kinematics - Question about $a = v mathrm dv/mathrm dx$

Consider $\vec{v}$ Now differentiating this w.r.t time,

$$\vec{a} = d/dt( \vec{v}) = \vec{v}(d\vec{v}/dx)$$ Now this multiplication of vectors obviously makes no sense. This along with the fact that on integrating we get $v^2$ leads me to believe that this 'formula' is only applicable when $v$ is really $|v|$ (speed) and $a$ is really the rate of change of speed.

But while studying one-dimensional motion, I remember countless times getting an acceleration function like $a(x) =\text{something }\hat{i}$ and applying this same result and integrating which gave me the velocity. Why is this?

Answer

Well, let me present my version as well: By definition we have

$$\vec a =\frac{\mathrm d}{\mathrm d t }\vec v = \frac{\mathrm d}{\mathrm d t }\left(v_x,v_y,v_z\right)^\mathrm{T} =\left(\frac{\mathrm d}{\mathrm d t} v_x,\frac{\mathrm d}{\mathrm d t} v_y,\frac{\mathrm d}{\mathrm d t} v_z\right)^\mathrm{T}$$ as in 1D $$\frac{\mathrm d}{\mathrm d t}v=\frac{\mathrm d x}{\mathrm d x}\frac{\mathrm d}{\mathrm d t}v=\frac{\mathrm d x}{\mathrm d t}\frac{\mathrm d}{\mathrm d x}v=v\frac{\mathrm d v}{\mathrm d x}$$ that is

$$\vec a =\left(v_x\frac{\mathrm d v_x}{\mathrm d x},v_y \frac{\mathrm d v_y}{\mathrm d y},v_z\frac{\mathrm d v_z}{\mathrm d z}\right)^\mathrm{T}$$

and we easily can just look at one component. Now concerning the integration: From

$$a=v \frac{\mathrm d v}{\mathrm dx}$$ follows $$a \mathrm d x=v \mathrm d v$$ i.e. $$\int_{x_1}^{x_2} a \mathrm d x=\int_{v_1}^{v_2}v \mathrm d v =\frac 1 2 (v_2^2-v_1^2)$$ and if $a=\mathrm{const}$

$$a(x_2-x_1)=\frac 1 2 (v_2^2-v_1^2)$$ which is one of the "SUVAT" equations. This should work fine if generalized to 3D.

EDIT Due to some misleading concepts and some comments, let me give a sort of interpretation and an extreme example.

Interpretation

Usually we have acceleration a change of speed over a certain time. There is no reason why you could not "measure" it as a change of speed over a certain distance. The problem here is: if the accelerated object is already very fast, the distance required to see an according change needs to be very large. The other way around it works like this: If over the same distance two objects change there speed by the same amount but one was much faster than the other in the beginning, its acceleration must have been larger as well. That is the multiplication with $v$ of $\frac{\mathrm d v}{\mathrm dx}$

Example

Lets take a charged particle in a magnetic field (suggested by a comment) Without losing generality we may look only at 2 dimensions. In principle we must be careful as there is no one-to-one mapping neither for $x$ and $t$ nor for $y$ and $t$, but it turns out that this is no problem here. Otherwise use the implicit function theorem. We can simplify this to a circular motion:

$$\vec x = R \binom {\sin \omega t}{\cos \omega t}, \vec v = R \omega \binom {\cos \omega t}{-\sin \omega t},\vec a = R\omega^2 \binom {-\sin \omega t}{-\cos \omega t}$$ Hence $$v_x=\omega\sqrt{R^2 -R^2 \sin \omega t}= \omega\sqrt{R^2-x^2}$$ Therefore $$\frac{\mathrm d v_x}{\mathrm d x}=-\frac{x\omega}{\sqrt(R^2-x^2)}=-\frac{x \omega}{\frac{v_x}{\omega}}= -\frac{x \omega^2}{v_x}$$ So $$v_x \frac{\mathrm d v_x}{\mathrm d x}= -x \omega^2$$ And it is easy to see that $a_x=-\omega^2 x$ as well. The same way you do it for $y$. So if you treat the vectors correctly, $a=v \mathrm d v/\mathrm d x$ is true even for a motion where the acceleration in $x$ depends on the velocity in $y$ and vice-verse, and where there is no one-to-one global mapping from time to coordinate.