Exercise 8d

The position vector \(\vec{r}(t)\) is given by:

\[\vec{r}(t) = \left( \begin{matrix} c\, (t- \sin t) \\ c\, (1 - \cos t) \end{matrix} \right)\]

The velocity is the derivative of the position vector with respect to time: \(\vec{v}(t) = \dot{\vec{x}}(t)\)

\[\vec{v}(t) = \left( \begin{matrix} c\, (1- \cos t) \\ c\, \sin t \end{matrix} \right)\] \[ \left| \vec{v} \right| = c\, \sqrt{(1- \cos t)^2 + \sin^2 t } = c\, \sqrt{ 2(1-\cos t)}\]

The acceleration is the derivative of the velocity vector with respect to time: \(\vec{a}(t) = \dot{\vec{v}}(t)\)
\[\vec{a}(t) = \left( \begin{matrix} c\, \sin t \\ c\, \cos t \end{matrix} \right)\]

Animation

position vector: red
velocitiy vector: blue
accelaration vector: green

Dimensions: 800 x 500 pixels, c = 75 Pixel, t in seconds (real time). The duration of the animation is one periode, this is \(2 \pi\) seconds. At \(t=0\) the position vector starts at the origin. The x component grows, the y component oscillates between 2c and 0. The intial velocity is 0. Please note the acceleration vector is moving on a circle with radius c.