Chapter 2
where
v
B
is driver B’s constant velocity, and
k >
0 is a positive constant. Separating variables
we get
dv
A
v
2
A
=
−
k dt
⇒
dv
A
v
2
A
=
−
k dt
⇒
1
v
A
(
t
)
=
kt
+
C .
From the initial condition we find
1
v
B
=
1
v
A
(0)
=
k
·
0 +
C
=
C
⇒
C
=
1
v
B
.
Thus
v
A
(
t
) =
1
kt
+ 1
/v
B
=
v
B
v
B
kt
+ 1
.
The function
s
(
t
) therefore satisfies
ds
dt
=
v
B
v
B
kt
+ 1
,
s
(0) = 0
.
Integrating we obtain
s
(
t
) =
v
B
v
B
kt
+ 1
dt
=
1
k
ln (
v
B
kt
+ 1) +
C
1
.
To find
C
1
we use the initial condition:
0 =
s
(0) =
1
k
ln (
v
B
k
·
0 + 1) +
C
1
=
C
1
⇒
C
1
= 0
.
So,
s
(
t
) =
1
k
ln (
v
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Constant of integration

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