= Solution
{c}
For this <homogeneous motion>, suppose the <matrix> $\mathbf A=\ddot{\mathbf F}\mathbf F^{-1}$ is symmetric. Put $\mathbf y=\mathbf x-\mathbf c(t)$. The <acceleration potential of a homogeneous motion> is
$$
\boxed{\xi(\mathbf x,t)
=-\frac12\mathbf y^T\mathbf A(t)\mathbf y
-\ddot{\mathbf c}(t)\cdot\mathbf y+g(t),}
$$
where $g(t)$ is arbitrary. When taking the spatial <gradient>, $t$, $\mathbf c$, $\ddot{\mathbf c}$, and $\mathbf A$ are fixed. Symmetry gives
$$
\nabla\xi=-\frac12(\mathbf A+\mathbf A^T)\mathbf y-\ddot{\mathbf c}
=-\mathbf A\mathbf y-\ddot{\mathbf c}
=-\ddot{\mathbf x}.
$$
This verifies the potential directly and also proves sufficiency of the <symmetric matrix> condition within the static-family setting. Its dimension is $\mathsf L^2\mathsf T^{-2}$, so the <pressure> correction $-\rho\xi$ has the dimension of <stress>. Combining with part (b)(ii), the corresponding dynamic <Cauchy stress tensor> is
$$
\mathbf T=\mathbf T_0(t)+
\left[\frac\rho2\mathbf y^T\mathbf A\mathbf y
+\rho\ddot{\mathbf c}\cdot\mathbf y+\widetilde C(t)\right]\mathbf1,
$$
where $\widetilde C(t)=C(t)-\rho g(t)$. Substituting its <divergence> into the <Cauchy momentum equation> reproduces $\rho[\mathbf A\mathbf y+\ddot{\mathbf c}]$, as required.
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