# Effective Acceleration for Non-Constant Acceleration Motion

This question uses the same symbols as "Effective Acceleration" is Distance-Averaged Acceleration?.

One of the kinematics formulas for constant acceleration is: $\Delta x=v_0*\Delta t+\frac{1}{2}a\Delta t^2$.

Now, in a scenario of non-constant acceleration, given $v_0$, $\Delta x$, and $\Delta t$, we can find an effective acceleration $a_{eff}$ by solving $\Delta x=v_0*\Delta t+\frac{1}{2}a_{eff}\Delta t^2$.

Is $a_{eff}$ an average of some kind? In particular, is it the time-averaged acceleration, distance-averaged acceleration, or neither?

I was not able to immediately determine the answer from the formula's derivation here: Find Distance Function from Acceleration Function.

Since $v(t) = v_0 + \int_0^t a(\tau) d \tau$, we have $x(t)-x_0 = \int_0^t v(\tau) d \tau = v_0 t + \int_0^t \int_0^\tau a(s) ds d \tau$.
Then the equivalent constant acceleration $a_\text{eff}$ that would produce the same change of position is given by solving $\int_0^t \int_0^\tau a_\text{eff}\, ds d \tau = {1 \over 2} a_\text{eff} t^2 = \int_0^t \int_0^\tau a(s) ds d \tau$.
Assuming that $a$ is integrable, we have $\int_0^t \int_0^\tau a(s) ds d \tau= \int_0^t \int_s^t a(s) d \tau ds = \int_0^t (t-s)a(s) ds$, so we see that $a_\text{eff} = { \int_0^t (t-s)a(s) ds \over \int_0^t (t-s) ds}$ and we can interpret $a_\text{eff}$ as a weighted average of some sort.
• In fact $a_\textrm{eff}$ is a time-averaged value weighted by the "time to go" (that is, the amount of time remaining before we have to mark the "final" position). For example, the amount of acceleration $10$ seconds before the end has ten times as much effect as the amount of acceleration $1$ second before the end, because the velocity change caused by the earlier acceleration has that much more time in which to change the particle's position. – David K Jan 24 '15 at 3:14
• I think it's also the first moment of $a(t)$ about $t$. – David K Jan 24 '15 at 13:14