Why isn't acceleration always zero whenever velocity is zero, such as the moment a ball bounces off a wall? The answer to this homework problem is $\left(\text{D}\right) :$



I understand why we rule out options $\left(\text{A}\right)$ and $\left(\text{B}\right)$ first. However, I don't get why $\left(\text{D}\right)$ is the answer.
Won't the ball's velocity at one point be $0$ when it comes in contact with the ceiling? (The same way as its velocity momentarily is $0$ when it is thrown downwards and comes in contact with the ground before bouncing back upwards). That should give zero acceleration at that point, but none of the acceleration profiles include $0 .$
 A: 
Won't the ball's velocity at one point be 0 when it comes in contact with the ceiling? (The same way as its velocity momentarily is 0 when it is thrown downwards and comes in contact with the ground before bouncing back upwards).

Yes. Its velocity will momentarily be zero.

That should give zero acceleration at that point[...]

No! Velocity being zero says nothing about the acceleration. Just like your position being zero (i.e. when returning home after a day at work) says nothing about your velocity (you could be running in high speed when arriving at your starting position, or you could reach it slowly, or you could stand still at the starting spot. Different velocities are possible where the position is zero; position and velocity are unrelated, and so are velocity and acceleration).
When throwing a ball up, the ball will also momentarily reach zero speed before falling back down to your hand. But gravity is there all the time causing a non-zero acceleration all the time - also when the speed is zero. So, acceleration can't be understood from the value of velocity, only from the change in the value of velocity.
And why the peak then in answer D? Because,


*

*while flying upwards, gravity causes a constant downwards acceleration. The ball slows down at a constant rate.

*When hitting the ceiling, the ceiling suddenly slows down the ball instantly. That requires a much larger downwards acceleration in that instant in order to reduce the speed to zero in very short time. Thus the peak on the graph.

A: It is often supposed that because a quantity (velocity in this case) is zero the rate of change of that quantity (acceleration in this case) is also zero.
If that were true a ball which is thrown upwards would stay at its greatest height above the ground when it was not moving and never come back to the ground.  
Just suppose that you were correct and at the instant the velocity of the ball was zero the acceleration $( = \frac{\text{change in velocity}}{\text{time}})$ was also zero.  
This means that at some instant if time the velocity of the ball does not change from being zero ie the ball is not moving and will stay in a position of rest on the ceiling for all time.
A: You could consider the question in terms of the forces acting on the ball.
During the entire time a downward force of gravity acts on it.  While it is in contact with the ceiling there is also a downwards normal force.  At no time are there balanced forces acting on the ball
This results in an acceleration is always downwards with a spike in magnitude while the ball is in contact with the ceiling.
