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
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Elementary algebra, T∗

Click to edit the document details