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

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?

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