The change in magnitude of centripetal acceleration When an object (e.g. racecar) moves around in circles with constant tangential velocity, constant centripetal acceleration is present.
What happens to the centripetal acceleration when the racecar is at rest, then increases its speed? I know that the tangential velocity increases due to the tangential acceleration, but what about the centripetal acceleration?
Since centripetal acceleration is tangential velocity squared divided by the radius, and the tangential velocity is increasing from rest, the centripetal acceleration must then be increasing as well.
How do you calculate the values for centripetal acceleration if it is changing? There doesn't seem to be a formula for it. And it seems that centripetal acceleration is changing, is there a term for the rate of change of it?
 A: As you have stated, the centripetal acceleration is given by:
$$a_c=\frac{v^2}{r}$$
where $v$ is the magnitude of the velocity (technically it is the magnitude of the tangential velocity, but I will assume we stay on a circle of radius $r$). 
Therefore, if the velocity is a function of time $v=v(t)$, then the centripetal acceleration will be
$$a_c(t)=\frac{v(t)^2}{r}$$
What determines $v(t)$ is the tangential acceleration $a_T$ according to
$$v(t)=v(0)+\int_0^t a_T(t')\ \text d t'$$
(Note this is because $a_T=\frac{\text d v}{\text d t}$. It is not derived from the above equations).
What determines these acceleration components is, of course, the centripetal and tangential components of the net force, but if you know what the tangential force is, then you could determine what centripetal force is required to keep the object moving in a circle of radius $r$ using the equations above.
A: In polar coordinates, the position vector from the center of the circle to the particle is $$\mathbf{r}=r\mathbf{i_{r}}(\theta)$$where r is the radius of the circular path and $\mathbf{i_{r}}(\theta)$ is the unit vector in the radial direction at polar angle $\theta$.  
The velocity of the particle is the derivative of the position vector with respect to time, and is thus given by:$$\mathbf{v}=\frac{d\mathbf{r}}{dt}=r\frac{d\mathbf{i_{r}}(\theta)}{dt}=r\frac{d\mathbf{i_{r}}(\theta)}{d\theta}\frac{d\theta}{dt}=r\frac{d\theta}{dt}\mathbf{i_{\theta}}(\theta)$$where $\mathbf{i_{\theta}}(\theta)$ is the unit vector in the $\theta$ direction at polar angle $\theta$.
The acceleration of the particle is the derivative of the velocity vector with respect to time, and is thus given by:$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=r\frac{d^2\theta}{dt^2}\mathbf{i_{\theta}}(\theta)+r\frac{d\theta}{dt}\frac{d\mathbf{i_{\theta}}(\theta)}{d\theta}\frac{d\theta}{dt}=r\frac{d^2\theta}{dt^2}\mathbf{i_{\theta}}(\theta)-r\left(\frac{d\theta}{dt}\right)^2\mathbf{i_{r}}(\theta)$$So, the tangential component of acceleration is $r\frac{d^2\theta}{dt^2}=\frac{dv}{dt}$ and the centripetal component of acceleration is $r\left(\frac{d\theta}{dt}\right)^2=r\omega^2=\frac{v^2}{r}$, where $\omega$ is the instantaneous angular velocity and v is the magnitude of the instantaneous tangential velocity.
A: The rate of change of acceleration is call "jerk":
$$ \vec j = \frac{d\vec a}{dt} = \frac{d^3\vec x}{dt^3} $$
I supposes it too can be broken down into "centripetal jerk" and "tangential jerk", although I haven't heard those term ever used.
Jerk is most certainly experienced by off road racers, as they bounce around their 5 point restraints is response to changing g-forces.
The derivative of jerk is called jounce.
A: The change in magnitude of centripetal acceleration
Your question is: I want to find out how centripetal acceleration changes over time as the tangential velocity changes over time
First i will calculate the equations of Motion for this case
\begin{align*}
&\text{The components of the position vector in polar coordinate are:}\\\\
  &\vec{R}=
  \begin{bmatrix}
    r(t)\cos(\varphi(t)) \\
    r(t)\sin(\varphi(t)) \\
  \end{bmatrix}&(1)\\
&\text{because the velocity changes over time, the kreis radius $r$ change over time}\\\\
& \Rightarrow\\
&\vec{\dot{R}}=
  \begin{bmatrix}
    \dot{r}\cos(\varphi)-r\sin(\varphi)\dot{\varphi}) \\
    \dot{r}\sin(\varphi)+r\cos(\varphi)\dot{\varphi}) \\
  \end{bmatrix}&(2)\\\\
  &\text{so the kinetic $T$ energy is:}\\
  &T=\frac{1}{2}\,m\,\dot{R}^2=\frac{1}{2}\,m\left(\dot{r}^2+r^2\,\dot{\varphi}^2\right)\\
  &\text{with euler lagrange approach we get the equations of motion :} \\\\
  &\ddot{r}=\dot{\varphi}^2\,r&(3)\\
   &\ddot{\varphi}=-2\,\frac{\dot{\varphi}\,\dot{r}}{r}&(4)
\end{align*}
The solutions of the EoM's with the initial conditions 
$\varphi(t=0)=0\,,\dot{\varphi}(t=0)=\omega$ and
$r(t=0)=r_0\,,\dot{r}(t=0)=0$ are:
The solutions of the EoM's with the initial conditions \ $\varphi(t=0)=0\,,\dot{\varphi}(t=0)=\omega$ and
$r(t=0)=r_0\,,\dot{r}(t=0)=0$ are:
\begin{align*}
  &r(t)=r_{{0}}\sqrt {\omega}{\frac {1}{\sqrt {{\frac {\omega}{1+{\omega}^{2}{
t}^{2}}}}}}
\\
&\varphi(t)=\arctan \left( \omega\,t \right) \\\\
&\Rightarrow\\
&\text{The centrifugal force:}\\
&F_z(t)=r\,\dot{\varphi}^2={\omega}^{5/2}r_{{0}} \left( 1+{\omega}^{2}{t}^{2} \right) ^{-2}{
\frac {1}{\sqrt {{\frac {\omega}{1+{\omega}^{2}{t}^{2}}}}}}\\
&\text{for $t=0$ we get $F_{z0}=r_0\,\omega^2$}\\
&\Rightarrow\\
&\text{Change of the Zentrifugal force over time:}\\\\
&\boxed{\Delta F_z=F_z(t)-F_{z0}}
\end{align*}

