Definition
The acceleration of a body is defined as the change in its velocity per time interval. Analogous to the definition of velocity, the average acceleration, or acceleration $a$ (from English acceleration), of a uniformly accelerated motion is:
$$a = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$$
and the instantaneous acceleration for non-uniformly accelerated motion is:
$$a = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{\mathrm{d}v}{\mathrm{d}t} = \dot{v}$$
The unit of acceleration is m/s${}^2$, which can be illustrated as follows: During accelerated motion, the velocity of a body changes its value per elapsed time. An acceleration of $1\,\mathrm{m/s^2}$ means that the velocity increases by $1\,\mathrm{m/s}$ every second. Acceleration can also have a minus sign. In this case, it is referred to as negative acceleration or deceleration, as the velocity decreases.
Acceleration is also a vector quantity with magnitude and direction. The same applies to movement in three-dimensional space.
$$\vec{v} =\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\begin{pmatrix}\dfrac{\mathrm{d}v_x}{\mathrm{d}t}\\ \dfrac{\mathrm{d}v_y}{\mathrm{d}t}\\ \dfrac{\mathrm{d}v_z}{\mathrm{d}t}\end{pmatrix}$$
The three velocity components are denoted here and below by $v_x$, $v_y$ and $v_z$. Inserting the equation above into this relationship yields:
$$\vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2}$$
The acceleration is therefore identical to the second derivative of the distance or position vector with respect to time.
Acceleration
The acceleration of an object is in general defined as the first derivative of $\vec{v}$ and the second derivative of $\vec{r}$.
$$\vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2}$$
Distance Calculation
If an object accelerates with a constant acceleration, the increase in speed is given as
$$\Delta v = a \Delta t$$
The covered distance can be calculated according to the average speed $\Delta v$ within this time interval:
$$\Delta s = \frac{1}{2} \Delta v (t_2 + t_1)$$
Now we can insert $\Delta v$ which results in
$$\Delta s = \frac{1}{2} a (t_2 - t_1) (t_2 + t_1)$$
Using the third binomial formula gives us
$$\Delta s = \frac{1}{2} a (t_2^2 - t_1^2)$$
In case, the start time $t_1$ is equal to 0, we can directly write
$$s(t) = \frac{1}{2} a t^2$$
As we can see, we can have a direct dependency between the covered distance and the time.
We can obtain the same result by integrating $v(t)$ with respect to $t$:
$$s(t) = \int_0^t v(t)\,\mathrm{d}t = \int_0^t at\,\mathrm{d}t = \frac{1}{2}at^2$$
which makes sense, as the integral is simply the area under the $v$-$t$-diagram.
Distance Calculation
If an object is accelerated with a constant acceleration $a$, its position after the time $t$ can be calculated according to the following quadratic relationship:
$$s(t) = \frac{1}{2}at^2$$
Experiment
A glider on an air-cushion track is connected to various weights via a thread running over a pulley, which are accelerated due to gravity. The position of the glider can be measured at different times using light barriers. If the measured values are transferred to a distance-time diagram, a quadratic relationship between distance and time results.