Beginning with the position vector, \(\mathbf{r}\), we can define the velocity, \(\mathbf{\dot{r}}\), as the time derivative of position.
\[ \mathbf{\dot{r}} = \frac{d\mathbf{r}}{dt}\,. \]Recall that the dot notation means the time-derivative.
We define the acceleration, \(\mathbf{\ddot{r}}\), as the time-derivative of velocity.
\[ \mathbf{\ddot{r}} = \frac{d\mathbf{\dot{r}}}{dt} = \frac{d}{dt}\frac{d\mathbf{r}}{dt} = \frac{d^2\mathbf{r}}{dt^2}\,.\]In polar coordinates, the position vector is given by,
\[ \mathbf{r} = r\cdot \hat{u_r}\,, \]where \(\hat{u_r}\) is the position unit vector, and both are functions of \(t\). Taking the time-derivative of both sides, we need to apply the chain rule.
\[\begin{aligned} \mathbf{\dot{r}} &= \frac{dr}{dt}\hat{u_r} + r\frac{d}{dt}\hat{u_r} \\ &= (\dot{r})\hat{u_r} + (r\dot{\theta})\hat{u_\theta} \\ \end{aligned}\]Once again taking the time-derivative of both times yields the acceleration in polar coordinates.
\[\begin{aligned} \mathbf{\ddot{r}} &= (\ddot{r} - r\dot{\theta}^2)\hat{u_r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{u_\theta}\\ \end{aligned}\]