration is the time derivative of the velocity:
$$ a = \frac{dv}{dt} $$
so if the velocity does not change with time the acceleration is necessarily zero. Since in your example the velocity is constant during the interval that means $dv/dt = 0$ and therefore that $a = 0$ during the interval.
The velocity doesn't have to be zero. Any constant non-zero value will also give $a = 0$.
You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time:
Acceleration is defined as:
$$ a = \frac{dv}{dt} $$
so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is constant. The gradient is just the gravitational acceleration $9.81$ m/s$^2$.
The point is that the gradient, and hence the acceleration, does not depend on $v$ at all. So it is the same value of $9.81$ m/s$^2$ when $v = 0$ just as it is at all other values of $v$.
Best Answer
If you by velocity mean instantaneous velocity, then the question makes no sence. The corresponding acceleration will as well be instantaneous (and the answer would be yes.) $$\lim_{\Delta t \to 0}a_{av}=a_{inst}$$
If you mean average velocity, then the answer is no. Average acceleration doesn't take into account what happen in between; only the end points are interesting: $$a_{av}=\frac{v_2-v_1}{\Delta t}$$ If average acceleration is non-zero, then $v_1 \neq v_2$ and the average of these is surely non-zero as well.