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Mechanics » Kinematics » Acceleration »

Acceleration

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.

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.

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