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
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Force, Mass, constant horizontal force, water resistance, hard blowing wind

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