Is the second derivative ($\ddot{r}$) equivalent to the first derivative squared ($\dot{r}^2$)?

Question

Is $\ddot{r}$ equivalent to $\dot{r}^2$?

When using Leibniz's notation I would write this as:

$ \begin{align} \ddot{r} &= \frac{d^2 r}{dt^2} \\ \dot{r}^2 &= \left(\frac{dr}{dt}\right)^2 = \frac{d r^2}{dt^2}\\ \end{align} $

So at first glance these are not in fact equal. Is this correct?

Answer 1

Only certain functions $r(t)$ have the two equal. These are precisely the solutions of the ODE $$\ddot{r}=\dot{r}^2, $$ which are $$r(t)=C_1-\log(t+C_2). $$

Answer 2

Take $r(t)=t^2$.

Then $\dot{r}^2 = 4t^2$ but $\ddot{r} = 2.$