fisica1-youn-e-freedman-exercicios-resolvidos

2
4.56: a) The equation of motion, − Cv = m dv
dt
cannot be integrated with respect to time,
as the unknown function v (t)
is part of the integrand. The equation must be separated
before integration; that is,
C dv
− dt =
2
m v
Ct 1 1
− = − + ,
m v v0
where v
0
is the constant of integration that gives v = v0
at t = 0 . Note that this form
shows that if v
0
= 0 , there is no motion. This expression may be rewritten as
dx ⎛ 1 Ct ⎞
v = =
⎜ +
⎟ ,
dt ⎝ v0
m ⎠
which may be integrated to obtain
m ⎡ Ctv0
⎤
x − x0 = ln
⎢
1 +
⎥
.
C ⎣ m ⎦
To obtain x as a function of v, the time t must be eliminated in favor of v; from the
Ctv 0
expression obtained after the first integration, 0 v
− 1, so
b) By the chain rule,
m
= v
−1
m ⎛ v0
⎞
x − x0 = ln ⎜ ⎟.
C ⎝ v ⎠
dv dv dv
=
dt dx dt
and using the given expression for the net force,
=
dv
v,
dx
−
2 ⎛ dv ⎞
− Cv = ⎜v
⎟m
⎝ dx ⎠
C dv
− dx =
m v
C
⎛ v ⎞
( x − x =
⎜
⎟
0
) ln
m
⎝ v0
⎠
m ⎛ v0
⎞
x − x0 = ln ⎜ ⎟.
C ⎝ v ⎠