If the velocity $v=\|\mathbf{v}\|$ of a point having position $\mathbf{x}(t)$ at time $t$ is never null, then acceleration $\mathbf{a}:=\frac{d^2\mathbf{x}}{dt^2}$ can be written as$$\mathbf{a}=\frac{d(v\mathbf{t})}{dt}=\frac{dv}{dt}\mathbf{t}+v\frac{d\mathbf{t}}{dt}$$where $\mathbf{t}:=v^{-1}\mathbf{t}$ is a unit vector, therefore such that $\frac{d\mathbf{t}}{dt}\perp\mathbf{t}$.
$\frac{dv}{dt}\mathbf{t}$ is called tangential component of acceleration and $v\frac{d\mathbf{t}}{dt}$ is called normal component of acceleration.
I have been told by a friend of mine that is a graduated engineer that, at least under the hypothesis that a Frenet basis exists a decomposition into a normal and a tangential component exists for $\mathbf{a}(t)$ even when $v(t)=0$, but I cannot prove it to myself and my friend cannot either.
The fact that the path is piecewise parametrisable by $\mathbf{r}:[a,b]\to\mathbb{R}^3$ -where $\mathbf{r}$ is not the same as $\mathbf{x}$ in general-, $\mathbf{r}\in C^2[a,b]$, $\forall t\in[a,b]\quad$$\mathbf{r}'(t)\ne\mathbf{0}$, if I am not wrong, may be sufficient to guarantee the existence of the three following vectors
$$\mathbf{T}(t):=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|},\quad\mathbf{N}(t):=\frac{d}{dt}\bigg(\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}\bigg),\quad\mathbf{B}(t):=\mathbf{T}(t)\times\mathbf{N}(t)$$which are a basis of $\mathbb{R}^3$. Therefore acceleration $\mathbf{a}(t)$ can be expressed as $$\mathbf{a}(t)=\lambda(t)\mathbf{T}(t)+ \mu(t)\mathbf{N}(t)+ \nu(t)\mathbf{B}(t)$$but is it true and, if it is, how can we prove that $\nu(t)=0$ and $ \lambda(t)=v(t)$ even when $v(0)=0$?