In this following, I'll assume that the drop maintains a spherical shape. Otherwise, the math will get horrendous, and one probably can't get an analytical solution.
The surface area of a sphere is given by:
S = 4*pi*r^2
The mass of a sphere of density D is simply the volume times the density. The volume of a sphere with radius r is given by:
V = 4/3 * pi * r^3,
so the mass is given by:
M = 4/3 * pi * r^3 * D.
We know that the drop gains mass according to:
dM/dt = k*S = k*4*pi*r^2
Substitute the expression for the mass as a function of radius into this differential equation:
4/3*pi*D*d(r^3)/dt = k*4*pi*r^2
Simplifying, we get:
d(r^3)/dt = 3*k*r^2/D
let u = r^3
du/dt = 3*k*u^(2/3)/D
u^(-2/3) du = 3*k/D dt
Integrate this to obtain
3(u^1/3 -u0^1/3) = 3*k*(t-t0)/D
where u0 and t0 are the constants of integration, and in this problem, we can let t0 = 0.
Substituting back for u and simplifying:
r - r0 = k*t/D
r(t) = k*t/D + r0
This gives the radius as a function of time, which is the desired result.
In the absence of friction and neglecting the change in gravitational acceleration as a function of height, the drop would have a constant acceleration of g (9.807 m/s^2). If we include air friction (and continue to assume a spherical raindrop) then the net acceleration is:
a = g - 0.5 * C * D * A * v^2/M
where C is the drag coefficient (= 1/2 for a sphere),
A is the cross sectional area
M is the mass
v is the velocity
D is the density
Substituting expressions for the mass and area in terms of the radius, we get:
a = g - (0.5 * C * D * pi * r^2 * v^2)/(4/3 * pi * r^3 * D)
a = g - 3/8 * C * v^2/r
if C = 0.5 (for a sphere)
a(t) = dv(t)/dt = g - (3*v(t)^2)/(16*r(t))
From here, one would need to substitute the expression derived previously for r as a function of t, solve the differential equation to find v(t), then take the derivative of v(t) with respect to t to get a(t) in terms of t. This is straightforward, but involves some tedious algebra that I'm not inclined to do at the moment!