Force-Driven Mass Reconsidered

Why not include position $x(t)$ as well as velocity $v(t)$ in the state-space model for the force-driven mass?

$$\left[\begin{array}{c} \dot{x}(t) \\ [2pt] \dot{v}(t) \end{array}\right] \eqsp \left[\begin{array}{cc} 0 & 1 \\ [2pt] 0 & 0 \end{array}\right]\left[\begin{array}{c} x(t) \\ [2pt] v(t) \end{array}\right] + \left[\begin{array}{c} 0 \\ [2pt] 1/m \end{array}\right] f(t)$$

We might expect this because we know from before that the complete physical state of a mass consists of its velocity $v$ and position $x$ !