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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.

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  • $\begingroup$ "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? $\endgroup$ – lucas Jul 11 '16 at 5:58
  • $\begingroup$ @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. $\endgroup$ – rb612 Jul 11 '16 at 5:58
  • $\begingroup$ 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? $\endgroup$ – WillO Jul 11 '16 at 5:59
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    $\begingroup$ 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. $\endgroup$ – John Rennie Jul 11 '16 at 5:59
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    $\begingroup$ 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. $\endgroup$ – knzhou Jul 11 '16 at 6:11
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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.

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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.

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