= Solution
Let $m=4\pi\rho a^3/3$ be the drop's <mass>, with constant <mass density> $\rho$. Its volume-growth rule gives $4\pi a^2\dot a=c\pi a^2v$, hence $\dot a=cv/4$ and $\dot m=\rho c\pi a^2v$. The incoming mist is stationary, so the <variable-mass system> momentum balance is
$$
\frac d{dt}(mv)=mg-k\rho\pi a^2v^2.
$$
Expanding the <derivative> and substituting the accretion rate gives the <accreting raindrop with quadratic drag> equations
$$
\boxed{\dot a=\frac c4v,\qquad
\dot v=g-\frac34(c+k)\frac{v^2}a.}
$$
The extra $c$ contribution is the <momentum> needed to accelerate collected matter from rest; omitting it would treat the growing drop as a constant-mass particle.
Set $q=v^2/a$ and $B=7c/8+3k/4$. The <chain rule> gives
$$
\begin{aligned}
\dot q
&=\frac{2v\dot v}a-\frac{v^2\dot a}{a^2}\\
&=\frac va\left[2g-\left(\frac74c+\frac32k\right)q\right]
=\boxed{\frac{2v}a(g-Bq)}.
\end{aligned}
$$
To justify the limiting value as well as its stability sign, use $c>0$, as appropriate to volume growth. Divide by $\dot a=cv/4$ and integrate the resulting first-order equation:
$$
\boxed{q(a)=\frac gB+\left(q(a_0)-\frac gB\right)
\left(\frac{a_0}a\right)^{7+6k/c}.}
$$
Thus $q$ remains between two positive constants, its initial value and $g/B$. Since $d\sqrt a/dt=c\sqrt q/8$, the radius is defined for all increasing time and tends to infinity. The correction vanishes, proving the <speed-radius attractor of an accreting raindrop>:
$$
\boxed{\frac{v^2}a\longrightarrow\frac g{7c/8+3k/4},\qquad
\dot v\longrightarrow g-\frac{3(c+k)}4\frac gB
=\frac{cg}{7c+6k}<\frac g7.}
$$
The strict inequality uses $k>0$. The growing radius prevents a finite terminal <velocity>; instead the <acceleration> approaches this positive constant.
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