# Is acceleration an average?

The acceleration of an object at a given time $$t$$ is, in fact, defined as the average acceleration of that object for a small interval around that time in the limit that the interval becomes "infinitely small." The mathematical concept of the limit makes the notion of "infinitely small" precise.

Suppose $$v(t)$$ represents the velocity of the object moving along a straight line at time $$t$$. If we want to find its acceleration at that time, what we do is to first determine its average acceleration for some interval of time between time $$t$$ and $$t+\Delta t$$ where $$\Delta t >0$$. This is just given by the change in velocity divided by the change in time over that interval; $$a_\mathrm{average}(t,t+\Delta t) = \frac{v(t+\Delta t) - v(t)}{\Delta t}$$ Next, we imagine taking that interval infinitely small; the result is the instantaneous acceleration at time $$t$$. This operation is made notationally precise by the following mathematical notation, and the precise mathematical notion can be defined in terms of the "delta-epsilon" definition found in standard calculus and real analysis texts: $$a(t) = \lim_{\Delta t \to 0} a_\mathrm{average}(t, t+\Delta t).$$ Putting our two equations together gives the definition of instantaneous acceleration: $$a(t) = \lim_{\Delta t\to 0} \frac{v(t+\Delta t)-v(t)}{\Delta t}$$ The right hand side of this expression is mathematically referred to as the derivative of the velocity function and is often denoted in physics with an overdot; $$\dot v(t) = \lim_{\Delta t\to 0} \frac{v(t+\Delta t)-v(t)}{\Delta t}.$$ This allows us to restate the definition of acceleration succinctly as $$a(t) = \dot v(t)$$ Or said in words, acceleration is the time derivative of velocity.