# Instantaneous acceleration

#### Prerequisites

Average acceleration tells us the change in velocity over some time interval. If we want to know the acceleration at a particular time, we have to make that time interval small. We still need two times in order to see a change in velocity, but if the time change is small enough, we can identify the acceleration as belonging to the time in the middle of the very small time interval.

When our $Δt$ is small enough, we identify the acceleration at that (central) time as the *instantaneous acceleration* and as the derivative of the velocity:

$$\overrightarrow{a}(t) = \frac{\overrightarrow{v}(t + ½ \Delta t) - \overrightarrow{v}(t - ½ \Delta t)}{\Delta t}$$

(for $\Delta t$ very small). Since we know that the instantaneous velocity is the derivative of the position vector, this makes the instantaneous acceleration equal to the second derivative of the position vector:

$$\overrightarrow{a}(t) = \frac{d\overrightarrow{v}}{dt} = \frac{d^2\overrightarrow{r}}{dt^2}$$

When we "open up" the vectors, we see that this vector equation stands for two statements about the acceleration: the x-acceleration is the derivative of the x-velocity, and the y-acceleration is the derivative of the y-velocity.

In terms of components, these can be written

$$\overrightarrow{a}(t) =a_x(t)\hat{i} + a_y(t) \hat{j}$$

$$\overrightarrow{v}(t) =v_x(t)\hat{i} + v_y(t) \hat{j}$$

We can put these together and identify the \hat{i} and \hat{j} parts of the vector equation:

$$\overrightarrow{a}(t) = \frac{d}{dt}\overrightarrow{v}(t) = \frac{d}{dt}(v_x(t)\hat{i} + v_y(t) \hat{j})$$

$$a_x(t)\hat{i} + a_y(t) \hat{j} = \frac{dv_x(t)}{dt}\hat{i} + \frac{dv_y(t)}{dt} \hat{j}$$

So matching the $\hat{i}$ and the $\hat{j}$ parts of the equation we get

$$a_x(t) = \frac{dv_x(t)}{dt} \; \; \; \; \; \; \; \; a_y(t) = \frac{dv_y(t)}{dt}$$

The x-part of the acceleration is the time derivative of the x-part of the velocity and similarly for y.

Joe Redish 9/10/11

#### Follow-on

Last Modified: August 5, 2018