What is an object's velocity with zero acceleration after positive acceleration? I am not 100% sure I'm understanding this graph correctly. Here are my 2 theories.
An object has constant +acceleration and is moving towards some direction.

If the object's acceleration is cut off to 0:

1) Would it mean the object is still in motion from its previous force?
 (example:   after car stops accelerating it is just rolling)

Proof: If so its moving at constant velocity and then Δv = 0 so a = Δv / Δt will be zero ?
2) Or does it mean the object immediately stops motion?
 (example:   a car stops accelerating immediately because it brakes)

Proof: If so mathematically this would be velocity V = a*t  which is V = 0*t and you get V = 0 m/s


 A: You have a fundamental misunderstanding about the relationship between velocity and acceleration.


*

*Velocity is the change in distance over time ($m/s$)

*Acceleration is the change in velocity over time ($m/s^2$)


Perhaps answering the questions you posed will help you understand this relationship better:

1) Would it mean the object is not accelerating but still in motion from the left over force? (example; after car stops accelerating it is just rolling)
  In this case there seems like there would be velocity just slowing down. 

Yes, if acceleration dropped to $0$ the object would still be in motion, but not because of some 'left over force', because the velocity of the object would not longer be changing. Look back to the relationship I described above, acceleration is the change in velocity over time, not just the velocity over time. If acceleration is $0$, the velocity is not changing. If the velocity is constant ($0$ acceleration) then the object will continue without slowing down or speeding up.

2) Or does this graph describe an object who is accelerating @ 0s-10s and stopping all motion @ 10s-15s? In this case mathematically makes sense because if you solve for velocity V = a*t you get 0 m/s

The graph describes an object which is accelerating from $0s$ to $10s$, and not accelerating from $10s$ to $15s$. As in the last question, the fact the the object is not accelerating does not mean that the velocity is $0$, it means that the velocity remains constant at whatever value it was just before acceleration became $0$.
Additionally, $V \ne a*t$, because the relationship is not between velocity and acceleration, but change in velocity and acceleration. The proper formula would be $\Delta V = a*\Delta t$, where $\Delta t$ would in this case be $15s - 10s = 5s$ This would mean that $\Delta V = (0m/s^2)*(5s) = 0$, so velocity would not change, but would remain constant.
A: 
Would it mean the object is still in motion from its previous force? (example; after car stops accelerating it is just rolling) 

Yes. As soon as acceleration disappears, the object just continues with whichever speed it has. The acceleration changes speed; is there no acceleration, then speed is constant.
Speed is how many extra metres you are moving every second. Acceleration is how many extra metres per second you are moving with every second. No acceleration then just means that no more metres per second are added or subtracted. The object just keeps what it has. 

Proof: If so its moving at constant velocity and then Δv = 0 so a = Δv / Δt will be zero ?

Sure.

Or does this graph describe an object who is accelerating @ 0s-10s and stopping all motion @ 10s-15s? 

Nope. That would be the case, if the speed was cut off to 0 suddenly.

Proof: If so mathematically this would be velocity V = a*t which is V = 0*t and you get V = 0 m/s

True. But only if that was the case; which is wasn't.
