If change in position over time is average velocity, why doesn't change in position over time squared equal average acceleration? For example, let's say a car is experiencing an acceleration of $1$m/s$^2$, for $6$ seconds so it goes $18$m. Now the average velocity is found through dividing $18$m by $6$s which is in line with the formula $v_\text{avg} = \frac{\Delta x}{\Delta t}$. And indeed, the average velocity is $3$m/s.
Acceleration has units of distance divided by time squared, however the average acceleration is not $18/6^2 = 0.5$m/s$^2$, the average acceleration is $1$m/s! So I have two questions from this:


*

*What exactly is that $.5m/s$ signifying? I know the kinematic equations including $\Delta x = v_0t+\frac{1}{2}at^2$ and this would allow us to find acceleration. 

*Aren't the units a bit deceiving on acceleration? Maybe I'm just not super comfortable visualizing second derivatives yet but if I have an $m/s^2$ I feel like I should be able to plug in meters and seconds and get the average acceleration. And I feel like it's part how we define the units also. Because:
$$v = \frac{\Delta x}{\Delta t}$$
$$a = \frac{\Delta v}{\Delta t}$$
Substitution then gives us:
$$a = \frac{\Delta\frac{\Delta x}{\Delta t}}{\Delta t}$$
Which is different than the units seem to imply, from my perception.
 A: The glitch in your logic is that you supposed acceleration to be distance/time squared while using your formula: $$ \Delta x = v_0t+\frac{1}{2}at^2  $$ 
The above formula gives $$  a= \frac{2\Delta x}{t^2}$$ supposing that vo is equal to zero.
This makes the acceleration equal to the average acceleration in your case since the car is moving in one direction and with a constant acceleration.
A: Why your computed average acceleration is wrong?
the average acceleration is defined as: 
$\overline a=(v_2-v_1)/(t_2-t_1)$ 
where the $v$'s are instantaneous speeds. If you start with zero initial speed you can simplify it to 
$\overline a= v /t $ 
$v$ is still the instantaneous speed at $t$. For a constant acceleration $v=at$ so in this case you recover the fact that $\overline a=a $, as expected.
But if instead of using the instant velocity $v$ you now use the average velocity $\overline v$, then you will get the wrong result:
$\overline a_{new} = \overline v /t=(x/t)/t=x/t^2 $, where again, $x$ is the instantaneous position. For constant acceleration $x=at^2/2$, and so you get   $\overline a_{new} =a/2$.
So the short answer is that by calculating average acceleration as $x/t^2$ you are not really using the correct definition of average acceleration because at some point you replaced an instantaneous speed by an average speed.
