= Solution
{c}
Treat both sides of $a$ at once: for $\sigma\in\{-1,1\}$ set $w(t)=v(a+\sigma t)$, $0\leq t\leq h$. Nonnegativity gives
$$
\left|\int_a^{a+\sigma t}v(s)\,ds\right|=\int_0^t w(r)\,dr,
\qquad w(t)\leq ct+A\int_0^t w(r)\,dr.
$$
Define $W(t)=ct+A\int_0^t w(r)\,dr$. By the <fundamental theorem of calculus>, $W'(t)=c+Aw(t)\leq c+AW(t)$ and $W(0)=0$. The <integrating factor> $e^{-At}$ gives
$$
\frac{d}{dt}(e^{-At}W(t))\leq ce^{-At},\qquad
W(t)\leq\frac{c}{A}(e^{At}-1).
$$
Since $w\leq W$, this proves the <affine Gronwall estimate>
$$
\boxed{v(x)\leq\frac{c}{A}\bigl(e^{A|x-a|}-1\bigr),\qquad |x-a|\leq h.}
$$
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