The instant an accelerating object has zero speed, is it speeding up, slowing down, or neither? 
This problem is from Khan Academy. Specifically for the blue point circled in red, the answer is that at this blue point, the object is neither speeding up nor slowing down. When I think about the rule about the signs of velocity and acceleration and what this means for change in speed, this makes sense: if velocity and acceleration and have the same sign, the object is speeding up, and if velocity and acceleration have opposite signs, the object is slowing down. At the blue point, the instantaneous velocity is zero and because zero is neither positive nor negative, the object is neither speeding up nor slowing down. 
However, intuitively, this doesn't make sense. At the blue point circled in red, the velocity is zero so the speed must be zero. Also, the slope of the function at this blue point is negative so acceleration is negative, meaning that velocity is decreasing every second. If velocity is decreasing every second, then right after 6 seconds, the velocity will turn negative but the speed will have increased. So at the blue point, the object, even though it has zero speed, is in the process of speeding up. Why is this thinking wrong?
 A: At the first point prior to the one you circled, the object is slowing down. Its velocity instantaneously is dropping and approaching the zero axis. However, at $0 \, \frac{m}{s}$, the instantaneous velocity has ceased dropping (because it has now reached zero, it can't slow down more than $0\, \frac{m}{s}$) but hasn't yet begun speeding up in the negative direction.
You may wonder, "but how can it not be speeding up or slowing down if its acceleration is not zero?" As we know, acceleration is the slope of the graph. At the circled point acceleration is non-zero because the object is changing directions from the positive to the negative direction, not because it is speeding up/slowing down.
Positive and negative signs here don't refer to slowing down and speeding up; they refer to two directions - the positive direction and negative direction. If this was a position vs. time graph then negative would refer to a negative position relative to the zero position and vice versa for positive. Positive and negative as signs are used here to give you a 1 dimensional line along which you can move in two directions with the origin being an arbitrary point we call zero.
A good physical one dimensional analogy to this question (although its velocity curve would be linear and not curved) is a ball that is dropped vertically. After impacting  the ground it is moving $0 \,\frac{m}{s}$ and has lost all its downward speed but hasn't gained any upwards speed yet in that instant (it's in the "in between stage of speeding up and slowing down"). The mathematical explanation for this is that the derivative of the magnitude of $v$ (which determines if the object is speeding up/slowing down) is undefined. At that instant acceleration (the derivative of $v$, not the magnitude of $v$) is non zero and pointing upwards, acting to change the ball's direction of motion.
A: Something doesn’t seem correct. For a curve of velocity versus time the acceleration at any point on the curve is the derivative of the function, that is, the instantaneous slope of the curve at the point. 
At the circled point the slope is negative and not zero, indicating negative acceleration. So while the velocity at the circled point is zero it is still changing, in this case changing direction.
The slope of the curve corresponding to t=2 seconds, on the other hand, appears to be zero. That’s where the acceleration is zero. 
Hope this helps.
A: The tricky part of this question is that you are given a graph of velocity but asked a question about speed.
Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:

The above is the graph of $$ y = \left \lvert 4 - \left ( \frac{x - 2}{2} \right ) ^2 \right \rvert \text{,}$$ which is just the absolute value of the velocity graph in your screenshot.
This represents the fact that speed is the absolute value of velocity.
We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?
This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says, 

a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.
A: Speed is the magnitude of the velocity and hence always a positive quantity.  
If the terms “speeding up” and “slowing down” refer to the speed then at the time indicated on the graph the speed is zero and having reached a minimum (zero) the speed would have been increasing in the future.
The problem is that prior to that time the speed was deceasing to eventually reach zero and that means that a graph of speed against time is discontinuous at the time in question ie the gradient of the graph is not defined at that time.
So perhaps that is why the response "neither" was given as being correct?  
The rate of change of velocity (acceleration), the gradient of the velocity against time graph which is well defined, is negative and so at that time the rate of change of velocity (acceleration) is negative.
What was happening was that the direction of motion of the body changed at that time from moving in the positive direction to moving in the negative direction.
So one could say that the component of velocity, in the direction which was chosen to be positive, changed from being positive to being negative.
The label on the graph "velocity" is an abbreviation for "component of velocity in a chosen direction".  
If one looked at the speedometer (a devise which measures speed) just before the 
that the speed was decreasing and just after that time the speed was increasing but at the instant of time in question which one of the two options do you choose?
A: I will provide a bit more formal of an answer. "Speeding up" or "slowing down" typically refers to whether the speed of an object is increasing or decreasing. Imagine you are in a boat speeding down a canal (so that you may only move in one dimension - the canal is very narrow). At $t=2$, you flip a switch and your engine starts running in reverse. Here there is an instant where your acceleration is 0 before becoming negative, and this corresponds to the maximum on your velocity-time plot. Now, your engine is running in reverse and your boat is "slowing down" in the traditional sense. This corresponds to $2<t<6$ in your plot. There comes an instant where you have killed off all of your velocity, and you start running in reverse. After you kill off your velocity, your speed begins to increase (you are "speeding up"). 
In math, we can explain this as follows. The speed of an object is defined as the magnitude of its velocity. In 1 dimension, this is saying
$$
s = |v|
$$
i.e. speed is the absolute value of velocity. If we are interested in whether or not you are speeding up or slowing down, we want to find $ds/dt$. This can be done using the chain rule of ordinary calculus. First, we note that:
$$
|v| \equiv \sqrt{v^2} = (v^2)^{1/2}
$$
as a definition. We take the derivative now:
$$
\begin{align}
\frac{d|v|}{dt} &= \frac{d(v^2)^{1/2}}{dt}\\
&=\frac{1}{2}(v^2)^{-1/2}\cdot 2v\cdot \frac{dv}{dt} \\
&= \frac{v}{\sqrt{v^2}}\cdot \frac{dv}{dt} \\
&=\frac{v}{\sqrt{v^2}}\cdot a
\end{align}
$$
This final expression tells us a few things. First, we recover the rule that you are familiar with: namely, that if $v$ and $a$ have the same sign, then $ds/dt$ will be positive. If they have different signs, it will be negative. However, we also note that we have a discontinuity at $v=0$, which is the situation considered here. At zero, $v/\sqrt{v^2}$ jumps from $-1$ to $1$ and the derivative $ds/dt$ does not exist - the speed is formally undefined. This is known as the sign function $sgn(v)$, which returns the sign of the argument. Since the derivative of the speed does not exist at $v=0$ in one dimension, we are justified in saying that we are neither speeding up or slowing down. However, the velocity is decreasing this whole time, as evidence by the constant negative acceleration.
A: The object is neither speeding up nor slowing down, or both speeding up and slowing down. The graph you show plots velocity vs time, which can be converted to speed by taking the absolute value. The rate of change in speed is the derivative of the speed function. On the graph of speed, we can see a discontinuous corner at t=6 where the graph touches the x-axis and leaves again. The derivative at this point is undefined, making its interpretation rather nebulous - it's not negative, it's not non-negative, it's not positive, it's not non-positive, it's not zero, it's simply undefined.
We can argue that this undefined derivative isn't positive (i.e. speeding up), nor is it negative (i.e. slowing down), so the object is neither speeding up nor slowing down. But we can equally well argue that the derivative is not negative and not zero (i.e. speeding up) and that the derivative is not positive and not zero (i.e. slowing down), so the object is both speeding up and slowing down.
But really, the rate change of the object's speed is undefined at t=6. We cannot say anything meaningful about the rate of change in the object's speed at t=6, as we literally cannot define it. Even the reasoning I use the previous paragraph is rather specious - there isn't really any basis to claim that an undefined quantity is not positive or non-zero. Undefined > 0 and Undefined < 0 aren't false or true statements, they simply cannot be evaluated. It's like asking if a sandwich is positive or negative - the term just doesn't apply.
A: In my opinion khan academy is majorly correct . You may think that the acceleration at $$v=0$$ 
 that is the derivative at the point where the velocity is zero is negative then it must be slowing down.but the term speeding up or speeding down means that if the modulus of velocity is increasing or decreasing.
Think physically
 this graph shows the body speeding down in a direction then stopping and then reversing it's direction to speed up in an opposite direction.
 so before the moment it stops it speeds down.
 after the moment it stops it speeds up
 and at the moment it stops it neither speeds up or speeds down.
That is $$d/dy{|v|}=0$$
Edit:
Thanks to dmckee to suggest that actually  the graph of |v| would have kink at v=0
And thus the graph is non differentiable at 0
So I'd want to add
I didn't thought in mathematical way but physical way and in any real physical system kinks in graph or undefined limits are not possible
various cases
.so yeah I should not use the word absolutely .I think the question was made to think about the change in velocity in more physical way not mathematical one . But we should suggest khan to modify the graph such that there would not be any kinks in |v|
A: This is the chart of the movement of an airplane which transitions into  "beta" or thrust reverse at time 2, and steadily increases thrust reverse from then onward. The aircraft slows to a stop, but does not deselect thrust reverse but continues advancing it. So the instant it stops, it heels back and starts rolling backwards down the runway.
At speed 0, the circled point, the reverse thrust levers have not been zeroed and continue to be advanced steadily.
That's pretty clear.  acceleration is a linear function (straight line) here, crossed zero at time 2, and it’s definitely nonzero at time 6.  
The only thing that makes time 6 interesting is this is when velocity crosses the zero line.   That, plus $6, will get you a small coffee at Starbucks, but it has no bearing on acceleration.  At time 6/velocity 0, it is definitely accelerating. 
The plane is stopped, the thrust reversers are howling, and the tower is wondering what the pilot intends. (Missed exit?)   So is velocity increasing at the circled instant? It is zero and in the next instant his velocity will be more, so yes.  Sounds like increasing to me. 
No, you say? Then the question of "speed increasing" boils down to semantics.  It seems like a trick/dodge question. 
A: Am assuming one dimensional motion so that when the speed is $+ve$ it is moving away from some fixed point and when it is $-ve$ it is moving in the opposite direction, back towards the fixed point.
Between $t = 0 s$ and $t = 2 s$ the slope of speed versus time is $+ve$ so that the particle is increasing its speed. At $t= 2 s$ the speed is $+ 4 m/s$ but it's acceleration is zero. Its speed is neither increasing or decreasing at this time.
For $t > 2 s$ the slope is negative so that the speed of the body is decreasing. It comes to a momentarily stop at $t = 6 s$ but its velocity appears to become more $-ve$ as time increases. Remember, velocity is "speed + direction" so that it's actual speed (magnitude of velocity) is getting larger.
