In the kinematic equations $\vec{v}^2 = \vec{v}_0^2 + 2\vec{a}\cdot \Delta \vec{x}$, what does squaring a vector mean? One of the kinematic equations, $\vec{v}^2 = \vec{v}_0^2 + 2\vec{a}\cdot \Delta \vec{x}$, involves squaring the initial and final velocities. My questions is, if velocity is a vector, what meaning does the squaring have? Does it signify taking a dot product with itself?
 A: The vector form of the kinematic equation is
$$ 2\boldsymbol{a}\cdot\left(\boldsymbol{x}_{1}-\boldsymbol{x}_{0}\right) = \boldsymbol{v}_{1}\cdot\boldsymbol{v}_{1}-\boldsymbol{v}_{0}\cdot\boldsymbol{v}_{0} $$ where in bold are vector quantities, and $\cdot$ is the dot product.

Proof
Take the vector form of the change in velocity
$$ \boldsymbol{v}_{1}-\boldsymbol{v}_{0}=\boldsymbol{a}\,\Delta t \tag{1} $$
and change in displacement below. Now any time the expression $\boldsymbol{a}\,\Delta t $ shows up it can be replaced with $(\boldsymbol{v}_{1}-\boldsymbol{v}_{0})$.
$$ \begin{align}\boldsymbol{x}_{1}-\boldsymbol{x}_{0} & =\boldsymbol{v}_{0}\Delta t+\tfrac{1}{2}\boldsymbol{a}\,\left(\Delta t\right)^{2}\\
 & =\boldsymbol{v}_{0}\Delta t+\tfrac{1}{2}\left(\boldsymbol{v}_{1}-\boldsymbol{v}_{0}\right)\,\left(\Delta t\right)\\
 & =\tfrac{1}{2}\left(\boldsymbol{v}_{1}+\boldsymbol{v}_{0}\right)\Delta t
\tag{2} \end{align} $$
The above makes total sense as the displacement vector is proportional to the average velocity in vector form.
Now take (2) and multiply both sides with $2\boldsymbol{a}$, and by multiply I mean take the dot product
$$ \begin{align}2\boldsymbol{a}\cdot\left(\boldsymbol{x}_{1}-\boldsymbol{x}_{0}\right) & =\boldsymbol{a}\Delta t\cdot\left(\boldsymbol{v}_{1}+\boldsymbol{v}_{0}\right)\\
 & =\left(\boldsymbol{v}_{1}-\boldsymbol{v}_{0}\right)\cdot\left(\boldsymbol{v}_{1}+\boldsymbol{v}_{0}\right)\\
 & =\boldsymbol{v}_{1}\cdot\boldsymbol{v}_{1}-\boldsymbol{v}_{0}\cdot\boldsymbol{v}_{0} \tag{3}
\end{align}  $$
A: The kinematic equations are scalar equations in general. They only include the magnitudes of vector quantities, not the vectors themselves. They are equations that apply along a bound path or direction given a sign so that the directionality is covered for before applying them.
Otherwise not only the $v$'s but also $a$ and $\Delta x$ ought to have been marked as vectors in your formula.
A: To answer your questions directly:

*

*Squaring a vector is meaningless.

*However, in this case, the equation works if you interpret the squaring as a dot product.

John Alexiou’s answer gives a nice proof of this.
