# 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:

1. 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.

2. 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.

• "let's say a car is experiencing an acceleration of $\frac{1m}{s^2}$, for 6 seconds" When you talk about constant acceleration, how the average acceleration can be less than the constant value? – lucas Jul 11 '16 at 5:58
• @knzhou You're right. It doesn't make sense, because instead of being "change in change in meters over change in time1 over change in time2, we simplify it to $m/s^2$, but I don't quite understand how we can do this when the change in time1 and change in time2 are different. – rb612 Jul 11 '16 at 5:58
• Presumably there is some reasoning that has convinced you that average velocity is $\Delta x/\Delta t$. Why don't you try carefully applying the same reasoning to calculating the average acceleration and see what happens? – WillO Jul 11 '16 at 5:59
• Hi rb612. I've corrected a couple of typos and I hope clarified what you are asking. Feel free to back out my changes if you don't like them. – John Rennie Jul 11 '16 at 5:59
• The thing is, there are a million things that are seconds. Like the age of the Sun, or the amount of time I brushed my teeth for, or 3 times the amount of time the car was accelerated. What stops me from dividing by $\pi (\Delta t)^2$? That has the right units too. – knzhou Jul 11 '16 at 6:11

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.

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.