Exercises 3.4
Solving this linear equation and using the initial condition,
v
(A)
1
(0) = 2, we get
v
(A)
1
(
t
)=
65
8
−
49
8
e
−
4
t/
3
,
and so
x
(A)
1
(
t
t
Z
0
±
65
8
−
49
8
e
−
4
s/
3
²
ds
=
65
8
t
−
147
32
(
e
−
4
t/
3
−
1
)
.
The boat A will have the velocity 5 m/sec at
t
=
t
∗
satisfying
65
8
−
49
8
e
−
4
t
∗
/
3
=5
⇒
t
∗
=
−
3ln(25
/
49)
4
≈
0
.
5 (sec)
,
and it will be
x
(A)
1
(
t
∗
65
8
t
∗
−
147
32
(
e
−
4
t
∗
/
3
−
1
)
≈
1
.
85 (m)
away from the starting point or, equivalently, 500
−
1
.
85 = 498
.
15 meters away from the Fnish.
Similarly to (3.12), resetting the time, we obtain an equation of the motion of the boat A
starting from the moment when its velocity reaches 5 m/sec. Denoting by
x
(A)
2
(
t
) the distance
passed by the boat A and by
v
(A)
2
(
t
) its velocity, we get
x
(A)
2
(0) = 0,
v
(A)
2
(0) = 5, and
m
dv
(A)
2
dt
= 650
−
b
2
v
(A)
2
⇒
dv
(A)
2
dt
=
65
6
−
v
(A)
2
⇒
v
(A)
2
(
t
65
6
−
35
6
e
−
t
⇒
x
(A)
2
(
t
t
Z
0
±
65
6
−
35
6
e
−
s
²
ds
=
65
6
t
+
35
6
(
e
−
t
−
1
)
.
Solving the equation
x
(A)
2
(
t
) = 498
.
15, we Fnd the time (counting from the moment when the
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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