Does constant acceleration equal a parabolic graph similar to $y=x^2$, while jerk follows the path of $y=x^3$? I was just thinking about this possibility. We are always told in physics class that acceleration is constant or jerk is constant but is there a way to figure that out without being told so?

I'll give an example.

1s = .833m

2s = 6.666m

3s = 22.5m

If I don't tell you whether it's constant acceleration or constant jerk, can you tell me which one it is and how you can tell? In addition, how much information do you need to know in order to make this distinction?


Suppose you take any quadratic equation:

$$ y(t) = At^2 + Bt + C $$

for constant $A$, $B$ and $C$. The velocity is $dy/dt$ giving:

$$ v = \frac{dy}{dt} = 2At + B $$

and the acceleration is given by differentiating again:

$$ a = \frac{d^2y}{dt^2} = 2A $$

and since $A$ is a constant that means the acceleration has the constant value $a=2A$ i.e. any quadratic equation corresponds to constant acceleration. The constants $B$ and $C$ are related to the initial conditions: $B$ is the velocity when $t=0$ and $C$ is the displacement when $t=0$.

Although I'll skip the details, a similar argument applies to any cubic equation:

$$ y(t) = Jt^3 + At^2 + Bt + C $$

The jerk is $d^3y/dt^3$ and grinding through the algebra gives us:

$$ j = \frac{d^3y}{dt^3} = 6J $$

so again any cubic equation corresponds to motion with constant jerk:

You give an example of three pairs of points $(t, y)$ but it is always possible to fit a quadratic equation to any three points, so there is always some trajectory with constant acceleration that passes through your points. You need a minimum of four points to be sure whether the acceleration is constant or not. Likewise you need a minimum of five points to be sure whether the jerk is constant or not.

  • $\begingroup$ Perfect, thank you. Out of curiosity, why do you need 4 points to determine constant acceleration and 5 points for constant jerk? $\endgroup$ – rb612 May 12 '16 at 5:58
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    $\begingroup$ @rb612: suppose you have motion with non-constant acceleration. If you only provide three points it is always possible to fit a quadratic to those three points, and a quadratic has constant acceleration. So with three points you can't tell the difference between constant and non-constant acceleration. If you provide a fourth point then it isn't possible to fit a quadratic to just any four points. You can only fit a quadratic to four (or more) points if the trajectory really is a quadratic. $\endgroup$ – John Rennie May 12 '16 at 6:08

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