PROBLEM 11.8
The
motion
of
a
particle
is
defined
by
the
relation
x
:=
2t
3
18t
2
+
48t
16, where
x
and
t
are expressed in millimeters
and seconds, respectively. Determine
(a)
when the velocity is zero,
(b)
the position and the total distance traveled when the acceleration is
zero.
SOLUTION
Position:
x
=
2t
3

18t
2
+
48t
 16 mm
dx
2
mmI
Velocity:
v
= =
6t
36t
+
48
s
dt
dv
Acceleration:
12t
 36
mmls
2
a
=
dt
(a)
When
v
=
0,
6t
2

36t
+
48
=
0
Solving the quadratic equation,
t
=
2 s and
t
4 s .
...
For
0
~
t
<
2 s,
v
is positive,
and
x
is increasing.
2s~t~4s,
v
is negative,
and
x
is decreasing.
t:2:4s,
v
is positive,
and
x
is increasing.
(b)
When
a
=
0,
12t
 36
=
0
t
=
3 s
When
t
=
3 s,
X3
(2)(3)3  (18)(3)2
+
(48)(3)
16
X3
=
20mm .
...
When
t
=
0,
Xo
16 mm
When
t
=
2 s,
X2
=
(2)(2)3
(18)(2)2
+
(48)(2)
16
=
24 mm
Distance traveled over 0
~
t
~
2 s:
d
l
=
I
x2
Xo
I
=
124  (16)
I
:=
40 mm
Distance traveled over 2 s
~
t
~
3 s:
d
2
:=
I
X3

x21
=
120
241:= 4 mm
Total distance traveled:
d
l
+
d
2
=
40
+
4
d=44mm .
...
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PROBLEM 11.29
The acceleration due to gravity of a particle falling toward the earth is
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 Fall '11
 STAFF
 Acceleration, Velocity, Trigraph

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