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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
=
−
kdt
⇒
Z
dv
A
v
2
A
=
−
Z
⇒
1
v
A
(
t
)
=
kt
+
C.
From the initial condition we fnd
1
v
B
=
1
v
A
(0)
=
k
·
0+
C
=
C
⇒
C
=
1
v
B
.
Thus
v
A
(
t
)=
1
+1
/v
B
=
v
B
v
B
.
The ±unction
s
(
t
) there±ore satisfes
ds
dt
=
v
B
v
B
,s
(0) = 0
.
Integrating we obtain
s
(
t
Z
v
B
v
B
dt
=
1
k
ln (
v
B
+1)+
C
1
.
To fnd
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
B
+1)
.
At the moment
t
=
t
1
when driver A’s speed was halved, i.e.,
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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