Summary: Problem 41 on page 55.
Suppose 0 < T ; m and v are continuous [0, T); m is positive and nondecreasing on (0, T); v is
differentiable on (0, T); and
(G)
d
dt
(mv) = gm, 0 < t < T.
Let
M(t) =
t
0
m() d.
This is equivalent to saying M = m and M(0) = 0. Integrating (G) from 0 to t (0, T) we obtain
mv = m0v0 + gM
where we have set m0 = m(0) and v0 = v(0). Dividing by m we obtain
v =
m0v0
m
+ g
M

Source: Allard, William K. - Department of Mathematics, Duke University

Collections: Mathematics