Newton's first axiom states that a body on which no forces act moves in a straight line with constant speed (magnitude of velocity vector), the second axiom states that $\vec{F_{tot}}=\frac{d\vec{p}}{dt}$. My question is, is the first axiom superflous? Does it follow directly from the first?
The case relevant to this question is $\vec{F}=\vec{0}$. Only with the second law one can derive that bodies with no fores acting on them move in straight lines.
$F=\frac{d\vec{p}}{dt}=\frac{dm}{dt}\vec{v}+m\frac{d\vec{v}}{dt}=\vec{0} \rightarrow \frac{dm}{dt}\vec{v}=-m\frac{d\vec{v}}{dt}$
This can only be true if $\vec{v} \space$and $\frac{d\vec{v}}{dt} \space$ are parallel, meaning acceleration only happens parallel to the direction of v, so the particle only moves in a line. The fact that $\vec{v} \parallel \frac{d\vec{v}}{dt}$ leads to motion on a line can be shown more rigorously as well. The linear part of the first axiom is contained within the second axiom.
What about the constant speed $\frac{d||\vec{v}||}{dt}=0 \space$ part?
$\vec{v}=||v||\hat{v}, \frac{d\vec{v}}{dt}=||\frac{d\vec{v}}{dt}||\hat{(\frac{d\vec{v}}{dt})}$
$\vec{v} \parallel \frac{d\vec{v}}{dt} \rightarrow \hat{(\frac{d\vec{v}}{dt})}=\pm\hat{v}, \frac{dm}{dt}||v||\hat{v}=-m||\frac{d\vec{v}}{dt}||\hat{(\frac{d\vec{v}}{dt})}$
$\frac{dm}{dt}||v||=\mp m||\frac{d\vec{v}}{dt}||$
I am able to get this far, but am unsure if this implies $\frac{d||\vec{v}||}{dt}=0$.
