This preview shows page 1. Sign up to view the full content.
Exercises 3.4
⇒
v
(
t
)=9
.
5
g
√
2
(
1
−
e
−
t/
20
)
.
Since
x
(0) = 0, integrating the above equation, we obtain
x
(
t
)=
t
Z
0
v
(
s
)
ds
=
t
Z
0
9
.
5
g
√
2
(
1
−
e
−
s/
20
)
ds
=9
.
5
g
√
2
(
s
+20
e
−
s/
20
)
±
±
±
s
=
t
s
=0
.
5
g
√
2
(
t
e
−
t/
20
−
20
)
≈
131
.
8
t
+ 2636
e
−
t/
20
−
2636
.
The object reaches the end of the inclined plane when
x
(
t
) = 131
.
8
t
+ 2636
e
−
t/
20
−
2636 = 10
⇒
t
≈
1
.
768 (sec)
.
21.
In this problem there are two forces acting on a sailboat: A constant horizontal force due to
the wind and a force due to the water resistance that acts in opposition to the motion of the
sailboat. All of the motion occurs along a horizontal axis. On this axis, we choose the origin
to be the point where the hard blowing wind begins and
x
(
t
) denotes the distance the sailboat
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

Click to edit the document details