Proof that $\vec {r(t)}=\vec r_0 + \vec v_0 t + \dfrac{1}{2} \vec a t^2$ for uniformly accelerated motion

Question

Displacement of a particle moving through $ x $ axis is given by $$ x(t)= x_0 + v_0 t + \dfrac{1}{2} at^2 $$ Can we deduce from it that $$ \vec r(t)=\vec r_0 + \vec v_0 t + \dfrac{1}{2} \vec a t^2$$ true too? Is there a nice way to see why is it true?

Answer 1

Yes this is true.

Consider all three coordinates - we know that the scalar equation is true for each:

\begin{align} x(t)&=x_0+v_{0,x}t+\frac{1}{2}a_xt^2\\ y(t)&=y_0+v_{0,y}t+\frac{1}{2}a_yt^2\\ z(t)&=z_0+v_{0,z}t+\frac{1}{2}a_zt^2\\ \end{align}

Converting to a vector equation, one can write

\begin{equation} \begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix}=\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}+\begin{pmatrix}v_{0,x}\\v_{0,y}\\v_{0,z}\end{pmatrix}t+\frac{1}{2}\begin{pmatrix}a_x\\a_y\\a_z\end{pmatrix}t^2 \end{equation} which can then be written in the desired form: \begin{equation} \boxed{\vec{r}(t)=\vec{r}_0+\vec{v}_0t+\frac{1}{2}\vec{a}t^2} \end{equation}

Answer 2

It is true because it is correct for every coordinate separately.