Just like velocity is the rate of change of position, acceleration is the rate of change of velocity. We will need to calculate acceleration since we want to not only measure motion, but also answer the question "What is responsible for motion?" It turns out that acceleration is the quantity that gives you the most insight into the causes of motion. (As you will see in future readings, for example: Newton's second law).

Since velocity requires a comparison of two positions, which means paying attention to the position at two different times, acceleration really requires paying attention to the velocity at two different times, which means considering the position at three different times. 

To see why we need three positions to construct an acceleration, consider a video of a ball being thrown upward by a juggler as shown in the video clip at the right.

Let's try to figure out the acceleration at frame 37 of the video. To do that, we need the velocity a little before and a little after. By comparing the positions at frames 36 and 37 we can infer a velocity at frame 36.5 (halfway between the two frames). By comparing the positions at frames 37 and 38 we can infer an average velocity in that time interval between frame 37 and frame 38. We can then also indicate that this is the velocity at frame "37.5" (the quotation marks are used since such a frame of course does not exist). By comparing the velocities at frames "36.5" and "37.5" we can infer an acceleration halfway between them  at frame 37. This is actually simpler in equations than in words. Instead of considering position, velocity, and acceleration as functions of time, let's use frame number to specify the time. So $y(37)$ means the y-coordinate of the ball in frame 37; $v(36.5)$ means the velocity of the ball halfway between frames 36 and 37. Using $v= Δy/Δt$ and $a = Δv/Δt$, and taking the time between one frame and the next to be $Δt$, we have

$$v(36.5) = \frac{y(37) - y(36)}{Δt}$$

$$v(37.5) = \frac{y(38) - y(37)}{Δt}$$

$$a(37) = \frac{v(37.5) - v(36.5)}{Δt} = \frac{y(38) - 2y(37) + y(36)}{Δt}$$

It's interesting to think about why it comes out like that. If the velocity were constant (zero acceleration), then how would the position change between frames 36-37 compare with the change from 37-38? What would that mean for the frame differences for a?

Just as with velocity, we will introduce two different kinds of acceleration — the average acceleration and the instantaneous acceleration.  The average acceleration is what we use when we are explicitly paying attention to the time interval; the instantaneous acceleration is what we use when the time interval we use to calculate the acceleration from the position (or the velocity) is very small compared to any times we want to pay attention to.

In either case, the acceleration answers the questions: how fast are you changing your velocity?

Read the two follow-ons for the details of average and instantaneous acceleration.

Joe Redish and Wolfgang Losert 9/7/2012


Article 328
Last Modified: July 12, 2019