CHAPTER 11
PROBLEM 11.1
The motion of a particle is defined by the relation x = 1.5t 4 − 30t 2 + 5t + 10, where x and t are expressed in
meters and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle
when t = 4 s.
SOLUTION
Given:
x = 1.5t 4 − 30t 2 + 5t + 10
dx
= 6t 3 − 60t + 5
dt
dv
= 18t 2 − 60
a=
dt
v=
Evaluate expressions at t = 4 s.
x = 1.5(4) 4 − 30(4) 2 + 5(4) + 10 = −66 m
x = −66.0 m �
v = 6(4)3 − 60(4) + 5 = 149 m/s
v = 149.0 m/s �
a = 18(4) 2 − 60 = 228 m/s 2 �
a = 228.0 m/s 2 ��
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3
3
PROBLEM 11.2
The motion of a particle is defined by the relation x = 12t 3 − 18t 2 + 2t + 55, where x and t are expressed in
meters and seconds, respectively. Determine the position and the velocity when the acceleration of the particle
is equal to zero.
SOLUTION
Given:
x = 12t 3 − 18t 2 + 2t + 5
dx
= 36t 2 − 36t
36t + 2
dt
dv
a=
= 72t − 36
dt
v=
Find the time for a = 0.
72t − 36 = 0 � t = 0
0.5 s
Substitute into above expressions.
x = 12(0.5
12(0.5)
12
(0.5))3 − 18(0.5) 2 + 2(0.
2(0.5)
2(
0.5)
5) + 5 = 3
v = 36(0.5
36(0.5)
36
(0.5)) 2 − 36(0.5)) + 2
= −7 m/s
x = 3.00 m �
v = −7.00 m/s ��
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4
PROBLEM 11.3
5 3 5 2
t − t − 30t
30t + 8 x, where x and t are expressed in
3
2
meters and seconds, respectively. Determine the time, the position, and the acceleration when v = 0.
The motion of a particle is defined by the relation x =
SOLUTION
We have
5
5
x = t 3 − t 2 − 30t + 8
3
2
Then
v=
dx
= 5t 2 − 55tt − 30
dt
and
a=
dv
= 10t − 5
dt
When v = 0:
or
At t = 3 s:
5t 2 − 5t − 30 = 5(t 2 − t − 6) = 0
t = 3 s and
and t = −2 s (Reject)
t = 3.00 s
5 3 5 2
x3 = (3
(3)) − ((3)
3) − 30(3)) + 8
3
2
or
a3 = 10(3) − 5
or a3 = 25.0 m/s2
x3 = −59.5 m
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5
PROBLEM 11.4
The motion of a particle is defined by the relation x = 6t 2 − 8 + 40 cos π t , where x and t are expressed in
meters and seconds, respectively. Determine the position, the velocity, and the acceleration when t = 6 s.
SOLUTION
We have
x = 6t 2 − 8 + 40 cos π t
Then
v=
dx
= 12t − 40π sin π t
dt
and
a=
dv
= 12 − 40π 2 cos π t
dt
At t = 6 s:
x6 = 6(6) 2 − 8 + 40 cos 6π
or
x6 = 248 m
v6 = 12(6) − 40π sin 6π
or
v6 = 72.0 m/s
a6 = 12 − 40π 2 cos 6π
or
a6 = −383 m/s2
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6
6
PROBLEM 11.5
The motion of a particle is defined by the relation x = 6t 4 − 22tt 3 − 12t 2 + 3t + 33, where x and t are expressed in
meters and seconds, respectively. Determine the time,
e, the position, and the velocity when a = 0.
SOLUTION
We have
x = 6t 4 − 2t 3 − 12t
12t 2 + 3t + 3
Then
v=
dx
= 24t 3 − 6t 2 − 224
4t + 3
dt
and
a=
dv
= 72t 2 − 12t
12t − 24
dt
When a = 0:
72t 2 − 12t − 24 = 12(6
12(6t
(6t 2 − t − 2) = 0
(3t − 2)(2
2)(2t + 1)
1) = 0
or
t=
or
At t =
2
s:
3
2
1
s and
and t = − s (Reject
Re )
Reject
3
2
4
3
2
�2�
�2�
�2�
�2�
x2/3 = 6 � � − 2 � � − 12 � � + 3 � � + 3
3
3
3
� �
� �
� �
�3�
3
t = 0.667 s �
or
x2/
2/3 = 0.259 m �
2
�2�
�2�
�2�
v2/3 = 24 � � − 6 � � − 2244 � � + 3
3
3
� �
� �
�3�
or v2/3 = −8.56 m/s �
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7
PROBLEM 11.6
The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x is expressed in meters
and t in seconds. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when
the acceleration is zero.
SOLUTION
x = 2t 3 − 15t 2 + 24t + 4
dx
= 6t 2 − 30t + 24
v=
dt
dv
= 12t − 30
a=
dt
(a)
v = 0 when
6t 2 − 30t + 24 = 0
6(t − 1)(t − 4) = 0
(b)
a = 0 when
For t = 2.5 s:
12t − 30 = 0
t = 1.000 s or t = 4.00 s
�
t = 2.5 s
x2.5 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4
x2.5 = +1.500 m �
To find total distance traveled, we note that
v = 0 when t = 1 s:
x1 = 2(1)3 − 15(1)2 + 24(1) + 4
x1 = +15 m
For t = 0,
x0 = +4 m
Distance traveled
From t = 0 to t = 1 s:
From t = 1 s to t = 2.5 s:
x1 − x0 = 15 − 4 = 11 m →
x2.5 − x1 = 1.5 − 15 = 13.5 m ←
Total distance traveled = 11 m + 13.5 m = 24.5 m
�
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8
8
PROBLEM 11.7
The motion of a particle is defined by the relation x = t 3 − 6t 2 − 36t − 40, where x and t are expressed in meters
and seconds, respectively. Determine (a) when the velocity is zero, (b) the velocity, the acceleration, and the
total distance traveled when x = 0.
SOLUTION
We have
x = t 3 − 6t 2 − 36t − 40
Then
v=
dx
= 3t 2 − 12t − 36
dt
and
a=
dv
= 6t − 12
dt
(a)
3t 2 − 12t − 36 = 3(t 2 − 4t − 12) = 0
When v = 0:
(t + 2)(t − 6) = 0
or
t = −2 s (Reject) and t = 6 s
or
(b)
t 3 − 6t 2 − 36t − 40 = 0
When x = 0:
Factoring
(t − 10)(t + 2)(t + 2) = 0 or t = 10 s
Now observe that
0
6s
and at t = 0:
t = 6 s:
t = 6.00 s
t
t
6 s:
v
0
10 s:
v
0
x0 = −40 m
x6 = (6)3 − 6(6)2 − 36(6) − 40
= −256 m
t = 10 s:
Then
v10 = 3(10) 2 − 12(10) − 36
or v10 = 144.0 m/s
a10 = 6(10) − 12
or
a0 = 48.0 m/s2
| x6 − x0 | = | −256 − (−40) | = 216 m
x10 − x6 = 0 − (−256) = 256 m
Total distance traveled = (216 + 256) m = 472 m
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9
9
PROBLEM 11.8
The motion of a particle is defined by the relation x = t 3 − 9t 2 + 24t − 8, where x and t are expressed in meters
and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance
traveled when the acceleration is zero.
SOLUTION
We have
x = t 3 − 9t 2 + 24t − 8
Then
v=
dx
= 3t 2 − 18t + 24
dt
and
a=
dv
= 6 t − 18
dt
(a)
3 t 2 − 18t + 24 = 3(t 2 − 6t + 8) = 0
When v = 0:
(t − 2)(t − 4) = 0
or
or t = 2.00 s and t = 4.00 s
(b)
When a = 0:
6t − 18 = 0 or t = 3 s
x3 = (3)3 − 9(3)2 + 24(3) − 8
At t = 3 s:
First observe that 0
t
2 s:
2s
t
3 s:
v
0
v
0
or
x3 = 10.00 m
Now
At t = 0:
x0 = −8 m
At t = 2 s:
x2 = (2)3 − 9(2) 2 + 24(2) − 8 = 12 m
Then
x2 − x0 = 12 − (−8) = 20 m
| x3 − x2 | = |10 − 12| = 2 m
Total distance traveled = (20 + 2) m = 22.0 m
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10
10
PROBLEM 11.9
The acceleration of a particle is defined by the relation a = −8 m/s 2 . Knowing that x = 20 m when t = 4 s and
that x = 4 m when v = 16 m/s, determine (a) the time when the velocity is zero, (b) the velocity and the total
distance traveled when t = 11 s.
SOLUTION
dv
= a = −8 m/s 2
dt
We have
� dv = � −8 dt + C
Then
C = constant
v = −8t + C (m/s)
or
dx
= v = −8t + C
dt
Also
x
t
At t = 4 s, x = 20 m:
�
or
x − 20 = [−4t 2 + Ct ]t4
20
dx =
� (−8t + C ) dt
4
x = −4t 2 + C (t − 4) + 84 (m)
or
When v = 16 m/s, x = 4 m:
16 = −8t + C � C = 16 + 8t
4 = −4t 2 + C (t − 4) + 84
Combining
Simplifying
0 = −4t 2 + (16 + 8t )(t − 4) + 80
t 2 − 4t + 4 = 0
t=2s
or
C = 32 m/s
v = −8t + 32 (m/s)
and
x = −4t 2 + 32 t − 44 (m)
(a)
When v = 0:
(b)
Velocity and distance at 11 s.
−8t + 32 = 0
or t = 4.00 s �
v11 = −(8)(11) + 32
At t = 0:
v11 = −56.0 m/s �
x0 = −44 m
t = 4 s:
x4 = 20 m
t = 11 s:
x11 = −4(11) 2 + 32(11) − 44 = −176 m
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11
11
PROBLEM 11.9 (Continued)
Now observe that
Then
0 � t � 4 s:
v�0
4 s � t � 11 s:
v�0
x4 − x0 = 20 − ( −44) = 64 m
| x11 − x4 | = | − 176 − 20| = 196 m
Total distance traveled = (64 + 196) m = 260 m
��
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12
12
PROBLEM 11.10
The acceleration of a particle is directly proportional to the square of the time t. When t = 0, the particle is
at x = 24 m. Knowing that at t = 6 s, x = 96 m and v = 18 m/s, express x and v in terms of t.
SOLUTION
a = kt 2
We have
k = constant
dv
= a = kt 2
dt
Now
v
t
At t = 6 s, v = 18 m/s:
�
or
1
v − 18 = k (t 3 − 216)
3
18
dv =
� kt dt
2
6
1
v = 18 + k (t 3 − 216)(m/s)
3
or
dx
1
= v = 18 + k (t 3 − 216)
dt
3
Also
x
t�
1
�
or
1 �1
�
x − 24 = 18t + k � t 4 − 216t �
3 �4
�
24
dx =
� ��18 + 3 k (t
�
− 216) � dt
�
At t = 0, x = 24 m:
0
3
Now
At t = 6 s, x = 96 m:
or
Then
or
and
or
1 �1
�
96 − 24 = 18(6) + k � (6) 4 − 216(6) �
3 �4
�
k=
1
m/s 4
9
1 � 1 �� 1
�
x − 24 = 18t + � �� t 4 − 216t �
3 � 9 �� 4
�
x(t ) =
1 4
t + 10t + 24
108
�
1�1�
v = 18 + � � (t 3 − 216)
3� 9�
v(t ) =
1 3
t + 10
27
�
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13
13
PROBLEM 11.11
The acceleration of a particle is directly proportional to the time t. At t = 0, the velocity of the particle
is v = 16 m/s. Knowing that v = 15 m/s and that x = 20 m when t = 1 s, determine the velocity, the position,
and the total distance traveled when t = 7 s.
SOLUTION
a = kt
We have
dv
= a = kt
dt
Now
At t = 0, v = 16 m/s:
or
v
dv =
16
v − 16 =
or
kt dt
1 2
kt
2
1 2
k t (m/s)
2
1
2
k(1 s )
2
k = −2 m/ s3
and
v = 16 − t 2
dx
= v = 16 − t 2
dt
Also
or
0
15 m/s = 16 m/s +
or
At t = 1 s, x = 20 m:
t
v = 16 +
or
At t = 1 s, v = 15 m/s:
k = constant
x
20
dx =
t
1
(16 − t 2 ) dt
1
x − 20 = 16t − t 3
3
t
1
1
13
x = − t 3 + 16t + (m)
3
3
Then
At t = 7 s:
When v = 0:
v7 = 16 − (7) 2
or v7 = −33.0 m/s
1
13
x7 = − (7)3 + 16(7) +
3
3
or
x7 = 2.00 m
16 − t 2 = 0 or t = 4 s
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14
14
PROBLEM 11.11 (Continued)
At t = 0:
x0 =
13
3
1
13
x4 = − (4)3 + 16(4) + = 47 m
3
3
t = 4 s:
Now observe that
Then
0
t
4 s:
v
0
4s
t
7 s:
v
0
13
= 42.67 m
3
| x7 − x4 | = |2 − 47| = 45 m
x4 − x0 = 47 −
Total distance traveled = (42.67 + 45) m = 87.7 m
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15
15
PROBLEM 11.12
The acceleration of a particle is defined by the relation a = kt 2 . (a) Knowing that v = −32 m/s when t = 0
and that v = +32 m/s when t = 4 s, determine the constant k. (b) Write the equations of motion, knowing also
that x = 0 when t = 4 s.
SOLUTION
a = kt 2
dv
= a = kt 2
dt
(1)
t = 0, v = −32 m/s and t = 4 s, v = +32 m/s
32
(a)
−32
dv =
4
0
kt 2 dt
1
32 − (−32) = k (4)3
3
(b)
k = 3.00 m/s4
Substituting k = 3 m/s4 into (1)
dv
= a = 3t 2
dt
t = 0, v = −32 m/s:
v
−32
dv =
t
0
a = 3t 2
3t 2 dt
1
v − (−32) = 3(t )3
3
v = t 3 − 32
dx
= v = t 3 − 32
dt
t = 4 s, x = 0:
x
0
dx =
t
4
t
(t 3 − 32) dt ; x =
1 4
t − 32t
4
4
1 4
1
t − 32t − (4) 4 − 32(4)
4
4
1
x = t 4 − 32t − 64 + 128
4
x=
x=
1 4
t − 32t + 64
4
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16
16
PROBLEM 11.13
The acceleration of a particle is defined by the relation a = A − 6 t 2 , where A is a constant. At t = 0, the
particle starts at x = 8 m with v = 0. Knowing that at t = 1 s, v = 30 m/s, determine (a) the times at which the
velocity is zero, (b) the total distance traveled by the particle when t = 5 s.
SOLUTION
a = A − 6t 2
We have
dv
= a = A − 6t 2
dt
Now
At t = 0, v = 0:
�
v
0
t
� ( A − 6t
0
2
)dt
30 = A(1) − 2(1)3
At t = 1 s, v = 30 m/s:
A = 32 m/s 2
or
and v = 32t − 2t 3
dx
= v = 32t − 2t 3
dt
Also
At t = 0, x = 8 m:
�
x
8
dx =
t
� (32t − 2t )dt
3
0
1
x = 8 + 16t 2 − t 4 (m)
2
or
When v = 0:
32t − 2t 3 = 2t (16 − t 2 ) = 0
t = 0 and t = 4.00 s
or
(b)
dv =
v = At − 2t 3 (m/s)
or
(a)
A = constant
At t = 4 s:
1
x4 = 8 + 16(4)2 − (4)4 = 136 m
2
t = 5 s:
1
x5 = 8 + 16(5) 2 − (5)4 = 95.5 m
2
�
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17
17
PROBLEM 11.13 (Continued)
Now observe that
Then
0 � t � 4 s:
v�0
4 s � t � 5 s:
v�0
x4 − x0 = 136 − 8 = 128 m
| x5 − x4 | = |95.5 − 136| = 40.5 m
Total distance traveled = (128 + 40.5) m = 168.5 m
�
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18
18
PROBLEM 11.14
It is known that from t = 2 s to t = 10 s the acceleration of a particle is inversely proportional to the cube
of the time t. When t = 2 s, v = −15 m/s, and when t = 10 s, v = 0.36 m/s. Knowing that the particle is twice
as far from the origin when t = 2 s as it is when t = 10 s, determine (a) the position of the particle
when t = 2 s, and when t = 10 s, (b) the total distance traveled by the particle from t = 2 s to t = 10 s.
SOLUTION
a=
We have
k
t3
k = constant
dv
k
=a= 3
dt
t
Now
At t = 2 s, v = −15 m/s:
�
v
−15
dv =
�
t
v − (−15) = −
or
v=
or
At t = 10 s, v = 0.36 m/s:
0.36 =
k
2 t3
k2
2
dt
�1
1 �
� 2−
�
(2) 2 �
�t
k�1 1
−
2 �� 4 t 2
k�1
1
− 2
�
2 � 4 10
or
k = 128 m ⋅ s
and
v =1−
(a)
�
� − 15
�
64
(m/s)
t2
64
dx
= v =1− 2
dt
t
We have
64 �
dt + C
2 �
�
�
� dx = � ��1 − t
Then
x=t+
or
Now x2 = 2 x10 :
�
� − 15 (m/s)
�
2+
C = constant
64
+ C (m)
t
64
64
�
�
+ C = 2 �10 +
+C�
2
10
�
�
or
C = 1.2 m
and
x=t+
64
+ 1.2 (m)
t
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19
19
PROBLEM 11.14 (Continued)
At t = 2 s:
x2 = 2 +
t = 10 s:
64
+ 1.2
2
x10 = 10 +
64
+ 1.2
10
Note: A second solution exists for the case x2 � 0, x10 � 0. For this case, C = −22
x2 = 11
and
(b)
When v = 0:
1−
x2 = 35.2 m �
or
x10 = 17.60 m �
4
m
15
11
13
m, x10 = −5 m
15
15
64
= 0 or t = 8 s
t2
At t = 8 s:
x8 = 8 +
Now observe that 2 s � t � 8 s:
v�0
8 s � t � 10 s:
v�0
Then
or
64
+ 1.2 = 17.2 m
8
| x8 − x2 | = |17.2 − 35.2| = 18 m
x10 − x8 = 17.6 − 17.2 = 0.4 m
Total distance traveled = (18 + 0.4) m = 18.40 m
�
Note: The total distance traveled is the same for both cases.
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20
20
PROBLEM 11.15
The acceleration of a particle is defined by the relation a = −k/x. It has been experimentally determined that
v = 15 m/s when x = 0.6 m and that v = 9 m/s when x = 1.2 m. Determine (a) the velocity of the particle
when x = 1.5 m, (b) the position of the particle at which its velocity is zero.
SOLUTION
a=
vdv − k
=
x
dx
Separate and integrate using x = 0.6 m, v = 15 m/s.
v
15
vdv = − k
x
0.6
v
dx
x
1 2
= − k ln x
v
2 15
x
0.6
x
1 2 1
v − (15) 2 = − k ln
2
2
0.6
(1)
When v = 9 m/s, x = 1.2 m
1 2 1
1.2
(9) − (15) 2 = −k ln
2
2
0.6
Solve for k.
k = 103.874 m2/s2
(a)
Substitute
2 2
k = 103.874 m /s and x = 1.5 m into (1).
1 2 1
1.5
v − (15) 2 = −103.874 ln
2
2
0.6
v = 5.89 m
(b)
For v = 0,
1
x
0 − (15) 2 = −103.874 ln
2
0.6
ln
x
= 1.083
0.6
x = 1.772 m
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21
21
PROBLEM 11.16
A particle starting from rest at x = 1 m is accelerated so that its velocity doubles in magnitude between
x = 2 m and x = 8 m. Knowing that the acceleration of the particle is defined by the relation a = k[ x − (A/x)],
determine the values of the constants A and k if the particle has a velocity of 29 m/s when x = 16 m.
SOLUTION
We have
v
v
When x = 1 m, v = 0:
0
dv
A
=a=k x−
x
dx
vdv =
x
1
k x−
A
dx
x
1 2
1
v = k x 2 − A ln x
2
2
or
=k
or
1 2
1
x − A ln x −
2
2
1 2
1
1
v8 = k (8) 2 − A ln 8 −
= k (31.5 − A ln 8)
2
2
2
x = 8 m:
or
1
1 2
1
1
3
v2 = k (2) 2 − A ln 2 −
=k
− A ln 2
2
2
2
2
At x = 2 m:
Now
x
1 2
v
2 8
1 2
v
2 2
v8
= 2:
v2
= (2) 2 =
k (31.5 − A ln 8)
k ( 32 − A ln 2 )
6 − 4 A ln 2 = 31.5 − A ln 8
25.5 = A(ln 8 − 4 ln 2) = A(ln 8 − ln 24 ) = A ln
1
2
A = −36.8 m2
or
When x = 16 m, v = 29 m/s:
Noting that
We have
1
1
25.5
1
(29) 2 = k (16) 2 −
ln(16) −
1
2
2
2
ln ( 2 )
ln(16) = 4 ln 2 and ln
841 = k 236 −
1
= − ln(2)
2
ln 25.5
= 4 ln(2) − 1
− ln(2)
k = 1.832 s −2
or
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22
22
PROBLEM 11.17
A particle oscillates between the points x = 40 mm and x = 160 mm with an acceleration a = k(100 – x),
where a and x are expressed in mm/s2 and mm, respectively, and k is a constant. The velocity of the particle
is 18 mm/s when x = 100 mm and is zero at both x = 40 mm and x = 160 mm. Determine (a) the value of k,
(b) the velocity when x = 120 mm.
SOLUTION
(a)
v
We have
When x = 40 mm, v = 0:
�
v
0
dv
= a = k (100 − x)
dx
vdv =
�
x
40
k (100 − x) dx
x
or
1 2
1 �
�
v = k �100 x − x 2 �
2
2 � 40
�
or
1 2
1
�
�
v = k �100 x − x 2 − 3200 �
2
2
�
�
When x = 100 mm, v = 18 mm/s:
1
1
�
�
(18) 2 = k �100(100) − (100) 2 − 3200 �
2
2
�
�
k = 0.0900 s −2 �
or
(b)
When x = 120 mm:
1 2
1
�
�
v = 0.09 �100(120) − (120) 2 − 3200 � = 144
2
2
�
�
v = ±16.97 mm/s ��
or
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23
23
PROBLEM 11.18
A particle starts from rest at the origin and is given an acceleration a = k/( x + 4)2 , where a and x are expressed
in mm/s2 and m, respectively, and k is a constant. Knowing that the velocity of the particle is 4 m/s when
x = 8 m, determine (a) the value of k, (b) the position of the particle when v = 4.5 m/s, (c) the maximum
velocity of the particle.
SOLUTION
(a)
We have
When x = 0, v = 0:
or
When x = 8 m, v = 4 m/s:
v
�
v
0
dv
k
=a=
dx
( x + 4)2
vdv =
�
x
0
k
dx
( x + 4) 2
1 2
1�
� 1
v = −k �
− �
2
� x+4 4�
1 2
1�
� 1
(4) = −k �
− �
2
�8+4 4�
k = 48 m3 /s 2 �
or
(b)
When v = 4.5 m/s:
1
1�
� 1
(4.5) 2 = −48 �
− �
2
� x+4 4�
x = 21.6 m �
or
(c)
Note that when v = vmax , a = 0.
Now a → 0 as x → ∞ so that
1 2
1 �
�1
�1�
vmax = 48lnx →−1∞ � −
= 48 � �
�
2
�4 x+4�
�4�
vmax = 4.90 m/s ��
or
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24
24
PROBLEM 11.19
A piece of electronic equipment that is surrounded by packing
material is dropped so that it hits the ground with a speed of
4 m/s. After impact the equipment experiences an acceleration
of a = − kx, where k is a constant and x is the compression of
the packing material. If the packing material experiences a
maximum compression of 20 mm, determine the maximum
acceleration of the equipment.
SOLUTION
a=
vdv
= −k x
dx
Separate and integrate.
�
vf
v0
vdv = −
�
xf
0
k x dx
1 2 1 2
1
v f − v0 = − kx 2
2
2
2
xf
0
1
= − k x 2f
2
Use v0 = 4 m/s, x f = 0.02 m, and v f = 0. Solve for k.
1
1
0 − (4) 2 = − k (0.02) 2
2
2
k = 40,000 s −2
Maximum acceleration.
amax = −kxmax : ( −40,000)(0.02) = −800 m/s 2
a = 800 m/s 2
��
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25
25
PROBLEM 11.20
Based on experimental observations, the acceleration of a particle is defined by the relation a = −(0.1 + sin x/b),
where a and x are expressed in m/s2 and meters, respectively. Knowing that b = 0.8 m and that v = 1 m/s
when x = 0, determine (a) the velocity of the particle when x = −1 m, (b) the position where the velocity is
maximum, (c) the maximum velocity.
SOLUTION
v
We have
When x = 0, v = 1 m/s:
�
v
1
dv
x �
�
= a = − � 0.1 + sin
0.8 ��
dx
�
vdv =
�
x
0
x �
�
− � 0.1 + sin
� dx
0.8
�
�
x
x �
1 2
�
(v − 1) = − �0.1x − 0.8 cos
�
2
0.8
�
�0
or
x
1 2
v = −0.1x + 0.8 cos
− 0.3
2
0.8
or
(a)
When x = −1 m:
−1
1 2
v = −0.1(−1) + 0.8 cos
− 0.3
2
0.8
v = ± 0.323 m/s �
or
(b)
When v = vmax, a = 0:
or
(c)
x �
�
− � 0.1 + sin
=0
0.8 ��
�
x = −0.080 134 m
x = −0.0801 m �
When x = −0.080 134 m:
1 2
−0.080 134
vmax = −0.1(−0.080 134) + 0.8 cos
− 0.3
2
0.8
= 0.504 m 2 /s 2
vmax = 1.004 m/s �
or
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26
PROBLEM 11.21
Starting from x = 0 with no initial velocity, a particle is given an acceleration a = 0.8 v 2 + 49, where a and
v are expressed in m/s2 and m/s, respectively. Determine (a) the position of the particle when v = 24 m/s,
(b) the speed of the particle when x = 40 m.
SOLUTION
v
We have
v
dv
= a = 0.8 v 2 + 49
dx
vdv
�
or
� v 2 + 49 � = 0.8 x
�
�0
or
v 2 + 49 − 7 = 0.8 x
(a)
0
v 2 + 49
=
�
x
When x = 0, v = 0:
0
0.8 dx
v
When v = 24 m/s:
242 + 49 − 7 = 0.8x
x = 22.5 m �
or
(b)
When x = 40 m:
v 2 + 49 − 7 = 0.8(40)
v = 38.4 m/s �
or
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27
27
PROBLEM 11.22
The acceleration of a particle is defined by the relation a = − k v , where k is a constant. Knowing that x = 0
and v = 81 m/s at t = 0 and that v = 36 m/s when x = 18 m, determine (a) the velocity of the particle when
x = 20 m, (b) the time required for the particle to come to rest.
SOLUTION
(a)
We have
v
dv
= a = −k v
dx
v dv = − k dx
so that
v
�
or
2 3/ 2 v
[v ]81 = −kx
3
or
2 3/ 2
[v − 729] = − kx
3
When x = 18 m, v = 36 m/s:
v dv =
81
�
x
When x = 0, v = 81 m/s:
0
− k dx
2
(363/ 2 − 729) = − k (18)
3
k = 19 m/s 2
or
Finally
When x = 20 m:
2 3/ 2
(v − 729) = −19(20)
3
v3/ 2 = 159
or
(b)
We have
dv
= a = −19 v
dt
At t = 0, v = 81 m/s:
�
v
dv
81
v
=
t
� −19dt
0
or
v
2[ v ]81
= −19t
or
2( v − 9) = −19t
When v = 0:
v = 29.3 m/s �
2(−9) = −19t
t = 0.947 s ��
or
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28
PROBLEM 11.23
The acceleration of a particle is defined by the relation a = − 0.8 v, where a is expressed in m/s2 and v in m/s.
Knowing that at t = 0 the velocity is 40 m/s, determine (a) the distance the particle will travel before coming
to rest, (b) the time required for the particle to come to rest, (c) the time required for the particle to be reduced
by 50 percent of its initial value.
SOLUTION
a=
(a)
vdv
= −0.8v
dx
dv = −0.8dx
Separate and integrate with v = 40 m/s when x = 0.
v
40
dv = −0.8
x
0
dx
v − 40 = −0.8 x
Distance traveled.
For v = 0,
(b)
Separate.
x=
−40
−0.8
a=
dv
= −0.8v
dt
x = 50.0 m
x
dv
= − 0.8dt
40 v
0
ln v − ln 40 = −0.8t
v
ln
v
= −0.8t
40
t = 1.25 ln
40
v
For v = 0, we get t = ∞.
t=∞
(c)
For v = 0.5(40 m/s) = 20 m/s,
t = 1.25 ln
40
= 0.866 s
20
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29
29
PROBLEM 11.24
A bowling ball is dropped from a boat so that it strikes the
surface of a lake with a speed of 8 m/s. Assuming the ball
experiences a downward acceleration of a = 10 − 0.9v2 when
in the water, determine the velocity of the ball when it strikes
the bottom of the lake.
SOLUTION
v0 = 8 m/s,
x − x0 = 10 m
a = 10 − 0.9v 2 = k (c 2 − v 2 )
10
= 11.111 m2/s2
0.9
c = 3.3333 m/s
k = 0.9 m−1 and c 2 =
Where
Since v0
c, write
a=v
dv
= − k (v 2 − c 2 )
dx
vdv
= − k dx
v − c2
2
v
Integrating,
1
ln(v 2 − c 2 ) = − k ( x − x0 )
2
v0
ln
v2 − c2
= −2k ( x − x0 )
v02 − c 2
v 2 − c2
= e−2 k ( x − x0 )
v02 − c 2
(
)
v 2 = c 2 + v02 − c 2 e −2 k ( x − x0 )
= 11.111 + [(8) 2 − 11.111] e− (2)(0.9)(10)
= 11.111 + 8.05 × 10−7 = 11.111 m2/s2
v = 3.33 m/s
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30
30
PROBLEM 11.25
The acceleration of a particle is defined by the relation a = 0.4(1 − kv), where k is a constant. Knowing that
at t = 0 the particle starts from rest at x = 4 m and that when t = 15 s, v = 4 m/s, determine (a) the constant k,
(b) the position of the particle when v = 6 m/s, (c) the maximum velocity of the particle.
SOLUTION
(a)
dv
= a = 0.4(1 − kv)
dt
We have
dv
=
0 1 − kv
v
�
At t = 0, v = 0:
t
� 0.4dt
0
1
− [ln(1 − kv)]v0 = 0.4t
k
or
or
ln(1 − kv) = −0.4kt
At t = 15 s, v = 4 m/s:
ln(1 − 4k ) = −0.4k (15)
(1)
= −6k
k = 0.145703 s/m
Solving yields
k = 0.1457 s/m �
or
(b)
v
We have
vdv
=
0 1 − kv
�
When x = 4 m, v = 0:
v
�
x
4
0.4dx
v
1
1/k
=− +
1 − kv
k 1 − kv
Now
Then
dv
= a = 0.4(1 − kv)
dx
�
1
1
�− +
� dv =
0
� k k (1 − kv) �
�
v�
�
x
4
0.4 dx
v
or
� v 1
�
x
� − k − k 2 ln(1 − kv) � = 0.4[ x]4
�
�0
or
�v 1
�
− � + 2 ln(1 − kv) � = 0.4( x − 4)
�k k
�
When v = 6 m/s:
�
�
6
1
ln(1 − 0.145 703 × 6) � = 0.4( x − 4)
−�
+
2
� 0.145 703 (0.145 703)
�
or
0.4( x − 4) = 56.4778
x = 145.2 m �
or
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31
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PROBLEM 11.25 (Continued)
(c)
The maximum velocity occurs when a = 0.
a = 0: 0.4(1 − kvmax ) = 0
vmax =
or
1
0.145 703
vmax = 6.86 m/s �
or
An alternative solution is to begin with Eq. (1).
ln(1 − kv) = −0.4kt
v=
Then
Thus, vmax is attained as t
1
(1 − k −0.4 kt )
k
∞
vmax =
1
k
as above.�
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32
PROBLEM 11.26
A particle is projected to the right from the position x = 0 with an initial velocity of 9 m/s. If the acceleration
of the particle is defined by the relation a = −0.6v3/ 2 , where a and v are expressed in m/s2 and m/s,
respectively, determine (a) the distance the particle will have traveled when its velocity is 4 m/s, (b) the time
when v = 1 m/s, (c) the time required for the particle to travel 6 m.
SOLUTION
(a)
v
We have
When x = 0, v = 9 m/s:
v
�
9
dv
= a = −0.6v3/2
dx
v − (3/2) dv =
�
x
0
−0.6dx
or
2[v1/2 ]9v = −0.6 x
or
x=
1
(3 − v1/ 2 )
0.3
When v = 4 m/s:
x=
1
(3 − 41/ 2 )
0.3
x = 3.33 m �
or
(b)
dv
= a = −0.6v3/2
dt
We have
v
t
When t = 0, v = 9 m/s:
�
or
−2[v − (1/2) ]v9 = −0.6t
or
When v = 1 m/s:
9
v − (3/2) dv =
1
v
1
1
� −0.6dt
0
−
1
= 0.3t
3
−
1
= 0.3t
3
t = 2.22 s �
or
(c)
We have
(1)
1
v
−
1
= 0.3t
3
2
or
Now
9
� 3 �
v=�
=
�
(1 + 0.9t ) 2
� 1 + 0.9t �
dx
9
=v=
dt
(1 + 0.9t ) 2
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33
PROBLEM 11.26 (Continued)
�
At t = 0, x = 0:
x
0
dx =
9
dt
0 (1 + 0.9t ) 2
�
t
t
1 �
� 1
x = 9 �−
�
� 0.9 1 + 0.9t � 0
or
1 �
�
= 10 �1 −
�
� 1 + 0.9t �
9t
=
1 + 0.9t
When x = 6 m:
6=
9t
1 + 0.9t
t = 1.667 s �
or
An alternative solution is to begin with Eq. (1).
x=
1
(3 − v1/ 2 )
0.3
dx
= v = (3 − 0.3x) 2
dt
Then
Now
At t = 0, x = 0:
�
x
0
dx
=
(3 − 0.3 x) 2
t
� dt
0
x
or
1 � 1 �
x
=
t=
�
�
0.3 � 3 − 0.3x � 0 9 − 0.9 x
Which leads to the same equation as above.
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34
PROBLEM 11.27
Based on observations, the speed of a jogger can be approximated by the
relation v = 7.5(1 − 0.04x)0.3, where v and x are expressed in km/h and kilometers respectively. Knowing that x = 0 at t = 0, determine (a) the distance
the jogger has run when t = 1 h, (b) the jogger’s acceleration in m/s2 at t = 0,
(c) the time required for the jogger to run 6 km.
SOLUTION
(a)
dx
= v = 7.5(1 − 0.04 x)0.3
dt
We have
x
At t = 0, x = 0:
or
0
dx
=
(1 − 0.04 x)0.3
t
0
7.5dt
1
1
[(1 − 0.04 x)0.7 ]0x = 7.5t
−
0.7 0.04
1 − (1 − 0.04 x)0.7 = 0.21t
or
or
x=
1
[1 − (1 − 0.21t )1/0.7 ]
0.04
At t = 1 h:
x=
1
{1 − [1 − 0.21(1)]1/0.7 }
0.04
x = 7.15 km
or
(b)
We have
(1)
a=v
dv
dx
d
[7.5(1 − 0.04 x)0.3 ]
dx
2
0.3
= 7.5 (1 − 0.04 x) [(0.3)(−0.04)(1 − 0.04 x) −0.7 ]
= 7.5(1 − 0.04 x)0.3
= −0.675(1 − 0.04 x) −0.4
At t = 0, x = 0:
1h
a0 = −0.675 km/h ×
3600 s
2
× (1000 m)
a0 = −52.1 × 10 −6 m/s2
or
(c)
2
From Eq. (1)
t=
When x = 6 mi:
t=
1
[1 − (1 − 0.04 x)0.7 ]
0.21
1
{1 − [1 − 0.04(6)]0.7 }
0.21
= 0.83229 h
t = 49.9 min
or
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35
35
PROBLEM 11.28
Experimental data indicate that in a region downstream of a
given louvered supply vent, the velocity of the emitted air is
defined by v = 0.18v0 /x, where v and x are expressed in m/s and
meters, respectively, and v0 is the initial discharge velocity of
the air. For v0 = 3.6 m/s, determine (a) the acceleration of the air
at x = 2 m, (b) the time required for the air to flow from x = 1 to
x = 3 m.
SOLUTION
(a)
dv
dx
0.18v0 d � 0.18v0 �
=
x dx �� x ��
We have
a=v
=−
When x = 2 m:
a=−
0.0324v02
x3
0.0324(3.6) 2
(2)3
a = −0.0525 m/s 2 �
or
(b)
0.18v0
dx
=v=
x
dt
We have
From x = 1 m to x = 3 m:
�
3
1
xdx =
�
t3
t1
0.18v0 dt
3
or
�1 2 �
� 2 x � = 0.18v0 (t3 − t1 )
�
�1
or
(t3 − t1 ) =
1
2
(9 − 1)
0.18(3.6)
t3 − t1 = 6.17 s ��
or
PROPRIETARY MATERIAL.
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36
36
PROBLEM 11.29
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
a=
−9.81
[1 + ( y/6.37 × 106 )]2
where a and y are expressed in m/s2 and meters, respectively. Using this expression,
compute the height reached by a projectile fired vertically upward from the surface
of the earth if its initial velocity is (a) 540 m/s, (b) 900 m/s, (c) 12,000 m/s.
SOLUTION
We have
v
dv
=a=−
dy
y = 0,
When
9.81
(
y
6.37 × 106
1+
)
2
v = v0
y = ymax , v = 0
0
Then
v0
ymax
(1 +
0
−9.81
y
6.37 × 106
v02 = 124.98 × 106 1 −
or
ymax =
or
dy
v0 = 540 m/s:
ymax =
v0 = 900 m/s:
ymax
0
1
1+
ymax
6.37 × 106
v02
19.62 −
v02
6.37 × 106
(540)2
19.62 −
(540) 2
6.37 × 106
ymax = 14897 m
or
(b)
)
2
1
1
− v02 = −9.81 −6.37 × 106
2
1 + 6.37y× 106
or
(a)
v dv =
ymax =
(900)2
19.62 −
(900) 2
6.37 × 106
ymax = 41554 m
or
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37
37
PROBLEM 11.29 (Continued)
(c)
v0 = 12,000 m/s:
ymax =
(12 000)2
19.62 −
(12 000) 2
6.37 × 106
ymax = −4019 m
or
The velocity 12,000 m/s is approximately the escape velocity vR from the earth. For vR
ymax → ∞
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38
38
PROBLEM 11.30
The acceleration due to gravity of a particle falling toward the earth
is a = − gR 2 /r 2, where r is the distance from the center of the earth to
the particle, R is the radius of the earth, and g is the acceleration due to
gravity at the surface of the earth. If R = 6370 km, calculate the escape
velocity, that is, the minimum velocity with which a particle must be
projected vertically upward from the surface of the earth if it is not to
return to the earth. (Hint: v = 0 for r = ∞.)
SOLUTION
We have
v
dv
gR 2
=a=− 2
dr
r
r = R, v = ve
When
r = ∞, v = 0
Then
or
or
or
0
ve
vdv =
∞
R
−
gR 2
dr
r2
1
1
− ve2 = gR 2
2
r
∞
R
ve = 2 gR
= (2 × 0.00981 km/s2 × 6370 km)1/2
ve = 11,179 m/s
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39
39
PROBLEM 11.31
The velocity of a particle is v = v0 [1 − sin (π t/T )]. Knowing that the particle starts from the origin with an
initial velocity v0 , determine (a) its position and its acceleration at t = 3T , (b) its average velocity during the
interval t = 0 to t = T .
SOLUTION
(a)
�
dx
� π t ��
= v = v0 �1 − sin � � �
dt
� T ��
�
We have
At t = 0, x = 0:
�
x
0
dx =
t
�v
0
0
�
� π t ��
�1 − sin � � � dt
� T ��
�
t
or
� T
� π t ��
x = v0 �t + cos � � �
� T �� 0
� π
� T
� πt � T �
= v0 �t + cos � � − �
�T � π�
� π
At t = 3T :
�
T
� π × 3T � T �
xBT = v0 �3T + cos �
�− �
π
� T � π�
�
2T �
�
= v0 � 3T −
π ��
�
x3T = 2.36 v0T �
or
a=
Also
At t = 3T :
π
πt
dv d � �
� π t �� �
= �v0 �1 − sin � � � � = −v0 cos
dt dt � �
T
T
� T �� �
a3T = −v0
π
T
cos
π × 3T
T
a3T =
or
(b)
(1)
π v0
T
�
Using Eq. (1)
At t = 0:
T
T�
�
x0 = v0 �0 + cos(0) − � = 0
π
π�
�
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40
PROBLEM 11.31 (Continued)
At t = T :
Now
�
T
� πT
xT = v0 �T + cos �
π
� T
�
2T �
�
= v0 � T −
π ��
�
= 0.363v0T
vave =
� T�
�−π �
�
�
xT − x0 0.363v0T − 0
=
∆t
T −0
vave = 0.363v0 �
or
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41
41
PROBLEM 11.32
The velocity of a slider is defined by the relation v = v′ sin (ωn t + φ ). Denoting the velocity and the position
of the slider at t = 0 by v0 and x0 , respectively, and knowing that the maximum displacement of the slider
is 2 x0 , show that (a) v′ = v02 + x02ωn2 2 x0ωn , (b) the maximum value of the velocity occurs when
x = x0 [3 − (v0 /x0ωn ) 2 ]/2.
(
)
SOLUTION
(a)
v0 = v′ sin (0 + φ ) = v′ sin φ
At t = 0, v = v0 :
Then
2
cos φ = v′ − v02 v′
dx
= v = v′ sin (ωn t + φ )
dt
Now
x
t
At t = 0, x = x0 :
�
or
� 1
�
cos (ωn t + φ ) �
x − x0 = v′ � −
� ωn
�0
x0
dx =
� v′ sin (ω t + φ )dt
n
0
t
or
x = x0 +
v′
ωn
[cos φ − cos (ωn t + φ )]
Now observe that xmax occurs when cos (ωn t + φ ) = −1. Then
xmax = 2 x0 = x0 +
Substituting for cos φ
or
x0 =
v′
ωn
[cos φ − ( −1)]
2
2
�
v′ �� v′ − v0
�
+
1
�
ωn �
v1
�
�
x0ωn − v1 = v′2 − v02
Squaring both sides of this equation
x02ωn2 − 2 x0ωn + v′2 = v′2 − v02
or
v′ =
v02 + x02ωn2
2 x0ωn
Q. E. D.
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42
PROBLEM 11.32 (Continued)
(b)
First observe that vmax occurs when ωn t + φ = π2 . The corresponding value of x is
v′ �
� π ��
�cos φ − cos � � �
ωn �
� 2 ��
v′
cos φ
= x0 +
xvmax = x0 +
ωn
Substituting first for cos φ and then for v′
xvmax = x0 +
2
v′ − v02
v′
v′
ωn
1/ 2
2
�
�
1 �� v02 + x02ωn2 �
2
= x0 +
��
�� − v0 �
�
ωn �� 2 x0ωn �
�
�
1
v04 + 2v02 x02ωn2 + x04ωn4 − 4 x02ωn2 v02
= x0 +
2 x0ωn2
(
= x0 +
= x0 +
1
(
� x 2ω 2 − v 2
0
2 � 0 n
2 x0ωn �
)
1/ 2
1/ 2
2�
) ��
x02ωn2 − v02
2 x0ωn2
2
x0 � � v0 � �
= �3 − �
� �
2 � � x0ωn � �
�
�
Q. E. D.
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43
43
PROBLEM 11.33
A motorist enters a freeway at 45 km/h and accelerates
uniformly to 99 km/h. From the odometer in the car, the
motorist knows that she traveled 0.2 km while accelerating.
Determine (a) the acceleration of the car, (b) the time
required to reach 99 km/h.
SOLUTION
(a)
Acceleration of the car.
v12 = v02 + 2a( x1 − x0 )
a=
Data:
v12 − v02
2( x1 − x0 )
v0 = 45 km/h = 12.5 m/s
v1 = 99 km/h = 27.5 m/s
x0 = 0
x1 = 0.2 km = 200 m
a=
(b)
(27.5) 2 − (12.5) 2
(2)(200 − 0)
a = 1.500 m/s 2 �
Time to reach 99 km/h.
v1 = v0 + a (t1 − t0 )
v1 − v0
a
27.5 − 12.5
=
1.500
= 10.00 s
t1 − t0 =
t1 − t0 = 10.00 s ��
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44
PROBLEM 11.34
A truck travels 220 m in 10 s while being decelerated at a
constant rate of 0.6 m/s2. Determine (a) its initial velocity,
(b) its final velocity, (c) the distance traveled during the
first 1.5 s.
SOLUTION
(a)
x = x0 + v0t +
Initial velocity.
1 2
at
2
x − x0 1
− at
t
2
220 1
=
− (−0.6)(10)
10 2
v0 =
(b)
Final velocity.
v = v0 + at
v = 25.0 + ( −0.6)(10)
(c)
v0 = 25.0 m/s
vf = 19.00 m/s
Distance traveled during first 1.5 s.
x = x0 + v0t +
1 2
at
2
= 0 + (25.0)(1.5) +
1
(−0.6)(1.5) 2
2
x = 36.8 m
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45
45
PROBLEM 11.35
Assuming a uniform acceleration of 3 m/s2 and knowing
that the speed of a car as it passes A is 50 km/h, determine
(a) the time required for the car to reach B, (b) the speed of
the car as it passes B.
SOLUTION
(a)
Time required to reach B.
vA = 50 km/h = 13.89 m/s,
xA = 0,
xB = 50 m,
a = 3 m/s2
1 2
at
2
1
50 = 0 + 13.89t + (3) t2
2
xB = x A + v A t +
1.5t 2 + 13.89t − 50 = 0
t=
=
Rejecting the negative root.
(b)
(13.89)2 − (4)(1.5)(−50)
(2)(1.5)
−13.89 ±
−13.89 ± 22.2
3
t = 2.78 s
t = 2.78 s
Speed at B.
vB = vA + at = 13.89 + (3)(2.78) = 22.23 m/s
vB = 80 km/h
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46
PROBLEM 11.36
A group of students launches a model rocket in the vertical direction. Based on tracking
data, they determine that the altitude of the rocket was 89.6 m at the end of the powered
portion of the flight and that the rocket landed 16 s later. Knowing that the descent
parachute failed to deploy so that the rocket fell freely to the ground after reaching its
maximum altitude and assuming that g = 9.81 m/s2, determine (a) the speed v1 of the
rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.
SOLUTION
(a)
1 2
at
2
We have
y = y1 + v1t +
At tland ,
y=0
Then
0 = 89.6 m + v1(16 s)
+
1
(−9.81 m/s2)(16 s)2
2
v1 = 72.9 m/s
or
(b)
v 2 = v12 + 2a( y − y1 )
We have
At
y = ymax , v = 0
Then
0 = (72.9 m/s)2 + 2(−9.81 m/s2)(ymax − 89.6) m
ymax = 360 m
or
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47
47
PROBLEM 11.37
A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs
with constant velocity. If the sprinter’s time for the first 35 m in 5.4 s, determine
(a) his acceleration, (b) his final velocity, (c) his time for the race.
SOLUTION
0 � × � 35 m, a = constant
Given:
35 m � × � 100 m, v = constant
At t = 0, v = 0 when
x = 35 m, t = 5.4 s
Find:
(a)
a
(b)
v when x = 100 m
(c)
t when x = 100 m
(a)
We have
At t = 5.4 s:
or
x = 0 + 0t +
35 m =
1 2
at
2
0 � x � 35 m
for
1
a(5.4 s)2
2
a = 2.4005 m/s 2
a = 2.40 m/s 2 �
(b)
First note that v = vmax for 35 m � x � 100 m.
Now
(c)
v 2 = 0 + 2a( x − 0)
for
0 � x � 35 m
When x = 35 m:
2
vmax
= 2(2.4005 m/s 2 )(35 m)
or
vmax = 12.9628 m/s
We have
When x = 100 m:
vmax = 12.96 m/s �
x = x1 + v0 (t − t1 )
35 m � x � 100 m
for
100 m = 35 m + (12.9628 m/s)(t2 − 5.4) s
t2 = 10.41 s �
or
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48
48
PROBLEM 11.38
A small package is released from rest at A and
moves along the skate wheel conveyor ABCD.
The package has a uniform acceleration of
4.8 m/s 2 as it moves down sections AB and CD,
and its velocity is constant between B and C. If
the velocity of the package at D is 7.2 m/s,
determine (a) the distance d between C and D,
(b) the time required for the package to reach D.
SOLUTION
(a)
For
A→ B
and
C→D
v 2 = v02 + 2a( x − x0 )
we have
Then,
2
= 0 + 2(4.8 m/s 2 )(3 − 0) m
vBC
at B
= 28.8 m 2 /s 2
2
vD2 = vBC
+ 2aCD ( xD − xC )
and at D
d = 2.40 m �
or
For
A→ B
and
C → D,
v = v0 + at
we have
Then
d = xd − xC
(7.2 m/s)2 = (28.8 m 2 /s 2 ) + 2(4.8 m/s 2 )d
or
(b)
(vBC = 5.3666 m/s)
A→ B
5.3666 m/s = 0 + (4.8 m/s 2 )t AB
t AB = 1.11804 s
or
and
or
C→D
7.2 m/s = 5.3666 m/s + (4.8 m/s 2 )tCD
tCD = 0.38196 s
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49
49
PROBLEM 11.38 (Continued)
Now,
for
we have
B → C,
xC = xB + vBC t BC
or
3 m = (5.3666 m/s)t BC
or
t BC = 0.55901 s
Finally,
t D = t AB + t BC + tCD = (1.11804 + 0.55901 + 0.38196) s
t D = 2.06 s �
or
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50
50
PROBLEM 11.39
A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a
slow, constant speed. Believing that the driver of the automobile might be intoxicated, the officer starts his
car, accelerates uniformly to 90 km/h in 8 s, and, maintaining a constant velocity of 90 km/h, overtakes the
motorist 42 s after the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing
the motorist, determine (a) the distance the officer traveled before overtaking the motorist, (b) the motorist’s
speed.
SOLUTION
(vP )18 = 0
(a)
(vP )26 = 90 km/h = 25 m/s
(vP ) 42 = 90 km/h = 25 m/s
Patrol car:
For 18 s � t � 26 s:
At t = 26 s:
vP = 0 + aP (t − 18)
25 m/s = aP (26 − 18) s
or
aP = 3.125 m/s 2
Also,
xP = 0 + 0(t − 18) −
At t = 26 s:
For 26 s � t � 42 s:
At t = 42 s:
(xP ) 26 =
1
aP (t − 18) 2
2
1
(3.125 m/s 2 )(26 − 18) 2 = 100 m
2
xP = ( xP )26 + (vP ) 26 (t − 26)
(xP ) 42 = 100 m + (25 m/s)(42 − 26)s
= 500 m
( xP )42 = 0.5 km �
(b)
For the motorist’s car:
xM = 0 + vM t
At t = 42 s, xM = xP :
500 m = vM (42 s)
or
vM = 11.9048 m/s
vM = 42.9 km/h �
or
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51
51
PROBLEM 11.40
As relay runner A enters the 20-m-long exchange zone with a
speed of 12.9 m/s, he begins to slow down. He hands the baton to
runner B 1.82 s later as they leave the exchange zone with the
same velocity. Determine (a) the uniform acceleration of each of
the runners, (b) when runner B should begin to run.
SOLUTION
(a)
For runner A:
At t = 1.82 s:
x A = 0 + (v A )0 t +
1
a At 2
2
20 m = (12.9 m/s)(1.82 s) +
1
a A (1.82 s) 2
2
a A = −2.10 m/s 2 �
or
Also
At t = 1.82 s:
v A = (v A ) 0 + a A t
(v A )1.82 = (12.9 m/s) + ( −2.10 m/s 2 )(1.82 s)
= 9.078 m/s
For runner B:
vB2 = 0 + 2aB [ xB − 0]
When
xB = 20 m, vB = v A : (9.078 m/s) 2 + 2aB (20 m)
or
aB = 2.0603 m/s 2
aB = 2.06 m/s 2 �
(b)
For runner B:
vB = 0 + aB (t − t B )
where t B is the time at which he begins to run.
At t = 1.82 s:
or
9.078 m/s = (2.0603 m/s 2 )(1.82 − t B ) s
t B = −2.59 s
Runner B should start to run 2.59 s before A reaches the exchange zone.
�
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52
52
PROBLEM 11.41
Automobiles A and B are traveling in adjacent highway lanes
and at t = 0 have the positions and speeds shown. Knowing
that automobile A has a constant acceleration of 0.5 m/s2 and
that B has a constant deceleration of 0.3 m/s2, determine
(a) when and where A will overtake B, (b) the speed of each
automobile at that time.
SOLUTION
aA = +0.5 m/s2
aB = −0.3 m/s2
| v A |0 = 36 km/h = 10 m/s
A overtakes B
| vB |0 = 54 km/h = 15 m/s
Motion of auto A:
v A = (v A )0 + a At = 10 + 0.5t
(1)
1
1
xA = (xA)0 + (vA)0t + aA t2 = 0 + 10t + (0.5)t2
2
2
(2)
vB = (vB )0 + aB t = 15 − 0.3t
(3)
Motion of auto B:
x B = ( x B ) 0 + ( vB ) 0 t +
(a)
1
1
aB t 2 = 24 + 15t + (−0.3)t2
2
2
(4)
A overtakes B at t = t1 .
x A = xB : 10t + 0.25t12 = 24 + 15t1 − 0.15t12
0.4t12 − 5t1 − 24 = 0
t1 = −3.65 s and t1 = 16.2 s
Eq. (2):
x A = 10(16.2) + 0.25(16.2)2
t1 = 16.2 s
x A = 227.6 m
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53
53
PROBLEM 11.41 (Continued)
(b)
Velocities when
t1 = 16.2 s
Eq. (1):
vA = 10 + 0.5(16.2)
vA = 18.1 m/s
Eq. (3):
vA = 65.2 km/h
vB = 15 − 0.3(16.2)
vB = 10.14 m/s
vB = 36.5 km/h →
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54
54
PROBLEM 11.42
In a boat race, boat A is leading boat B by 36 m
and both boats are traveling at a constant speed
of 168 km/h. At t = 0, the boats accelerate at
constant rates. Knowing that when B passes A,
t = 8 s and v A = 216 km/h, determine (a) the
acceleration of A, (b) the acceleration of B.
SOLUTION
(a)
We have
v A = (v A ) 0 + a A t
(v A )0 = 168 km/h = 46.67 m/s
At t = 8 s:
Then
v A = 216 km/h = 60 m/s
60 m/s = 46.67 m/s + aA (8s)
a A = 1.67 m/s2
or
(b)
We have
x A = ( x A ) 0 + (v A ) 0 t +
and
xB = 0 + (vB )0 t +
At t = 8 s:
x A = xB
1
a At 2
2
1
aB t 2
2
( x A )0 = 36 m
(vB )0 = 46.67 m/s
1
(1.67 m/s2)(8s)2
2
1
= (46.67 m/s)(8s) + aB (8s)2
2
36 m + (46.67 m/s)(8 s) +
aB = 2.8 m/s2
or
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55
55
PROBLEM 11.43
Boxes are placed on a chute at uniform intervals of time t R and
slide down the chute with uniform acceleration. Knowing that as
any box B is released, the preceding box A has already slid 6 m
and that 1 s later they are 10 m apart, determine (a) the value
of t R , (b) the acceleration of the boxes.
SOLUTION
Let tS = 1 s be the time when the boxes are 10 m apart.
a A = aB = a; ( x A )0 = ( xB )0 = 0; (v A )0 = (vB )0 = 0.
Let
(a)
1 2
at
2
For
t > 0, x A =
For
t > t R , xB =
At
t = tR , xA = 6 m
At
t = t R + tS ,
1
a(t − t R ) 2
2
6=
1 2
at R
2
(1)
x A − xB = 10 m
1
1
a(t R + t S ) 2 − a(t R + tS − t R )2
2
2
1 2
1
1
= at R + at R tS + at S2 − atS2 = 6 + atRtS
2
2
2
10 − 6 4
at R =
= = 4 m/s
tS
1
10 =
(2)
Dividing Equation (1) by Eq. (2),
1
2
at R2
at R
(b)
=
1
6
tR =
2
4
t R = 3.00 s
Solving Eq. (2) for a,
a=
4
= 1.33 m/s2
3
a = 1.33 m/s2
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56
56
PROBLEM 11.44
Two automobiles A and B are approaching
each other in adjacent highway lanes. At
t = 0, A and B are 1 km apart, their speeds
are v A = 108 km/h and vB = 63 km/h, and
they are at Points P and Q, respectively.
Knowing that A passes Point Q 40 s after B
was there and that B passes Point P 42 s
after A was there, determine (a) the
uniform accelerations of A and B, (b) when
the vehicles pass each other, (c) the speed
of B at that time.
SOLUTION
(a)
x A = 0 + (v A )0 t +
We have
At t = 40 s:
1
a At 2
2
1000 m = (30 m/s)(40 s) +
(v A )0 = 108 km/h = 30 m/s
1
a A (40 s)2
2
a A = −0.250 m/s 2 �
or
x B = 0 + ( vB ) 0 t +
Also,
At t = 42 s:
1000 m = (17.5 m/s)(42 s) +
(vB )0 = 63 km/h = 17.5 m/s
1
aB (42 s) 2
2
aB = 0.30045 m/s 2
or
(b)
1
aB t 2
2
aB = 0.300 m/s 2 �
When the cars pass each other
x A + xB = 1000 m
Then
or
Solving
1
2
(−0.250 m/s)t AB
+ (17.5 m/s)t AB
2
1
2
+ (0.30045 m/s 2 )t AB
= 1000 m
2
(30 m/s)t AB +
2
2
0.05045t AB
+ 95t AB
− 2000 = 0
t = 20.822 s and t = −1904 s
t � 0 � t AB = 20.8 s �
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57
57
PROBLEM 11.44 (Continued)
(c)
We have
vB = ( vB ) 0 + a B t
At t = t AB :
vB = 17.5 m/s + (0.30045 m/s 2 )(20.822 s)
= 23.756 m/s
vB = 85.5 km/h �
or
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58
PROBLEM 11.45
Car A is parked along the northbound
lane of a highway, and car B is traveling
in the southbound lane at a constant
speed of 96 km/h. At t = 0, A starts and
accelerates at a constant rate a A , while
at t = 5 s, B begins to slow down with a
constant deceleration of magnitude a A /6.
Knowing that when the cars pass each
other x = 88 m and v A = vB , determine
(a) the acceleration a A , (b) when the
vehicles pass each other, (c) the distance d
between the vehicles at t = 0.
SOLUTION
For t
vA = 0 + aAt
0:
xA = 0 + 0 +
0
t
5 s:
1
aA t 2
2
xB = 0 + (vB )0 t (vB )0 = 90 km/h = 25 m/s
At t = 5 s:
xB = (25 m/s)(5 s) = 125 m
For t
1
vB = (vB )0 + aB (t − s ) aB = − a A
6
5 s:
xB = ( xB ) S + (vB )0 (t − s ) +
Assume t
1
aB (t − s ) 2
2
5 s when the cars pass each other.
At that time (t AB ),
v A = vB :
a At AB = (25 m/s) −
x A = 88 m:
88 m =
aA
(tAB − 5)
6
1
2
a At AB
2
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59
59
PROBLEM 11.45 (Continued)
a A( 76 t AB − 56 )
Then
1
2
=
25
88
12.5t 2AB − 102.7t AB + 73.3 = 0
or
t AB = 0.79 s and
Solving
(a)
a At A2 B
With t AB
5 s,
88 m =
tAB = 7.4 s
1
a A (7.4 s)2
2
a A = 3.2 m/s2
or
(b)
t AB = 7.00 s
From above
Note: An acceptable solution cannot be found if it is assumed that t AB
(c)
We have
5 s.
d = x + ( xB )t AB
= 88 m + 125 m + (25 m/s)(7.4 s) s
+
1 1
− × 3.2 m/s2 (7.4 s)2s2
2 6
d = 383 m
or
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60
60
PROBLEM 11.46
Two blocks A and B are placed on an incline as shown. At
t = 0, A is projected up the incline with an initial velocity
of 8 m/s and B is released from rest. The blocks pass each
other 1 s later, and B reaches the bottom of the incline
when t = 3.4 s. Knowing that the maximum distance from
the bottom of the incline reached by block A is 6 m and
that the accelerations of A and B (due to gravity and
friction) are constant and are directed down the incline,
determine (a) the accelerations of A and B, (b) the distance d,
(c) the speed of A when the blocks pass each other.
SOLUTION
(a)
We have
v A2 = (v A )02 + 2a A [ x A − 0]
When
x A = ( x A ) max , v A = 0
Then
or
0 = (8 m/s)2 + 2aA (6 m)
a A = −5.3 m/s2
a A = 5.3 m/s2
or
Now
x A = 0 + (v A )0 t +
and
xB = 0 + 0t +
1
a At 2
2
1
aB t 2
2
At t = 1 s, the blocks pass each other.
( x A )1 + ( xB )1 = d
At t = 3.4 s, xB = d :
Thus
or
( x A )1 + ( xB )1 = ( xB )3.4
(8 m/s)(1 s) +
+
or
1
( −5.3 m/s2)(1 s)2
2
1
1
aB (1 s) 2 = aB (3.4 s) 2
2
2
aB = 1.01 m/s2
a B = 1.01 m/s2
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61
61
PROBLEM 11.46 (Continued)
(b)
At t = 3.4 s, xB = d :
d=
1
(1.01 m/s2)(3.4 s)2
2
d = 5.8 m
or
(c)
We have
v A = (v A )0 + aA t
At t = 1 s:
v A = 8 m/s + (−5.3 m/s)(1 s)
v A = 2.7 m/s
or
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62
62
PROBLEM 11.47
Slider block A moves to the left with a constant velocity of 6 m/s.
Determine (a) the velocity of block B, (b) the velocity of portion D of
the cable, (c) the relative velocity of portion C of the cable with respect
to portion D.
SOLUTION
From the diagram, we have
x A + 3 yB = constant
Then
v A + 3vB = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Substituting into Eq. (1)
6 m/s + 3vB = 0
v B = 2 m/s ↑ �
or
(b)
From the diagram
yB + yD = constant
Then
vB + vD = 0
v D = 2 m/s ↓ ��
(c)
�
From the diagram
x A + yC = constant
Then
v A + vC = 0
Now
vC/D = vC − vD = (−6 m/s) − (2 m/s) = − 8 m/s
vC = −6 m/s
vC/D = 8 m/s ↑ ��
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63
63
PROBLEM 11.48
Block B starts from rest and moves downward with a constant
acceleration. Knowing that after slider block A has moved 400 mm its
velocity is 4 m/s, determine (a) the accelerations of A and B, (b) the
velocity and the change in position of B after 2 s.
SOLUTION
From the diagram, we have
x A + 3 yB = constant
Then
v A + 3vB = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Eq. (2): a A + 3aB = 0 and a B is constant and
positive � a A is constant and negative
Also, Eq. (1) and (vB )0 = 0 � (v A )0 = 0
v A2 = 0 + 2a A [ x A − ( x A )0 ]
Then
When |∆ x A | = 0.4 m:
(4 m/s) 2 = 2a A (0.4 m)
a A = 20 m/s 2 → ��
or
Then, substituting into Eq. (2):
−20 m/s 2 + 3aB = 0
aB =
or
20
m/s 2
3
a B = 6.67 m/s 2 ↓ ��
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64
PROBLEM 11.48 (Continued)
(b)
We have
vB = 0 + a B t
At t = 2 s:
� 20
�
m/s 2 � (2 s)
vB = �
� 3
�
v B = 13.33 m/s ↓ ��
or
yB = ( yB )0 + 0 +
Also
At t = 2 s:
or�
yB − ( yB )0 =
1
aB t 2
2
1 � 20
�
m/s 2 � (2 s) 2 �
2 �� 3
�
y B − (y B )0 = 13.33 m ↓ ��
�
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65
65
PROBLEM 11.49
The elevator shown in the figure moves downward with a constant
velocity of 4.5 m/s. Determine (a) the velocity of the cable C, (b) the
velocity of the counterweight W, (c) the relative velocity of the cable C
with respect to the elevator, (d) the relative velocity of the
counterweight W with respect to the elevator.
SOLUTION
Choose the positive direction downward.
(a)
Velocity of cable C.
yC + 2 yE = constant
vC + 2vE = 0
vE = 4.5 m/s
But,
or
(b)
vC = −2vE = −9 m/s
vC = 9 m/s ↑
Velocity of counterweight W.
yW + yE = constant
vW + vE = 0 vW = −vE = −9 m/s
(c)
vW = 9 m/s ↑
Relative velocity of C with respect to E.
vC/E = vC − vE = (−9 m/s) − (+4.5 m/s) = −13.5 m/s
vC/E = 13.5 m/s ↑
(d)
Relative velocity of W with respect to E.
vW/E = vW − vE = (−4.5 m/s) − (4.5 m/s) = −9 m/s
vW/E = 9 m/s ↑
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66
66
PROBLEM 11.50
The elevator shown starts from rest and moves upward with a constant
acceleration. If the counterweight W moves through 10 m in 5 s, determine
(a) the accelerations of the elevator and the cable C, (b) the velocity of the
elevator after 5 s.
SOLUTION
We choose Positive direction downward for motion of counterweight.
yW =
At t = 5 s,
1
aW t 2
2
yW = 10 m
10 m =
1
aW (5 s) 2
2
aW = 0.4 m/s2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW = 0.4 m/s2 ↓
yW + yE = constant vW + vE = 0, and aW + aE = 0
aE = − aW = −(0.4 m/s2),
a E = 0.4 m/s2 ↑
yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0
aC = −2aE = −2(−0.4 m/s2) = + 0.8 m/s2,
aC = 0.8 m/s2 ↓
Velocity of elevator after 5 s.
vE = (vE )0 + aE t = 0 + (−0.4 m/s2)(5 s) = −2 m/s
( v E )5 = 2 m/s ↑
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67
67
PROBLEM 11.51
Collar A starts from rest and moves upward with a constant acceleration. Knowing
that after 8 s the relative velocity of collar B with respect to collar A is 600 mm/s,
determine (a) the accelerations of A and B, (b) the velocity and the change in position
of B after 6 s.
SOLUTION
From the diagram
2 y A + yB + ( yB − y A ) = constant
Then
v A + 2 vB = 0
(1)
and
a A + 2aB = 0
(2)
(a)
Eq. (1) and (v A )0 = 0
(vB )0
Also, Eq. (2) and a A is constant and negative
a B is constant
and positive
v A = 0 + a At
Then
vB = 0 + a B t
vB/A = vB − v A = (aB − a A )t
Now
1
aB = − a A
2
From Eq. (2)
3
vB/A = − a At
2
So that
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68
PROBLEM 11.51 (Continued)
At t = 8 s:
3
600 mm/s = − a A (8 s)
2
a A = 50 mm/s2 ↑
or
and then
1
aB = − (−50 mm/s2)
2
a B = 25 mm/s2 ↑
or
(b)
At t = 6 s:
vB = (25 mm/s2)(6 s)
v B = 150 mm/s ↓
or
Now
At t = 6 s:
yB = ( yB )0 + 0 +
y B − ( y B )0 =
1
aB t 2
2
1
(25 mm/s2)(6 s)2
2
y B − (y B )0 = 450 mm ↓
or
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69
PROBLEM 11.52
In the position shown, collar B moves downward with a velocity of 300 mm/s. Determine
(a) the velocity of collar A, (b) the velocity of portion C of the cable, (c) the relative
velocity of portion C of the cable with respect to collar B.
SOLUTION
From the diagram
2 y A + yB = ( yB − y A ) = constant
Then
v A + 2 vB = 0
(1)
and
a A + 2aB = 0
(2)
(a)
Substituting into Eq. (1)
v A + 2(300 mm/s) = 0
v A = 600 mm/s ↑
or
(b)
From the diagram
2 y A + yC = constant
Then
2v A + vC = 0
2(−600 mm/s) + vC = 0
Substituting
vC = 1.2 m/s ↓
or
(c)
vC/B = vC − vB
We have
= (1200 mm/s) − (300 mm/s)
vC/B = 900 mm/s ↓
or
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70
70
PROBLEM 11.53
Slider block B moves to the right with a constant
velocity of 300 mm/s. Determine (a) the velocity
of slider block A, (b) the velocity of portion C of
the cable, (c) the velocity of portion D of the
cable, (d) the relative velocity of portion C of the
cable with respect to slider block A.
SOLUTION
From the diagram
xB + ( xB − x A ) − 2 x A = constant
Then
2vB − 3v A = 0
(1)
and
2aB − 3a A = 0
(2)
Also, we have
− xB − x A = constant
vD + v A = 0
Then
(a)
Substituting into Eq. (1)
2(300 mm/s) − 3v A = 0
v A = 200 mm/s → ��
or
(b)
From the diagram
Then
Substituting
(3)
xB + ( xB − xC ) = constant
2vB − vC = 0
2(300 mm/s) − vC = 0
vC = 600 mm/s → �
or
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71
PROBLEM 11.53 (Continued)
(c)
From the diagram
Then
Substituting
( xC − x A ) + ( xB − x A ) = constant
vC − 2v A + vD = 0
600 mm/s − 2(200 mm/s) + vD = 0
v D = 200 mm/s ← �
or
(d)
We have
vC/A = vC − v A
= 600 mm/s − 200 mm/s
vC/A = 400 mm/s → ��
or
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72
72
PROBLEM 11.54
At the instant shown, slider block B is moving
with a constant acceleration, and its speed is
150 mm/s. Knowing that after slider block A
has moved 240 mm to the right its velocity is
60 mm/s, determine (a) the accelerations of A
and B, (b) the acceleration of portion D of the
cable, (c) the velocity and change in position of
slider block B after 4 s.
SOLUTION
xB + ( xB − x A ) − 2 x A = constant
From the diagram
Then
2vB − 3v A = 0
(1)
and
2aB − 3a A = 0
(2)
(a)
First observe that if block A moves to the right, v A → and Eq. (1) � v B → . Then, using
Eq. (1) at t = 0
2(150 mm/s) − 3(v A )0 = 0
(v A )0 = 100 mm/s
or
Also, Eq. (2) and aB = constant � a A = constant
v A2 = (v A )02 + 2a A [ x A − ( x A )0 ]
Then
When x A − ( x A )0 = 240 mm:
(60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm)
or
aA = −
40
mm/s 2
3
a A = 13.33 mm/s 2 ← ��
or
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73
73
PROBLEM 11.54 (Continued)
Then, substituting into Eq. (2)
� 40
�
2aB − 3 � −
mm/s 2 � = 0
� 3
�
aB = −20 mm/s 2
or
(b)
a B = 20.0 mm/s 2 ← ��
From the solution to Problem 11.53
vD + v A = 0
Then
Substituting
aD + a A = 0
� 40
�
mm/s 2 � = 0
aD + � −
3
�
�
a D = 13.33 mm/s 2 → �
or
(c)
We have
v B = ( vB ) 0 + a B t
At t = 4 s:
vB = 150 mm/s + ( −20.0 mm/s 2 )(4 s)
v B = 70.0 mm/s → �
or
Also
At t = 4 s:
y B = ( y B ) 0 + ( vB ) 0 t +
1
aB t 2
2
yB − ( yB )0 = (150 mm/s)(4 s)
+
1
(−20.0 mm/s 2 )(4 s) 2
2
y B − (y B )0 = 440 mm → ��
or
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74
74
PROBLEM 11.55
Block B moves downward with a constant velocity of 20 mm/s.
At t = 0, block A is moving upward with a constant acceleration,
and its velocity is 30 mm/s. Knowing that at t = 3 s slider block C
has moved 57 mm to the right, determine (a) the velocity of slider
block C at t = 0, (b) the accelerations of A and C, (c) the change in
position of block A after 5 s.
SOLUTION
From the diagram
3 y A + 4 yB + xC = constant
Then
3v A + 4vB + vC = 0
(1)
and
3a A + 4aB + aC = 0
(2)
v B = 20 mm/s ↓ ;
Given:
( v A )0 = 30 mm/s ↑
(a)
Substituting into Eq. (1) at t = 0
3(−30 mm/s) + 4(20 mm/s) + (vC )0 = 0
vC = 10 mm/s
(b)
xC = ( xC )0 + (vC )0 t +
We have
At t = 3 s:
or
1
aC t 2
2
57 mm = (10 mm/s)(3 s) +
1
aC (3 s) 2
2
aC = 6 mm/s 2
�
( vC )0 = 10 mm/s → �
or
aC = 6 mm/s 2 → ��
v B = constant → aB = 0
Now
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75
75
PROBLEM 11.55 (Continued)
Then, substituting into Eq. (2)
3a A + 4(0) + (6 mm/s 2 ) = 0
a A = −2 mm/s 2
(c)
We have
At t = 5 s:
y A = ( y A )0 + (v A )0 t +
or
1
a At 2
2
y A − ( y A )0 = (−30 mm/s)(5 s) +
= −175 mm
a A = 2 mm/s 2 ↑ �
1
(−2 mm/s 2 )(5 s) 2
2
y A − (y A )0 = 175 mm ↑ �
or
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76
76
PROBLEM 11.56
Block B starts from rest, block A moves with a constant
acceleration, and slider block C moves to the right with a
constant acceleration of 75 mm/s 2 . Knowing that at t = 2 s the
velocities of B and C are 480 mm/s downward and 280 mm/s to
the right, respectively, determine (a) the accelerations of A and B,
(b) the initial velocities of A and C, (c) the change in position of
slider C after 3 s.
SOLUTION
From the diagram
3 y A + 4 yB + xC = constant
Then
3v A + 4vB + vC = 0
(1)
and
3a A + 4aB + aC = 0
(2)
(vB ) = 0,
Given:
a A = constent
(aC ) = 75 mm/s 2 →
v B = 480 mm/s ↓
At t = 2 s,
(a)
vC = 280 mm/s →
Eq. (2) and a A = constant and aC = constant � aB = constant
vB = 0 + a B t
Then
At t = 2 s:
480 mm/s = aB (2 s)
aB = 240 mm/s 2
�
or
a B = 240 mm/s 2 ↓ ��
or
a A = 345 mm/s 2 ↑ �
Substituting into Eq. (2)
3a A + 4(240 mm/s 2 ) + (75 mm/s 2 ) = 0
a A = −345 mm/s
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77
PROBLEM 11.56 (Continued)
(b)
vC = (vC )0 + aC t
We have
At t = 2 s:
280 mm/s = (vC )0 + (75 mm/s)(2 s)
vC = −130 mm/s
�
or
( vC )0 = 130 mm/s → ��
or
(v A )0 = 43.3 mm/s ↑ �
Then, substituting into Eq. (1) at t = 0
3(v A )0 + 4(0) + (130 mm/s) = 0
v A = −43.3 mm/s
(c)
We have
At t = 3 s:
xC = ( xC )0 + (vC )0 t +
1
aC t 2
2
xC − ( xC )0 = (130 mm/s)(3 s) +
= −728 mm
1
(75 mm/s 2 )(3 s) 2
2
or
xC − (xC )0 = 728 mm → ��
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78
78
PROBLEM 11.57
Collar A starts from rest at t = 0 and moves downward with a constant
acceleration of 140 mm/s2. Collar B moves upward with a constant acceleration,
and its initial velocity is 160 mm/s. Knowing that collar B moves through 400 mm
between t = 0 and t = 2 s, determine (a) the accelerations of collar B and
block C, (b) the time at which the velocity of block C is zero, (c) the distance
through which block C will have moved at that time.
SOLUTION
From the diagram
− y A + ( yC − y A ) + 2 yC + ( yC − yB ) = constant
Then
−2v A − vB + 4vC = 0
(1)
and
−2a A − aB + 4aC = 0
(2)
(v A ) 0 = 0
Given:
(a A ) = 140 mm/s2 ↓
( v B )0 = 160 mm/s ↑
a B = constant ↑
y − (yB )0 = 400 mm ↑
At t = 2 s
(a)
We have
At t = 2 s:
y B = ( y B ) 0 + ( vB ) 0 t +
1
aB t 2
2
−400 mm = (−160 mm/s)(2 s) +
aB = −40 mm/s2 or
1
aB (2 s)2
2
a B = 40 mm/s2 ↑
Then, substituting into Eq. (2)
−2(140 mm/s2) − (−40 mm/s2) + 4aC = 0
aC = 60 mm/s2 or
aC = 60 mm/s2 ↓
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PROBLEM 11.57 (Continued)
(b)
Substituting into Eq. (1) at t = 0
−2(0) − (−160 mm/s) + 4(vC )0 = 0 or (vC )0 = −40 mm/s
vC = (vC )0 + aC t
Now
(c)
When vC = 0:
0 = (−40 mm/s) + (60 mm/s2)t
or
t=
2
3
t = 0.667 s
yC = ( yC )0 + (vC )0 t +
We have
At t =
2
s
3
s:
yC − (yC )0 = (−40 mm/s)
1
aC t 2
2
2
1
2
s + (60 mm/s2)
s
3
3
2
= −13.3 mm
or
2
y C − (y C )0 = 13.3 mm ↑
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80
PROBLEM 11.58
Collars A and B start from rest, and collar A moves upward with an
acceleration of 60t 2 mm/s. Knowing that collar B moves downward with a
constant acceleration and that its velocity is 160 mm/s after moving 640 mm,
determine (a) the acceleration of block C, (b) the distance through which
block C will have moved after 3 s.
SOLUTION
From the diagram
− y A + ( yC − y A ) + 2 yC + ( yC − yB ) = constant
Then
−2v A − vB + 4vC = 0
(1)
and
−2a A − aB + 4aC = 0
(2)
(v A )0 = 0
( vB ) 0 = 0
Given:
a A = 3t 2 mm/s2 ↑
a B = constant ↓
y B − (y B )0 = 640 mm ↓,
When
v B = 160 mm/s
(a)
We have
vB2 = 0 + 2aB [ yB − ( yB )0 ]
When yB − ( yB )0 = 640 mm: (160 mm/s)2 = 2aB (640 mm)
aB = 20 mm/s2
or
Then, substituting into Eq. (2)
−2(−20t 2 mm/s2) − (20 mm/s2) + 4aC = 0
aC = 5(1 − 6t 2) mm/s2
or
(b)
Substituting into Eq. (1) at t = 0
−2(0) − (0) + 4(vC )0 = 0 or (vC )0 = 0
dvC
= aC = 5(1 − 6t 2)
dt
Now
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81
PROBLEM 11.58 (Continued)
vC
At t = 0, vC = 0:
dvC =
0
t
0
(5 − 30t 2)dt
or
vC = 5t − 10t 3
Thus,
vC = 0
5t(1 − 2t 3) = 0
At
t = 0,
or
t = 0.707 s
Therefore, block C initially moves downward (vC
yC
( yC )0
dyC =
t
0
(5t − 10t 3)dt
5 2 4
(t − t )
2
or
yC − ( yC )0 =
At t = 0.707 s
5
yC − ( yC )0 = [(0.707)2 − (0.707)4] = 0.625 mm
2
At t = 3 s:
5
yC − ( yC )0 = [(3) 2 − (3) 4 ] = −180 mm
2
Total distance traveled
0).
dyC
= vC = 5t − 10t 3
dt
Now
At t = 0, yC = ( yC )0 :
0) and then moves upward (vC
= (0.625) + −180 − 0.625 = 181.25 mm
= 181 mm
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82
82
PROBLEM 11.59
The system shown starts from rest, and each component moves with a constant
acceleration. If the relative acceleration of block C with respect to collar B is
60 mm/s 2 upward and the relative acceleration of block D with respect to block A
is 110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the
change in position of block D after 5 s.
SOLUTION
From the diagram
2 y A + 2 yB + yC = constant
Cable 1:
Then
2v A + 2vB + vC = 0
(1)
and
2a A + 2aB + aC = 0
(2)
( yD − y A ) + ( yD − yB ) = constant
Cable 2:
Then
and
(3)
− a A − aB + 2aD = 0
(4)
At t = 0, v = 0; all accelerations constant;
aC/B = 60 mm/s 2 ↑, aD/A = 110 mm/s 2 ↓
Given:
(a)
− v A − vB + 2 vD = 0
We have
aC/B = aC − aB = −60 or aB = aC + 60
and
aD/A = aD − a A = 110 or a A = aD − 110
Substituting into Eqs. (2) and (4)
Eq. (2):
2(aD − 110) + 2( aC + 60) + aC = 0
3aC + aD = 100
or
Eq. (4):
−(aD − 110) − ( aC + 60) + 2aD = 0
− aC + aD = −50
or
(5)
(6)
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PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and aD
aC = 40 mm/s 2
aD = −10 mm/s 2
Now
vC = 0 + aC t
At t = 3 s:
vC = (40 mm/s 2 )(3 s)
vC = 120 mm/s ↓ ��
or
(b)
We have
At t = 5 s:
yD = ( yD )0 + (0)t +
yD − ( yD )0 =
1
aD t 2
2
1
( −10 mm/s 2 )(5 s) 2
2
y D − (y D )0 = 125 mm ↑ ��
or
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84
84
PROBLEM 11.60*
The system shown starts from rest, and the length of the upper cord is adjusted so
that A, B, and C are initially at the same level. Each component moves with a
constant acceleration, and after 2 s the relative change in position of block C with
respect to block A is 280 mm upward. Knowing that when the relative velocity of
collar B with respect to block A is 80 mm/s downward, the displacements of A
and B are 160 mm downward and 320 mm downward, respectively, determine
(a) the accelerations of A and B if aB � 10 mm/s 2 , (b) the change in position of
block D when the velocity of block C is 600 mm/s upward.
SOLUTION
From the diagram
2 y A + 2 yB + yC = constant
Cable 1:
Then
2v A + 2vB + vC = 0
(1)
and
2a A + 2aB + aC = 0
(2)
Cable 2: ( yD − y A ) + ( yD − yB ) = constant
Then
− v A − vB − 2 vD = 0
(3)
and
− a A − aB + 2aD = 0
(4)
t =0
Given: At
v=0
(y A )0 = ( yB )0 = ( yC )0
All accelerations constant at t = 2 s
yC/A = 280 mm ↑
vB/A = 80 mm/s ↓
When
y A − ( y A )0 = 160 mm ↑
yB − ( yB )0 = 320 mm ↓
aB � 10 mm/s 2
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85
85
PROBLEM 11.60* (Continued)
(a)
We have
y A = ( y A )0 + (0)t +
1
a At 2
2
and
yC = ( yC )0 + (0)t +
1
aC t 2
2
Then
yC/A = yC − y A =
1
(aC − a A )t 2
2
At t = 2 s, yC/A = −280 mm:
−280 mm =
or
1
(aC − a A )(2 s) 2
2
aC = a A − 140
(5)
Substituting into Eq. (2)
2a A + 2aB + (a A − 140) = 0
or
1
a A = (140 − 2aB )
3
Now
vB = 0 + a B t
(6)
v A = 0 + a At
vB/A = vB − v A = (aB − a A )t
Also
When
�
yB = ( yB )0 + (0)t +
1
aB t 2
2
v B/A = 80 mm/s ↓ : 80 = (aB − a A )t
∆y A = 160 mm ↓ : 160 =
1
a At 2
2
∆yB = 320 mm ↓ : 320 =
1
aB t 2
2
(7)
1
(aB − a A )t 2
2
Then
160 =
Using Eq. (7)
320 = (80)t or t = 4 s
Then
160 =
1
a A (4)2
2
or
a A = 20 mm/s 2 ↓ �
and
320 =
1
aB (4) 2
2
or
a B = 40 mm/s 2 ↓ ��
Note that Eq. (6) is not used; thus, the problem is over-determined.
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86
PROBLEM 11.60* (Continued)
Alternative solution:
v A2 = (0) + 2a A [ y A − ( y A )0 ]
We have
vB2 = (0) + 2aB [ yB − ( yB )0 ]
Then
vB/A = vB − v A = 2aB [ yB − ( yB )0 ] − 2a A [ y A − ( y A )0 ]
When
v B/A = 80 mm/s ↓ :
80 mm/s = 2 � aB (320 mm) − a A (160 mm) �
�
�
or
20 = 2
(
200 B − 100 A
)
(8)
Solving Eqs. (6) and (8) yields a A and a B .
(b)
Substituting into Eq. (5)
aC = 20 − 140 = −120 mm/s 2
and into Eq. (4)
−(20 mm/s 2 ) − (40 mm/s 2 ) + 2aB = 0
or
aD = 30 mm/s 2
Now
vC = 0 + aC t
When vC = −600 mm/s:
− 600 mm/s = (−120 mm/s 2 )t
t =5s
or
yD = ( yD )0 + (0)t +
Also
At t = 5 s:
yD − ( yD )0 =
1
aD t 2
2
1
(30 mm/s 2 )(5 s)2
2
y D − (y D )0 = 375 mm ↓ ��
or
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87
87
PROBLEM 11.61
A
B
x
d
A subway car leaves station A; it gains speed at the rate of 2 m/s2 for 6 s
and then at the rate of 3 m/s2 until it has reached the speed of 24 m/s. The
car maintains the same speed until it approaches station B; brakes are then
applied, giving the car a constant deceleration and bringing it to a stop in
6 s. The total running time from A to B is 40 s. Draw the a−t, v−t, and x−t
curves, and determine the distance between stations A and B.
SOLUTION
Acceleration-Time Curve. Since the acceleration is either constant or zero,
the a−t curve is made of horizontal straight-line segments. The values of t2
and a4 are determined as follows:
a(m/s2)
4
3
0 , t , 6:
2
1
0
–1
34 40
6 t2
t(s)
a4
–2
–3
–4
Change in v 5 area under a – t curve
v6 2 0 5 (6 s)(2 m/s2) 5 12 m/s
6 , t , t2:
Since the velocity increases from 12 to 24 m/s,
Change in v 5 area under a – t curve
24 m/s 2 12 m/s 5 (t2 2 6)(3 m/s2)
t2 5 10 s
t2 , t , 34:
Since the velocity is constant, the acceleration is zero.
34 , t , 40:
Change in v 5 area under a – t curve
a4 5 24 m/s2
0 2 24 m/s 5 (6 s)a4
The acceleration being negative, the corresponding area is below the t axis;
this area represents a decrease in velocity.
Velocity-Time Curve. Since the acceleration is either constant or zero, the
v−t curve is made of straight-line segments connecting the points determined above.
v(m/s)
Change in x 5 area under v−t curve
24
12
0
6 10
34 40
t(s)
0
6
10
34
,
,
,
,
t
t
t
t
,
,
,
,
6:
10:
34:
40:
x6 2 0
x10 2 x6
x34 2 x10
x40 2 x34
5
5
5
5
1
2 (6)(12) 5 36
1
2 (4)(12 1 48)
m
5 72 m
(24)(24) 5 576 m
1
2 (6)(24) 5 72 m
Adding the changes in x, we obtain the distance from A to B:
d 5 x40 2 0 5 756 m
x(m)
d 5 756 m
756 m
0
6 10
34 40
◀
Position-Time Curve. The points determined above should be joined by
three arcs of parabola and one straight-line segment. In constructing the
x−t curve, keep in mind that for any value of t the slope of the tangent to
the x−t curve is equal to the value of v at that instant.
t(s)
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88
88
PROBLEM 11.62
For the particle and motion of Problem 11.61, plot the v −t and x −t curves for 0 t 20 s and determine
(a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate.
SOLUTION
(a)
t = 0, v0 = −3.6 m/s,
Initial conditions:
x0 = 0
Change in v equals area under a −t curve:
v0 = −3.6 m/s
0
t
4 s:
v4 − v0 = (0.6 m/s2)(4 s) = +2.4 m/s
v4 = −1.2 m/s
4s
t
10 s:
v10 − v4 = (1.2 m/s2)(6 s) = +7.2 m/s
v10 = +6 m/s
10 s
t
12 s:
v12 − v10 = (−1 m/s2)(2 s) = −2 m/s
v12 = +4 m/s
12 s
t
20 s:
v20 − v12 = (−1 m/s2)(8 s) = −8 m/s
v20 = −4 m/s
Change in x equals area under v −t curve:
x0 = 0
0
t
4 s:
4s
t
5 s:
5s
t
10 s:
12 s
t
10 s:
1
(−3.6 −1.2)(4) = −9.6 m
2
1
x5 − x4 = (−1.2)(1) = −0.6 m
2
1
x10 − x5 = (+6)(5) = +15 m
2
1
x12 − x10 = (+6 + 4)(2) = +10 m
2
x4 − x0 =
x4 = −9.6 m
x5 = −10.2 m
x10 = +4.8 m
x12 = +14.8 m
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90
PROBLEM 11.62 (Continued)
16 s
t
12 s:
20 s
t
16 s:
1
(+4)(4) = +8 m
2
1
x20 − x16 = (−4)(4) = −8 m
2
x16 − x12 =
x16 = +22.8 m
x20 = +14.8 m
From v −t and x −t curves, we read
(a)
At t = 10 s:
vmax = +6 m/s
(b)
At t = 16 s:
xmax = +22.8 m
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91
90
PROBLEM 11.63
A particle moves in a straight line with the velocity
shown in the figure. Knowing that x = −540 mm at t = 0,
(a) construct the a −t and x −t curves for 0 t 50 s,
and determine (b) the total distance traveled by the
particle when t = 50 s, (c) the two times at which x = 0.
SOLUTION
(a)
at = slope of v −t curve at time t
From t = 0 to t = 10 s:
v = constant
−20 − 60
= −5 mm/s2
26 − 10
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant
t = 41 s to t = 46 s:
a=
t = 46 s:
a=0
a=0
−5 − (−20)
= 3 mm/s2
46 − 41
v = constant
a=0
x2 = x1 + (area under v −t curve from t1 to t2 )
At t = 10 s:
x10 = −540 + 10(60) + 60 mm
Next, find time at which v = 0. Using similar triangles
tv = 0 − 10
60
At
t = 22 s:
t = 26 s:
t = 41 s:
t = 46 s:
t = 50 s:
=
26 − 10
80
or
tv = 0 = 22 s
1
(12)(60) = 420 mm
2
1
x26 = 420 − (4)(20) = 380 mm
2
x41 = 380 − 15(20) = 80 mm
x22 = 60 +
20 + 5
= 17.5 mm
2
x50 = 17.5 − 4(5) = −2.5 mm
x46 = 80 − 5
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PROBLEM 11.63 (Continued)
(b)
From t = 0 to t = 22 s: Distance traveled = 420 − (−540)
= 960 mm
t = 22 s to t = 50 s: Distance traveled = |− 2.5 − 420|
= 422.5 mm
Total distance traveled = (960 + 422.5) mm = 1382.5 mm
= 1.383 mm
(c)
Using similar triangles
Between 0 and 10 s:
(t x = 0 )1 − 0 10
=
540
600
(t x = 0 )1 = 9.00 s
or
Between 46 s and 50 s:
(t x = 0 )2 − 46 4
=
17.5
20
(t x =0 ) 2 = 49.5 s
or
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PROBLEM 11.64
A particle moves in a straight line with the velocity
shown in the figure. Knowing that x = −540 mm at
t = 0, (a) construct the a −t and x −t curves for
0 t 50 s, and determine (b) the maximum value
of the position coordinate of the particle, (c) the
values of t for which the particle is at x = 100 mm.
SOLUTION
(a)
at = slope of v −t curve at time t
From t = 0 to t = 10 s:
v = constant
−20 − 60
= −5 mm/s2
26 − 10
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant
t = 41 s to t = 46 s:
a=
t = 46 s:
a=0
a=0
−5 − ( −20)
= 3 mm/s2
46 − 41
v = constant
a=0
x2 = x1 + (area under v −t curve from t1 to t2 )
At t = 10 s:
x10 = −540 + 10(60) = 60 mm
Next, find time at which v = 0. Using similar triangles
tv = 0 − 10
60
At
t = 22 s:
t = 26 s:
t = 41 s:
t = 46 s:
t = 50 s:
=
26 − 10
80
or
tv = 0 = 22 s
1
(12)(60) = 420 mm
2
1
x26 = 420 − (4)(20) = 380 mm
2
x41 = 380 − 15(20) = 80 mm
x22 = 60 +
20 + 5
= 17.5 mm
2
x50 = 17.5 − 4(5) = −2.5 mm
x46 = 80 − 5
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PROBLEM 11.64 (Continued)
(b)
Reading from the x −t curve
(c)
Between 10 s and 22 s
xmax = 420 mm
100 mm = 420 mm − (area under v−t curve from t, to 22 s) mm
1
100 = 420 − (22 − t1 )(v1 )
2
or
or
(22 − t1 )(v1 ) = 640
Using similar triangles
v1
60
=
22 − t1 12
Then
or
v1 = 5(22 − t1 )
or
(22 − t1 )[5(22 − t1 )] = 640
t1 = 10.69 s and t1 = 33.3 s
We have
10 s
t1
22 s
t1 = 10.69 s
Between 26 s and 41 s:
Using similar triangles
41 − t2
15
=
20
300
or
t2 = 40 s
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94
PROBLEM 11.65
A parachutist is in free fall at a rate of 200 km/h when he opens his
parachute at an altitude of 600 m. Following a rapid and constant
deceleration, he then descends at a constant rate of 50 km/h from 586 m
to 30 m, where he maneuvers the parachute into the wind to further
slow his descent. Knowing that the parachutist lands with a negligible
downward velocity, determine (a) the time required for the parachutist
to land after opening his parachute, (b) the initial deceleration.
SOLUTION
Assume second deceleration is constant. Also, note that
200 km/h = 55.555 m/s,
50 km/h = 13.888 m/s
(a)
Now ∆x = area under v −t curve for given time interval
Then
or
� 55.555 + 13.888 �
(586 − 600) m = −t1 �
� m/s
2
�
�
t1 = 0.4032 s
(30 − 586) m = −t2 (13.888 m/s)
or
t2 = 40.0346 s
1
(0 − 30) m = − (t3 )(13.888 m/s)
2
or
t3 = 4.3203 s
ttotal = (0.4032 + 40.0346 + 4.3203) s
ttotal = 44.8 s �
or
(b)
We have
ainitial =
=
∆ vinitial
t1
[−13.888 − (−55.555)] m/s
0.4032 s
= 103.3 m/s 2
ainitial = 103.3 m/s 2 ↑ �
or
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95
96
PROBLEM 11.66
A machine component is spray-painted while it is mounted on a pallet that travels 4 m in 20 s. The pallet has
an initial velocity of 80 mm/s and can be accelerated at a maximum rate of 60 mm/s2. Knowing that the
paining process requires 15 s to complete and is performed as the pallet moves with a constant speed,
determine the smallest possible value of the maximum speed of the pallet.
SOLUTION
First note that (80 mm/s) (20 s) � 4000 mm, so that the speed of the pallet must be increased. Since
vpaint = constant, it follows that vpaint = vmax and then t1 � .5 s. From the v −t curve, A1 + A2 = 4000 mm
and it is seen that (vmax ) min occurs when
� v − 80 �
a1 � = max
� is maximum.
t1
�
�
Thus,
(vmax − 80)mm/s
= 60 mm/s
t1(s)
t1 =
or
and
� 80 + vmax
t1 �
2
�
(vmax − 80)
60
�
� + (20 − t1 ) (vmax ) = 4000
�
Substituting for t1
(vmax − 80) � 80 + vmax
�
60
2
�
vmax − 80 �
� �
vmax = 4000
� + � 20 −
60 ��
� �
2
vmax
− 2560vmax + 486400 = 0
Simplifying
Solving
vmax = 207 mm/s
and
For
vmax = 207 mm/s,
t1 � 5 s
vmax = 2353 mm/s,
t1 � 5 s
vmax = 2353 mm/s
(vmax )min = 207 mm/s �
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97
96
PROBLEM 11.67
A temperature sensor is attached to slider AB, which moves back
and forth through 60 cm. The maximum velocities of the slider
are 12 cm/s to the right and 30 cm/s to the left. When the slider is
moving to the right, it accelerates and decelerates at a constant
rate of 6 cm/s2; when moving to the left, the slider accelerates and
decelerates at a constant rate of 20 cm/s2. Determine the time
required for the slider to complete a full cycle, and construct the
v – t and x−t curves of its motion.
SOLUTION
The v −t curve is first drawn as shown. Then
ta =
vright
aright
=
12 cm/s
=2s
6 cm/s2
vleft 30 cm/s
=
aleft 20 cm/s
= 1.5 s
td =
A1 = 60 cm
Now
or
[(t1 − 2)s](12 cm/s) = 60 cm
t1 = 7 s
or
A2 = 60 cm
and
or
{[(t2 − 7) − 1.5] s}(30 cm/s) = 60 cm
or
t2 = 10.5 s
tcycle = t2
Now
tcycle = 10.5 s
We have xii = xi + (area under v −t curve from ti to tii )
At
t = 5 s:
1
(2) (12) = 12 cm
2
x5 = 12 + (5 − 2)(12)
t = 7 s:
= 48 cm
x7 = 60 cm
t = 2 s:
t = 8.5 s:
t = 9 s:
t = 10.5 s:
x2 =
1
x8.5 = 60 − (1.5)(30)
2
= 37.5 cm
x9 = 37.5 − (0.5)(30)
x10.5
= 22.5 cm
=0
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98
PROBLEM 11.68
A commuter train traveling at 40 km/h is 3 km
from a station. The train then decelerates so
that its speed is 20 km/h when it is 0.5 km from
the station. Knowing that the train arrives at the
station 7.5 min after beginning to decelerate
and assuming constant decelerations, determine (a) the time required for the train is travel
the first 2.5 km, (b) the speed of the train as it
arrives at the station, (c) the final constant
deceleration of the train.
SOLUTION
Given: At t = 0, v = 40 km/h x = 0; when x = 2.5 km, v = 20 km/h;
at t = 7.5 min, x = 3 km; constant decelerations
The v −t curve is first drawn is shown.
(a)
A1 = 2.5 km
We have
or
(t1 min)
40 + 20
1h
km/h ×
= 2.5 km
2
60 min
t1 = 5 min
or
(b)
A2 = 0.5 km
We have
or
(7.5 − 5) min ×
20 + v2
1h
km/h ×
= 0.5 km
2
60 min
v2 = 4 km/h
or
(c)
We have
afinal = a12
=
(4 − 20) km/h 1000 m 1 min
1h
×
×
×
60s 3600 s
(7.5 − 5) min
km
afinal = −0.03 m/s2
or
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99
98
PROBLEM 11.69
Two road rally checkpoints A and B are located on
the same highway and are 12 km apart. The speed
limits for the first 8 km and the last 4 km of the
section of highway are 100 km/h and 70 km/h,
respectively. Drivers must stop at each checkpoint,
and the specified time between Points A and B is 8
min 20 s. Knowing that a driver accelerates and
decelerates at the same constant rate, determine the
magnitude of her acceleration if she travels at the
speed limit as much as possible.
SOLUTION
Given:
(vmax ) AC = 100 km/h, (vmax )CB = 70 km/h;
v A = vB = 0; t AB = B min, 20 s;
| a | = constant; v = vmax as such as possible
The v −t curve is first drawn as shown, where the magnitudes of the slopes (accelerations) of the three
inclined lines are equal.
Note:
At
8 min 20 s =
5
h
36
t1 , x = 8 km
5
h, x = 12 km
36
Denoting the magnitude of the accelerations by a, we have
a=
100
ta
a=
30
tb
a=
70
tc
where a is in km/h2 and the times are in h.
Now A1 = 8 km:
1
1
(t1 )(100) − (ta )(100) − (tb )(30) = 8
2
2
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99
100
PROBLEM 11.69 (Continued)
Substituting
1 � 100 �
1 � 30 �
100t1 − �
(100) − � � (30) = 8
�
2� a �
2� a �
t1 = 0.08 +
or
Also A2 = 4 km:
Substituting
1
� 5
�
� 36 − t1 � (70) − 2 (tc )(70) = 4
�
�
1 � 70 �
� 5
�
� 36 − t1 � (70) − 2 � a � (70) = 4
�
�
� �
t1 =
or
Then
or
54.5
a
0.08 =
103 35
−
1260 a
54.5 103 35
−
=
a
1260 a
1000 m � 1 h �
a = 51.259 km/h ×
�
�
km
� 3600 s �
2
2
a = 3.96 m/s 2 �
or
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101
100
PROBLEM 11.70
In a water-tank test involving the launching of a small model boat,
the model’s initial horizontal velocity is 6 m/s, and its horizontal
acceleration varies linearly from −12 m/s 2 at t = 0 to −2 m/s 2 at
t = t1 and then remains equal to −2 m/s 2 until t = 1.4 s. Knowing
that v = 1.8 m/s when t = t1 , determine (a) the value of t1 , (b) the
velocity and the position of the model at t = 1.4 s.
SOLUTION
Given:
v0 = 6 m/s; for 0 � t � t , a × t ;
for
t1 � t � 1.4 s a = −2 m/s 2 ;
at
t = 0 a = −12 m/s 2 ; at t = t1
a = −2 m/s 2 , v = 1.8 m/s 2
The a −t and v −t curves are first drawn as shown.
(a)
We have
vt1 = v0 + A1
� 12 + 2 �
2
1.8 m/s = 6 m/s − (t1 s) �
� m/s
� 2 �
or
t1 = 0.6 s ��
or
(b)
We have
v1.4 = vt1 + A2
or
v1.4 = 1.8 m/s − (1.4 − 0.6)5 × 2 m/s 2
v1.4 = 0.20 m/s �
or
Now x1.4 = A3 + A4 , where A3 is most easily determined using
integrating. Thus,
for 0 � t � t1 :
a=
−2 − ( −12)
50
t − 12 = t − 12
0.6
3
dv
=a
dt
50
= t − 12
3
Now
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101
102
PROBLEM 11.70 (Continued)
At t = 0, v = 6 m/s:
or
�
v
6
dv =
t � 50
�
� �� 3 t − 12 �� dt
0
v=6+
25 2
t − 12t
3
We have
25
dx
= v = 6 − 12t + t 2
3
dt
Then
A3 =
�
xt1
0
dx =
�
0.6
0
(6 − 12t +
25 2
t )dt
3
0.6
25 �
�
= �6t − 6t 2 + t 3 � = 2.04 m
9 �0
�
Also
Then
� 1.8 + 0.2 �
A4 = (1.4 − 0.6) �
� = 0.8 m
2
�
�
x1.4 = (2.4 + 0.8) m
x1.4 = 2.84 m �
or
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103
102
PROBLEM 11.71
A car and a truck are both traveling at the constant
speed of 56 km/h; the car is 12 m behind the truck.
The driver of the car wants to pass the truck, i.e.,
he wishes to place his car at B, 12 m in front of the
truck, and then resume the speed of 56 km/h. The
maximum acceleration of the car is 1.5 m/s2 and
the maximum deceleration obtained by applying
the brakes is 6 m/s2. What is the shortest time in
which the driver of the car can complete the
passing operation if he does not at any time exceed
a speed of 80 km/h? Draw the v − t curve.
SOLUTION
Relative to truck, car must move a distance:
Allowable increase in speed:
∆x = 4.8 + 12 + 15 + 12 = 43.8 m
∆vm = 80 − 56 = 24 km/h = 6.7 m/s
Acceleration Phase:
t1 = 6.7/1.5 = 4.5 s
1
A1 = (6.7)(4.5) = 15.1 m
2
Deceleration Phase:
t3 = 6.7/6 = 1.1 s
1
A3 = (6.7)(1.1) = 7.4 m
2
But: ∆x = A1 + A2 + A3 :
43.8 m = 15.1 + (6.7)t2 + 7.4
ttotal = t1 + t2 + t3 = 4.5 s + 3.2 s + 1.1 s = 8.8 s
t2 = 3.2 s
t F = 8.8 s
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104
PROBLEM 11.72
Solve Problem 11.71, assuming that the driver of
the car does not pay any attention to the speed
limit while passing and concentrates on reaching
position B and resuming a speed of 56 km/h in
the shortest possible time. What is the maximum
speed reached? Draw the v − t curve.
SOLUTION
Relative to truck, car must move a distance:
∆x = 4.8 + 12 + 15 + 12 = 43.8 m
∆vm = 1.5t1 = 6t2;
∆x = A1 + A2 :
t2 =
43.8 m =
1
(∆vm )(t1 + t2 )
2
43.8 m =
1
1
(1.5t1 ) t1 + t1
2
4
1
t1
4
t12 = 46.72 t1 = 6.835 s t2 =
1
t1 = 1.709
4
ttotal = t1 + t2 = 6.835 + 1.709
t F = 8.54 s
∆vm = 1.5t1 = 1.5(6.835) = 10.35 m/s = 36.9 km/h
Speed vtotal = 56 km/h, vm = 56 km/h + 36.9 km/h
vm = 92.9 km/h
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104
PROBLEM 11.73
An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2
until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the
elevator begins to move, a man standing 12 m above the initial position of the
top of the elevator throws a ball upward with an initial velocity of 20 m/s.
Determine when the ball will hit the elevator.
SOLUTION
Given:
At t = 0 vE = 0; For 0 � vE � 7.8 m/s,
aE = 1.2 m/s 2 ↑; For vE = 7.8 m/s, aE = 0;
At t = 2 s, vB = 20m/s ↑
The v −t curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve
for the elevator is 1.2 m/s 2 , while the slope of the curve for the ball is − g (−9.81 m/s 2 ).
The time t1 is the time when vE reaches 7.8 m/s.
Thus,
vE = (0) + aE t
or
7.8 m/s = (1.2 m/s 2 )t1
or
t1 = 6.5 s
The time ttop is the time at which the ball reaches the top of its trajectory.
Thus,
or
or
vB = (vB )0 − g (t − 2)
0 = 20 m/s − (9.81 m/s 2 ) (t top − 2) s
ttop = 4.0387 s
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105
106
PROBLEM 11.73 (Continued)
Using the coordinate system shown, we have
0 � t � t1 :
�1
�
yE = −12 m + � aE t 2 � m
�2
�
At t = t top :
yB =
and
yE = −12 m +
At
and at t = t1 ,
1
(4.0387 − 2) s × (20 m/s)
2
= 20.387 m
1
(1.2 m/s 2 )(4.0387 s) 2
2
= −2.213 m
t = [2 + 2(4.0387 − 2)] s = 6.0774 s, yB = 0
yE = −12 m +
1
(6.5 s) (7.8 m/s) = 13.35 m
2
The ball hits the elevator ( yB = yE ) when ttop � t � t1 .
For t � t top :
�1
�
yB = 20.387 m − � g (t − t top ) 2 � m
2
�
�
Then,
when
y B = yE
1
20.387 m − (9.81 m/s 2 ) (t − 4.0387) 2
2
1
= −12 m + (1.2 m/s 2 ) (t s) 2
2
or
Solving
5.505t 2 − 39.6196t + 47.619 = 0
t = 1.525 s and t = 5.67 s
t = 1.525 s �
Choosing the smaller value
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107
106
PROBLEM 11.74
The acceleration record shown was obtained for a small
airplane traveling along a straight course. Knowing that x = 0
and v = 60 m/s when t = 0, determine (a) the velocity and
position of the plane at t = 20 s, (b) its average velocity
during the interval 6 s � t � 14 s.
SOLUTION
Geometry of “bell-shaped” portion of v −t curve
The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of v −t diagram is:
= ∆x = 6 m
(a)
v20 = 50 m/s �
When t = 20 s:
x20 = (60 m/s) (20 s) − (shaded area)
= 1200 m − 6 m
(b)
From t = 6 s to t = 14 s:
x20 = 1194 m �
∆t = 8 s
∆x = (60 m/s) (14 s − 6 s) − (shaded area)
= (60 m/s)(8 s) − 6 m = 480 m − 6 m = 474 m
vaverage =
∆x 474 m
=
8s
∆t
vaverage = 59.25 m/s �
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107
108
PROBLEM 11.75
Car A is traveling on a highway at a constant
speed (vA)0 = 95 km/h and is 115 m from the
entrance of an access ramp when car B
enters the acceleration lane at that point at a
speed (vB)0 = 25 km/h. Car B accelerates
uniformly and enters the main traffic lane
after traveling 60 m in 5 s. It then continues
to accelerate at the same rate until it reaches
a speed of 95 km/h, which it then maintains.
Determine the final distance between the
two cars.
SOLUTION
Given:
(v A )0 = 95 km/h, (vB )0 = 25 km/h; at t = 0,
( x A )0 = −115 m, (xB)0 = 0;
at t = 5 s,
xB = 60 m; for 25 km/h
95 km/h,
vB
aB = constant; for vB = 95 km/h,
First note
aB = 0
95 km/h = 26.4 m/s
25 km/h = 6.9 m/s
The v −t curves of the two cars are then drawn as shown.
Using the coordinate system shown, we have
at t = 5 s, xB = 60 m:
(5 s)
(6.9 + vB ) S
m/s = 60 m
2
(vB )S = 17.1 m/s
or
Then, using similar triangles, we have
(26.4 − 6.9) m/s (17.1 − 6.9) m/s
=
(= aB)
t1
5s
or
t1 = 9.6 s
Finally, at t = t1
xB/A = xB − x A = (9.6 s)
6.9 + 26.4
m/s
2
−[−115 m + (9.6 s)(26.4 m/s)]
xB/A = 21.4 m
or
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109
108
PROBLEM 11.76
Car A is traveling at 64 km/h when it enters
a 48 km/h speed zone. The driver of car A
decelerates at a rate of 4.8 m/s2 until reaching a speed of 48 km/h, which she then
maintains. When car B, which was initially
18 m behind car A and traveling at a
constant speed of 72 km/h, enters the
speed zone, its driver decelerates at a rate
of 6 m/s2 until reaching a speed of 45
km/h. Knowing that the driver of car B
maintains a speed of 45 km/h, determine
(a) the closest that car B comes to car A,
(b) the time at which car A is 21 m in front
of car B.
SOLUTION
(v A )0 = 64 km/h; For 48 km/h
Given:
vA
64 km/h;
aA = −4.8 m/s2; For vA = 48 km/h, aA = 0;
( x A/B )0 = 18 m, (vB)0 = 72 km/h; when
xB = 0, aB = −6 m/s2; for vB = 45 km/h
aB = 0
64 km/h = 17.8 m/s
72 km/h = 20 m/s
First note
48 km/h = 13.3 m/s
45 km/h = 12.5 m/s
At t = 0
The v −t curves of the two cars are as shown.
At
t = 0:
car A enters the speed zone
t = (t B )1 :
car B enters the speed zone
t = tA:
car A reaches its final speed
t = tmin :
v A = vB
t = (t B )2 :
car B reaches its final speed
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110
PROBLEM 11.76 (Continued)
(a)
aA =
We have
−4.8 m/s2 =
or
(v A )final − (v A )0
tA
(13.3 − 17.8) m/s
tA
t A = 0.938 s
or
Also
18 m = (t B )1 (vB )0
or
18 m = (t B )1 (20 m/s)
aB =
(vB )final − (vB )0
(t B ) 2 − (t B )1
−6 m/s2 =
(12.5 − 20) m/s
[(t B )2 − 0.9] s
and
or
(t B )1 = 0.9 s
or
(tB)2 = 2.15 s
Car B will continue to overtake car A while vB
v A . Therefore, ( x A/B ) min will occur when v A = vB ,
which occurs for
(t B )1
tmin
(tB ) 2
For this time interval
v A = 13.3 m/s
vB = (vB )0 + aB [t − (tB )1 ]
Then
13.3 m/s = 20 m/s + (−6 m/s2)(tmin − 0.9) s
at t = tmin :
tmin = 2.017 s
or
Finally ( x A/B ) min = ( x A )tmin − ( xB )tmin
= tA
(v A )0 + (v A )final
+ (tmin − t A )(v A )final
2
− ( xB )0 + (t B )1 (vB )0 + [tmin − (t B )1 ]
= (0.938 s)
(vB )0 + (v A )final
2
17.8 + 13.3
m/s + (2.017 − 0.938) s × (13.3 m/s)
2
− −18 m + (0.9 s)(20 m/s) + (2.017 − 0.9) s ×
20 + 13.3
m/s
2
= (14.586 + 14.351) m − (−18 + 18 + 18.598) m
= 10.339 m
or ( x A/B ) min = 10.3 m
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111
110
PROBLEM 11.76 (Continued)
(b)
Since ( x A/B )
[Note (t B ) 2
18 m for t
tmin , it follows that x A/B = 21 m for t
tmin ]. Then, for t
(t B )2
(t B )2
x A/B = ( x A/B ) min + [(t − tmin )(v A )final ]
− [(t B )2 − (tmin )]
or
(v A )final + (vB )final
+ [t − (t B ) 2 ](vB )final
2
21 m = 10.339 m + [(t − 2.017) s × (13.3 m/s)]
− (2.15 − 2.017) s ×
13.3 + 12.5
m/s + (t − 2.15) s × (12.5) m/s
2
t = 15.4 s
or
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111
112
PROBLEM 11.77
A car is traveling at a constant speed of
54 km/h when its driver sees a child run into
the road. The driver applies her brakes until
the child returns to the sidewalk and then
accelerates to resume her original speed of
54 km/h; the acceleration record of the car is
shown in the figure. Assuming x = 0 when
t = 0, determine (a) the time t1 at which the
velocity is again 54 km/h, (b) the position of
the car at that time, (c) the average velocity of
the car during the interval 1 s � t � t1.
SOLUTION
Given:
t = 0, x = 0, v = 54 km/h; for t = t1 ,
At
v = 54 km/h
First note
(a)
54 km/h = 15 m/s
vb = va + (area under a −t curve from ta to tb )
We have
Then
at
t = 2 s:
v = 15 − (1)(6) = 9 m/s
t = 4.5 s:
1
v = 9 − (2.5)(6) = 1.5 m/s
2
t = t1:
15 = 1.5 +
1
(t1 − 4.5)(2)
2
t1 = 18 s ��
or
(b)
Using the above values of the velocities, the v −t curve is drawn as shown.
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113
112
PROBLEM 11.77 (Continued)
Now
x at t = 18 s
x18 = 0 + Σ (area under the v −t curve from t = 0 to t = 18 s)
� 15 + 9 �
= (1 s)(15 m/s) + (1 s) �
� m/s
� 2 �
1
�
�
+ �(2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s) �
3
�
�
2
�
�
+ �(13.5 s)(1.5 m/s) + (13.5 s)(13.5 m/s) �
3
�
�
= [15 + 12 + (3.75 + 6.25) + (20.25 + 121.50)] m
= 178.75 m
(c)
First note
or
x18 = 178.8 m �
x1 = 15 m
x18 = 178.75 m
Now
vave =
∆x (178.75 − 15) m
= 9.6324 m/s
=
∆t
(18 − 1) s
vave = 34.7 km/h ��
or
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113
114
PROBLEM 11.78
As shown in the figure, from t = 0 to t = 4 s, the acceleration
of a given particle is represented by a parabola. Knowing that x
= 0 and v = 8 m/s when t = 0, (a) construct the v −t and x–t
curves for 0 � t � 4 s, (b) determine the position of the particle at t
= 3 s. (Hint: Use table inside the front cover.)
SOLUTION
Given:
(a)
At
t = 0, x = 0, v = 8 m/s
We have
v2 = v1 + (area under a −t curve from t1 to t2)
and
x2 = x1 + (area under v −t curve from t1 to t2)
Then, using the formula for the area of a parabolic spandrel, we have
1
at t = 2 s: v = 8 − (2)(12) = 0
3
1
t = 4 s: v = 0 − (2)(12) = −8 m/s
3
The v −t curve is then drawn as shown.
Note: The area under each portion of the curve is a spandrel of Order no. 3.
(2)(8)
=4m
3 +1
(2)(8)
t = 4 s: x = 4 −
=0
3 +1
Now at t = 2 s: x = 0 +
The x −t curve is then drawn as shown.
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114
PROBLEM 11.78 (Continued)
(b)
We have
At t = 3 s: a = −3(3 − 2)2 = −3 m/s 2
1
v = 0 − (1)(3) = −1 m/s
3
(1)(1)
x =4−
3 +1
x3 = 3.75 m ��
or
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115
116
PROBLEM 11.79
During a manufacturing process, a conveyor belt starts from rest and travels a total of 0.4 m before
temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ±1.6 m/s2
per second, determine (a) the shortest time required for the belt to move 0.4 m, (b) the maximum and
average values of the velocity of the belt during that time.
SOLUTION
Given:
(a)
At
t = 0, x = 0, v = 0; xmax = 0.4 m;
when
x = xmax , v = 0;
da
dt
= 1.6 m/s2
max
Observing that vmax must occur at t = 12 tmin , the a −t curve must have the shape shown. Note that the
magnitude of the slope of each portion of the curve is 1.6 m/s2/s.
We have
at t = ∆t :
t = 2∆t :
v =0+
1
1
(∆t )(amax ) = amax ∆t
2
2
vmax =
1
1
amax ∆t + ( ∆t )( amax ) = amax ∆t
2
2
Using symmetry, the v −t is then drawn as shown.
Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have
(2∆t )(vmax ) = xmax
vmax = amax ∆t
2amax ∆t 2 = xmax
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116
PROBLEM 11.79 (Continued)
Now
or
Then
amax
= 1.6 m/s2/s so that
∆t
2(1.6∆t m/s3) ∆t2 = 0.4 m
∆t = 0.5 s
tmin = 4∆t
tmin = 2.00 s
or
(b)
We have
vmax = amax ∆t
= (1.6 m/s2/s × ∆t)∆t
= 1.6 m/s2/s × (0.5 s)2
vmax = 0.4 m/s
or
Also
vave =
∆x
0.4 m
=
∆t total 2.00 s
vave = 0.2 m/s
or
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118
PROBLEM 11.80
An airport shuttle train travels between two terminals that are 2.6 km apart. To maintain passenger comfort,
the acceleration of the train is limited to ±1.2 m/s2, and the jerk, or rate of change of acceleration, is limited
to ±0.2 m/s2 per second. If the shuttle has a maximum speed of 32 km/h, determine (a) the shortest time for
the shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.
SOLUTION
xmax = 2.6 km; |amax | = 1.2 m/s2
Given:
da
dt
First note
(a)
= 0.2 m/s2/s; vmax = 32 km/h
max
32 km/h = 8.9 m/s
To obtain tmin, the train must accelerate and decelerate at the maximum rate to maximize the time
for which v = vmax. The time ∆t required for the train to have an acceleration of 1.2 m/s2 is found
from
da
dt
or
or
=
max
amax
∆t
1.2 m/s2
0.2 m/s2/s
∆t = 6 s
∆t =
Now,
da
= constant
dt
after 5 s, the speed of the train is
v5 =
1
(∆t )(amax )
2
or
v5 =
1
(6 s)(1.2 m/s2) = 3.6 m/s
2
since
Then, since v5 < vmax, the train will continue to accelerate at 1.2 m/s2 until v = vmax. The a−t curve
must then have the shape shown. Note that the magnitude of the slope of each inclined portion of
the curve is 0.2 m/s2/s.
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118
PROBLEM 11.80 (Continued)
Now
at t = (12 + ∆t1 ) s, v = vmax :
1
(6 s)(1.2 m/s2) + (∆t1 )(1.2 m/s2) = 8.9 m/s
2
2
or
∆t1 = 1.42 s
Then
at t = 6 s:
t = 7.42 s:
1
(6)(1.2) = 3.6 m/s
2
v = 3.6 + (1.42)(1.2) = 5.3 m/s
t = 13.42 s:
v = 5.3 +
v=0+
1
(6)(1.2) = 8.9 m/s
2
Using symmetry, the v −t curve is then drawn as shown.
Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have
2 (1.42 s)
3.6 + 5.3
m/s
2
+ (12 + ∆t2) s × (8.9 m/s) = 2600 m
or
∆t2 = 278.7 s
Then
tmin = 4(6 s) + 2(1.42 s) + 278.7 s
= 305.54 s
tmin = 5.09 min
or
(b)
We have
vave =
∆x 2.6 km 3600 s
=
×
∆t 305.54 s
1h
vave = 30.6 km/h
or
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120
PROBLEM 11.81
The acceleration record shown was obtained
during the speed trials of a sports car. Knowing
that the car starts from rest, determine by
approximate means (a) the velocity of the car at
t = 8 s, (b) the distance the car has traveled at
t = 20 s.
SOLUTION
Given: a −t curve; at
1.
t = 0, x = 0, v = 0
The a −t curve is first approximated with a series of rectangles, each of width ∆t = 2 s. The
area (∆t )(aave ) of each rectangle is approximately equal to the change in velocity ∆v for the
specified interval of time. Thus,
∆v � aave ∆t
where the values of aave and ∆v are given in columns 1 and 2, respectively, of the following table.
2.
Noting that v0 = 0 and that
v2 = v1 + ∆v12
where ∆v12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s
interval can be computed; see column 3 of the table and the v −t curve.
3.
The v −t curve is next approximated with a series of rectangles, each of width ∆t = 2 s. The
area (∆t )(vave ) of each rectangle is approximately equal to the change in position ∆x for the
specified interval of time.
Thus,
∆x � vave ∆t
where vave and ∆x are given in columns 4 and 5, respectively, of the table.
4.
With x0 = 0 and noting that
x2 = x1 + ∆x12
where ∆x12 is the change in position between times t1 and t2, the position at the end of each 2 s
interval can be computed; see column 6 of the table and the x −t curve.
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120
PROBLEM 11.81 (Continued)
(a)
At t = 8 s, v = 32.58 m/s
(b)
At t = 20 s
v = 117.3 km/h �
or
x = 660 m ��
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121
122
PROBLEM 11.82
Two seconds are required to bring the piston
rod of an air cylinder to rest; the acceleration
record of the piston rod during the 2 s is as
shown. Determine by approximate means
(a) the initial velocity of the piston rod,
(b) the distance traveled by the piston rod as
it is brought to rest.
SOLUTION
a −t curve; at
Given:
1.
t = 25, v = 0
The a −t curve is first approximated with a series of rectangles, each of width ∆t = 0.25 s. The area
(∆t)(aave) of each rectangle is approximately equal to the change in velocity ∆v for the specified
interval of time. Thus,
∆v � aave ∆t
where the values of aave and ∆v are given in columns 1 and 2, respectively, of the following table.
2.
v(2) = v0 +
Now
and approximating the area
�
2
0
�
2
0
a dt = 0
a dt under the a −t curve by Σaave ∆t ≈ Σ∆v, the initial velocity is then
equal to
v0 = −Σ∆v
Finally, using
v2 = v1 + ∆v12
where ∆v12 is the change in velocity between times t1 and t2, the velocity at the end of each 0.25
interval can be computed; see column 3 of the table and the v −t curve.
3.
The v −t curve is then approximated with a series of rectangles, each of width 0.25 s. The area
(∆t )(vave ) of each rectangle is approximately equal to the change in position ∆x for the specified
interval of time. Thus
∆x ≈ vave ∆t
where vave and ∆x are given in columns 4 and 5, respectively, of the table.
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PROBLEM 11.82 (Continued)
4.
With x0 = 0 and noting that
x2 = x1 + ∆x12
where ∆x12 is the change in position between times t1 and t2, the position at the end of each 0.25 s
interval can be computed; see column 6 of the table and the x −t curve.
(a)
We had found
(b)
At t = 2 s
v0 = 1.914 m/s �
x = 0.840 m ��
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123
124
PROBLEM 11.83
A training airplane has a velocity of 37.8 m/s
when it lands on an aircraft carrier. As the
arresting gear of the carrier brings the airplane
to rest, the velocity and the acceleration of the
airplane are recorded; the results are shown
(solid curve) in the figure. Determine by approximate means (a) the time required for the
airplane to come to rest, (b) the distance traveled
in that time.
SOLUTION
Given: a −v curve:
v0 = 37.8 m/s
The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on
the figure).
For uniformly accelerated motion
v22 = v12 + 2a( x2 − x1 )
v2 = v1 + a(t2 − t1 )
v22 − v12
2a
v2 − v1
∆t =
a
∆x =
or
For the five regions shown above, we have
Region
v1 , m/s
v2 , m/s
a, m/s2
∆ x, m
∆t , s
1
37.8
36
− 3.8
17.5
0.474
2
36
30
− 9.9
20
0.606
3
30
24
−13.7
11.8
0.438
4
24
12
−16.2
13.3
0.741
5
12
0
−17.4
4.1
0.690
66.7
2.949
Σ
(a)
From the table, when v = 0
t = 2.95 s
(b)
From the table and assuming x0 = 0, when v = 0
x = 66.7 m
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124
PROBLEM 11.84
Shown in the figure is a portion of the experimentally
determined v − x curve for a shuttle cart. Determine by
approximate means the acceleration of the cart (a) when
x = 0.25 m, (b) when v = 2 m/s.
SOLUTION
Given: v − x curve
First note that the slope of the above curve is
dv
.
dx
a=v
(a)
Now
dv
dx
When
x = 0.25 m, v = 1.4 m/s
Then
a = 1.4 m/s
1 m/s
0.34 m
a = 4.1 m/s2
or
(b)
When v = 2 m/s, we have
a = 2 m/s
1 m/s
0.7 m
a = 2.86 m/s2
or
Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary
that the same scale be used for the x and v axes (e.g., 1 cm = 0.5 m, 1 cm = 0.5 m/s). In the above
solution, ∆v and ∆x were measured directly, so different scales could be used.
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126
PROBLEM 11.85
Using the method of Section 11.8, derive the formula x = x0 + v0t + 12 at 2 for the position coordinate of a
particle in uniformly accelerated rectilinear motion.
SOLUTION
The a −t curve for uniformly accelerated motion is as shown.
Using Eq. (11.13), we have
x = x0 + v0t + (area under a −t curve) (t − t )
� 1 �
= x0 + v0t + (t × a) � t − t �
� 2 �
1 2
= x0 + v0 t + at
Q.E.D.
2
�
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126
PROBLEM 11.86
Using the method of Section 11.8, determine the position of
the particle of Problem 11.61 when t = 14.
PROBLEM 11.61 A particle moves in a straight line with
the acceleration shown in the figure. Knowing that it starts
from the origin with v0 = −18 m/s, (a) plot the v − t and x − t
curves for 0 t 20 s, (b) determine its velocity, its position,
and the total distance traveled after 12 s.
SOLUTION
x0 = 0
v0 = −18 m/s
When t = 14 s:
x = x0 + v0t + ΣA(t1 − t )
= 0 − (18 m/s)(14 s) + [(3 m/s2)(4 s)](12 s) + [(6 m/s2)(6 s)](7 s)
+ [(−5 m/s)(4 s)](2 s)
x14 = −252 m + 144 m + 252 m − 40 m
x14 = +104 m
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128
PROBLEM 11.87
While testing a new lifeboat, an accelerometer attached to the
boat provides the record shown. If the boat has a velocity of
2 m/s at t = 0 and is at rest at time t1, determine, using the
method of Section 11.8, (a) the time t1, (b) the distance
through which the boat moves before coming to rest.
SOLUTION
The area under the curve is divided into three regions as shown.
(a)
First note
ta 0.75
=
18 22.5
or
ta = 0.60 s
Now
v = v0 +
t
0
a dt
where the integral is equal to the area under the a −t curve. Then, with v0 = 2 m/s, vt1 = 0
(b)
We have
0 = 2 m/s +
or
t1 = 2.32 s
1
1
(0.6 s )(18 m/s2) − (0.15 s)(4.5 m/s2) − (t1 − 0.75) s × 4.5 m/s2)
2
2
t1 = 2.32 s
From the discussion following Eq. (11.13) and assuming x = 0, we have
x = 0 + v0 t1 + ΣA(t1 − t )
where A is the area of a region and t is the distance to its centroid. Then for t1 = 2.32 s
x = (2 m/s)(2.32 s) +
−
1
(0.6 s)(18 m/s2) (2.32 − 0.2) s
2
1
(0.15 s)(4.5 m/s2) (2.32 − 0.70) s
2
− [(1.57 s)(4.5 m/s2)] 2.32 − 0.75 +
1
× 1.57 s
2
= [4.64 + (11.448 − 0.547 − 5.546)] m
x = 10 m
or
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128
PROBLEM 11.88
For the particle of Problem 11.63, draw
the a −t curve and determine, using the
method of Section 11.8, (a) the position of
the particle when t = 52 s, (b) the
maximum value of its position coordinate.
PROBLEM 11.63 A particle moves in a
straight line with the velocity shown in the
figure. Knowing that x = −540 mm at t = 0,
(a) construct the a −t and x −t curves for
0 t 50 s, and determine (b) the total
distance traveled by the particle when
t = 50 s, (c) the two times at which x = 0.
SOLUTION
a=
We have
where
dv
dt
dv
dt
is the slope of the v −t curve. Then
t=0
from
to t = 10 s:
v = constant
t = 10 s to t = 26 s: a =
t = 26 s to t = 41 s:
t > 46 s:
−20 − 60
= −5 mm/s2
26 − 10
v = constant
t = 41 s to t = 46 s: a =
a=0
a=0
−5 − (−20)
= 3 mm/s2
46 − 41
v = constant
a=0
The a −t curve is then drawn as shown.
(a)
From the discussion following Eq. (11.13),
we have
x = x0 + v0t1 + ΣA(t1 − t )
where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s
x = −540 mm + (60 mm/s)(52 s) + {−[(16 s)(5 mm/s2)](52 − 18) s
+ [(5 s)(3 mm/s2)](52 − 43.5)s}
= [−540 + (3120) + (−2720 + 127.5)] mm
x = −12.50 mm
or
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130
PROBLEM 11.88 (Continued)
(b)
Noting that xmax occurs when v = 0 ( dx
= 0 ), it is seen from the v–t curve that xmax occurs for
dt
10 s
t 26 s. Although similar triangles could be used to determine the time at which x = xmax
(see the solution to Problem 11.63), the following method will be used.
For
10 s
t1
26 s, we have
x = −540 + 60t1
− [(t1 − 10)(5)]
1
(t1 − 10) mm
2
5
= −540 + 60t1 − (t1 − 10) 2
2
When x = xmax :
or
Then
dx
= 60 − 5(t1 − 10) = 0
dt
(t1 ) xmax = 22 s
5
xmax = −540 + 60(22) − (22 − 10)2
2
xmax = 420 mm
or
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130
PROBLEM 11.89
The motion of a particle is defined by the equations x = 4t 3 − 5t 2 + 5t and y = 5t 2 − 15t , where x and y are
expressed in millimeters and t is expressed in seconds. Determine the velocity and the acceleration when
(a) t = 1 s, (b) t = 2 s.
SOLUTION
x = 4t 3 − 5t 2 + 5t
dx
vx =
= 12t 2 − 10t + 5
dt
dvx
ax =
= 24t − 10
dt
(a)
y = 5t 2 − 15t
dy
vy =
= 10t − 15
dt
dv y
ay =
= 10
dt
When t = 15, vx = 7 mm/s, v y = −5 mm/s
v = 7 2 + (−5) 2 = 8.60 φ = tan −15/ 7 = 35.5°
v = 8.60 mm/s
ax = 14 mm/s 2
35.5° �
a y = 10 mm/s 2
a = 142 + 102 = 17.20 mm/s 2
� 10 �
φ = tan −1 � � = 35.5°
� 14 �
a = 17.20 mm/s 2
�
(b)
35.5° ��
When t = 25, vx = 33 mm/s, v y = 5 mm/s
v = 33.4 mm/s
8.62° �
ax = 38 mm/s 2
a y = 10 mm/s 2
a = 39.3 mm/s 2
�
14.74° ��
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131
132
PROBLEM 11.90
The motion of a particle is defined by the equations
x = 2 cos π t and y = 1 − 4 cos 2π t , where x and y are
expressed in meters and t is expressed in seconds. Show
that the path of the particle is part of the parabola shown,
and determine the velocity and the acceleration when
(a) t = 0, (b) t = 1.5 s.
SOLUTION
We have
x = 2 cos π t
Then
y = 1 − 4(2 cos 2 π t − 1)
� x�
= 5 − 8� �
�2�
y = 1 − 4 cos 2π t
2
y = 5 − 2x2
or
Q.E.D.
Now
vx =
dx
= −2π sin π t
dt
vy =
and
ax =
dvx
= −2π 2 cos π t
dt
ay =
(a)
At t = 0: vx = 0
dy
= 8π sin 2π t
dt
dv y
dt
= 16π 2 cos 2π t
v=0 �
vy = 0
ax = −2π 2 m/s 2
a y = 16π 2 m/s 2
a = 159.1 m/s 2
or
(b)
�
82.9° �
At t = 1.53: vx = −2π sin(1.5π ) = 2π m/s
v y = 8π sin(2π × 1.5) = 0
v = 6.28 m/s → �
or
ax = −2π 2 cos(1.5π ) = 0
a y = 16π 2 cos(2π × 1.5) = −16π 2
a = 157.9 m/s 2 ↓ ��
or
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PROBLEM 11.91
The motion of a particle is defined by the equations x = t 2 − 8t + 7 and y = 0.5t 2 + 2t − 4, where x and y are
expressed in meters and t in seconds. Determine (a) the magnitude of the smallest velocity reached by the
particle, (b) the corresponding time, position, and direction of the velocity.
SOLUTION
x = t 2 − 8t + 7
vx = dx/dt = 2t − 8
y = 0.5t 2 + 2t − 4
vy = t + 2
ax = dvx /dt = 2
a y = dv y /dt = 1
v 2 = vx2 + v 2y = (2t − 8) 2 + (t + 2) 2
When v is minimum, v2 is also minimum; thus, we write
d (v 2 )
= 2(2t − 8)(2) + 2(t + 2) = 2(4t − 16 + t + 2) = 0
dt
= 2(5t − 14) = 0
t = 2.80 s �
When t = 2.8 s :
x = (2.8)2 − 8(2.8) + 7
y = 0.5(2.8) 2 + 2(2.8) − 4
vx = 2(2.8) − 8 vx = −2.40 m/s �
�
v y = 2.8 + 2
v y = +4.80 m/s �
x = −7.56 m �
y = 5.52 m �
v = 5.37 m/s
63.4° ��
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134
PROBLEM 11.92
The motion of a particle is defined by the equations x = 4t − 2 sin t and y = 4 − 2 cos t, where x and y are
expressed in meters and t is expressed in seconds. Sketch the path of the particle, and determine (a) the
magnitudes of the smallest and largest velocities reached by the particle, (b) the corresponding times,
positions, and directions of the velocities.
SOLUTION
x = 4t − 2 sin t
We have
t, s
x, m
y, m
0
2.0
4.28
4.0
π
12.57
6.0
3π
2
20.8
4.0
2π
25.1
2.0
0
π
2
(a)
x = 4t − 2 sin t
We have
y = 4 − 2 cos t
y = 4 − 2 cos t
dx
= 4 − 2 cos t
dt
vy =
dy
= 2 sin t
dt
Then
vx =
Now
v 2 = vx2 + v 2y = (4 − 2 cos t ) 2 + (2 sin t ) 2
= 20 − 16 cos t
By observation,
for vmin , cos t = 1, so that
2
vmin
=4
or
vmin = 2 m/s
2
vmax
= 36
or
vmax = 6 m/s
cos t = 1
or
t = 2nπ s
for vmax , cos t = −1, so that
(b)
When v = vmin :
where n = 0, 1, 2, . . .
Then
x = 4(2nπ ) − 2 sin (2nπ )
or
x = 8nπ m
y = 4 − 2(1)
or
y=2m
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PROBLEM 11.92 (Continued)
Also,
vx = 4 − 2(1) = 2 m/s
v y = 2 sin (2nπ ) = 0
θv min = 0 →
When v = vmax :
or t = (2n + 1)π s
cos t = −1
where n = 0, 1, 2,
Then
x = 4(2n + 1)π − 2 sin (2n + 1)π
y = 4 − 2(−1)
Also,
or
x = 4(2n + 1)π m
or
y=6 m
vx = 4 − 2(−1) = 6 m/s
v y = 2 sin (2n + 1)π = 0
θvmax = 0 →
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PROBLEM 11.93
The motion of a particle is defined by the position vector
r = A(cos t + t sin t )i + A(sin t − t cos t ) j, where t is expressed in
seconds. Determine the values of t for which the position vector
and the acceleration are (a) perpendicular, (b) parallel.
SOLUTION
We have
r = A(cos t + t sin t )i + A(sin t − t cos t ) j
Then
v=
and
a=
(a)
dr
= A( − sin t + sin t + t cos t )i
dt
+ A(cos t − cos t + t sin t ) j
= A(t cos t )i + A(t sin t ) j
dv
= A(cos t − t sin t )i + A(sin t + t cos t ) j
dt
When r and a are perpendicular, r ⋅ a = 0
A[(cos t + t sin t )i + (sin t − t cos t ) j] ⋅ A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0
(b)
or
(cos t + t sin t )(cos t − t sin t ) + (sin t − t cos t )(sin t + t cos t ) = 0
or
(cos2 t − t 2 sin 2 t ) + (sin 2 t − t 2 cos 2 t ) = 0
or
1− t2 = 0
or
t =1 s �
or
t=0 �
When r and a are parallel, r × a = 0
A[(cos t + t sin t )i + (sin t − t cos t ) j] × A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0
or
Expanding
[(cos t + t sin t )(sin t + t cos t ) − (sin t − t cos t )(cos t − t sin t )]k = 0
(sin t cos t + t + t 2 sin t cos t ) − (sin t cos t − t + t 2 sin t cos t ) = 0
2t = 0
or
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PROBLEM 11.94
The damped motion of a vibrating particle is defined by
the position vector r = x1[1 − 1/(t + 1)]i + ( y1e−π t/ 2 cos 2π t ) j,
where t is expressed in seconds. For x1 = 30 mm and
y1 = 20 mm, determine the position, the velocity, and the
acceleration of the particle when (a) t = 0, (b) t = 1.5 s.
SOLUTION
We have
1 �
�
−π t/2
r = 30 �1 −
cos 2π t ) j
� i + 20(e
� t +1�
Then
v=
= 30
1
� π
�
i + 20 � − e −π t/2 cos 2π t − 2π e−π t/2 sin 2π t � j
2
(t + 1) 2
�
�
= 30
�
1
�1
��
i − 20π �e −π t/ 2 � cos 2π t + 2 sin 2π t � � j
2
(t + 1)
�2
��
�
a=
and
dr
dt
dv
dt
� π
�
2
�1
�
i − 20π � − e−π t/2 � cos 2π t + 2 sin 2π t � + e−π t/2 (−π sin 2π t + 4 cos 2π t ) � j
3
(t + 1)
�2
�
� 2
�
60
i + 10π 2 e−π t/2 (4 sin 2π t − 7.5 cos 2π t ) j
=−
(t + 1)3
= −30
(a)
At t = 0:
� 1�
r = 30 �1 − � i + 20(1) j
� 1�
r = 20 mm ↑ �
or
� �1
�1�
��
v = 30 � � i − 20π �(1) � + 0 � � j
�1�
��
� �2
or
60
a = − i + 10π 2 (1)(0 − 7.5) j
(1)
v = 43.4 mm/s
a = 743 mm/s 2
or
46.3° �
85.4° �
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138
PROBLEM 11.94 (Continued)
(b)
At t = 1.5 s:
1 �
�
−0.75π
r = 30 �1 −
(cos 3π ) j
� i + 20e
� 2.5 �
= (18 mm)i + ( −1.8956 mm) j
or
r = 18.10 mm
6.01° �
v = 5.65 mm/s
31.8° �
a = 70.3 mm/s 2
86.9° �
30
�1
�
i − 20π e−0.75π � cos 3π + 0 � j
2
(2.5) 2
�
�
= (4.80 mm/s)i + (2.9778 mm/s) j
v=
or
a=−
60
i + 10π 2 e−0.75π (0 − 7.5 cos 3π ) j
(2.5)3
= (−3.84 mm/s 2 )i + (70.1582 mm/s 2 ) j
or
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PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin
ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described
by the particle is a conic helix.)
SOLUTION
We have
r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k
Then
v=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
dt
(
)
= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i
(
)
+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k
(
)
(
)
= R −2ωn sin ωn t − ωn2 t cos ωn t i + R 2ωn cos ωn t − ωn2 t sin ωn t k
Now
v 2 = vx2 + v y2 + vz2
= [ R(cos ωn t − ωn t sin ωn t )]2 + (c)2 + [ R(sin ωn t + ωn t cos ωn t )]2
(
)
= R 2 � cos 2 ωn t − 2ωn t sin ωn t cos ωn t + ωn2 t 2 sin 2 ωn t
�
+ sin 2 ωn t + 2ωn t sin ωn t cos ωn t + ωn2 t 2 cos 2 ωn t � + c 2
�
(
(
)
)
= R 2 1 + ωn2 t 2 + c 2
(
Also,
)
v = R 2 1 + ωn2 t 2 + c 2 �
or
a 2 = ax2 + a 2y + az2
(
)
cos ω t − ω t sin ω t ) �
�
2
= � R −2ωn sin ωn t − ωn2 t cos ωn t � + (0) 2
�
�
(
(
2
2
+ � R 2ωn
n
n
n
�
= R 2 � 4ωn2 sin 2 ωn t + 4ωn3t sin ωn t cos ωn t + ωn4t 2 cos 2 ωn t
�
+ 4ωn2 cos 2 ωn t − 4ωn3t sin ωn t cos ωn t + ωn4 t 2 sin 2 ωn t �
�
(
(
= R 2 4ωn2 + ωn4 t 2
)
)
)
a = Rωn 4 + ωn2 t 2 �
or�
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139
140
PROBLEM 11.96*
The three-dimensional motion of a particle is defined by the position
vector r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k , where r and t are
expressed in meters and seconds, respectively. Show that the curve
described by the particle lies on the hyperboloid (y/A)2 − (x/A)2
− (z/B)2 = 1. For A = 3 and B = 1, determine (a) the magnitudes of the
velocity and acceleration when t = 0, (b) the smallest nonzero value
of t for which the position vector and the velocity are perpendicular
to each other.
SOLUTION
We have
r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k
or
x = At cos t
cos t =
Then
x
At
sin t =
or
y
A
2
Then
y
A
2
or
x
A
2
t2 =
x
A
2
−1 =
−
x
A
2
−
z = Bt sin t
z
Bt
t2 =
x
At
2
z
B
2
+
z
B
2
+
cos 2 t + sin 2 t = 1
Now
(a)
y = A t2 +1
z
B
+
z
Bt
y
A
2
−1
2
=1
2
=1
Q.E.D.
With A = 3 and B = 1, we have
v=
dr
t
= 3(cos t − t sin t )i + 3
j + (sin t + t cos t )k
2
dt
t +1
( )
t
and
t 2 + 1 − t t 2 +1
dv
= 3( − sin t − sin t − t cos t )i + 3
j
a=
dt
(t 2 + 1)
+ (cos t + cos t − t sin t )k
1
= −3(2 sin t + t cos t )i + 3 2
j + (2 cos t − t sin t )k
(t + 1)3/2
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140
PROBLEM 11.96* (Continued)
v = 3(1 − 0)i + (0) j + (0)k
At t = 0:
v = vx2 + v y2 + vz2
v = 3 m/s
or
a = −3(0)i + 3(1) j + (2 − 0)k
and
a 2 = (0) 2 + (3) 2 + (2) 2 = 13
Then
a = 3.61 m/s2
or
(b)
If r and v are perpendicular, r ⋅ v = 0
(
)
[(3t cos t )i + 3 t 2 + 1 j + (t sin t )k ] ⋅ [3(cos t − t sin t )i + 3
or
(3t cos t )[3(cos t − t sin t )] + (3 t 2 + 1) 3
Expanding
t
t2 +1
t
t2 + 1
j + (sin t + t cos t )k ] = 0
+ (t sin t )(sin t + t cos t ) = 0
(9t cos 2 t − 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0
10 + 8 cos 2 t − 8t sin t cos t = 0
or (with t ≠ 0)
7 + 2 cos 2t − 2t sin 2t = 0
or
Using “trial and error” or numerical methods, the smallest root is
t = 3.82 s
Note: The next root is t = 4.38 s.
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141
142
PROBLEM 11.97
An airplane used to drop water on brushfires is flying
horizontally in a straight line at 315 km/h at an altitude of
80 m. Determine the distance d at which the pilot should
release the water so that it will hit the fire at B.
SOLUTION
First note
v0 = 315 km/h
= 87.5 m/s
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t −
At B:
or
1 2
gt
2
1
−80 m = − (9.81 m/s 2 )t 2
2
t B = 4.03855 s
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At B:
d = (87.5 m/s)(4.03855 s)
d = 353 m ��
or
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143
142
PROBLEM 11.98
Three children are throwing
snowballs at each other. Child A
throws a snowball with a
horizontal velocity v 0 . If the
snowball just passes over the
head of child B and hits child C,
determine (a) the value of v0,
(b) the distance d.
SOLUTION
(a)
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t −
1 2
gt
2
1
−1 m = − (9.81 m/s 2 )t 2
2
At B:
or tB = 0.451 524 s
Horizontal motion (Uniform)
x = 0 + (v x ) 0 t
At B:
or
(b)
Vertical motion: At C:
or
7 m = v0 (0.451524 s)
v0 = 15.5031 m/s
v0 = 15.50 m/s �
1
−3 m = − (9.81 m/s 2 ) t 2
2
tC = 0.782062 s
Horizontal motion.
At C:
�
(7 + d ) m = (15.5031 m/s)(0.782062 s)
d = 5.12 m �
or
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143
144
PROBLEM 11.99
While delivering newspapers, a girl
throws a newspaper with a horizontal
velocity v0. Determine the range of
values of v0 if the newspaper is to
land between Points B and C.
SOLUTION
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t −
1 2
gt
2
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = v0 t
At B:
y:
x:
y:
or
1
−0.6 m = − (9.81 m/s2)t2
2
tC = 0.35 s
or
Then
2.1 m = (v0 ) B (0.452 s)
(v0 ) B = 4.65 m/s
or
At C:
1
(9.81 m/s2)t2
2
t B = 0.452 s
or
Then
−1 m = −
x:
3.7 m = (v0 )C (0.35 s)
(v0 )C = 10.57 m/s
4.65 m/s
v0
10.57 m/s
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145
144
PROBLEM 11.100
A baseball pitching machine “throws” baseballs with a horizontal velocity v0. Knowing that height h varies
between 0.8 m and 1 m, determine (a) the range of values of v0, (b) the values of a corresponding to h = 0.8 m
and h = 1 m.
SOLUTION
(a)
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t −
1 2
gt
2
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = v0 t
When h = 0.8 m, y = −0.7 m:
1
−0.7 m = − (9.81 m/s2)t2
2
t31 = 0.38 s
or
Then
12 m = (v0 )31 (0.38 s)
or
(v0 )31 = 31.58 m/s = 113.69 km/h
When h = 1 m, y = −0.5 m:
1
−0.5 m = − (9.81 m/s2)t2
2
t42 = 0.32 s
or
Then
12 m = (v0 ) 42 (0.32 s)
or
(v0 )42 = 37.5 m/s = 135 km/h
113.7 km/h
v0
135 km/h
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146
PROBLEM 11.100 (Continued)
(b)
For the vertical motion
v y = (0) − gt
Now
When h = 0.8 m:
tan α =
|(v y ) B |
(v x ) B
=
gt
v0
(9.81 m/s2)(0.38 s)
31.58 m/s
= 0.118043
tan α =
α 31 = 6.7°
or
When h = 1 m:
(9.81 m/s2)(0.32 s)
37.5 m/s
= 0.083712
tan α =
α 42 = 4.8°
or
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147
146
PROBLEM 11.101
A volleyball player serves the ball with
an initial velocity v0 of magnitude
13.40 m/s at an angle of 20° with the
horizontal. Determine (a) if the ball
will clear the top of the net, (b) how far
from the net the ball will land.
SOLUTION
First note
(vx )0 = (13.40 m/s) cos 20° = 12.5919 m/s
(v y )0 = (13.40 m/s) sin 20° = 4.5831 m/s
(a)
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At C
9 m = (12.5919 m/s) t or tC = 0.71475 s
Vertical motion. (Uniformly accelerated motion)
y = y0 + ( v y ) 0 t −
At C:
1 2
gt
2
yc = 2.1 m + (4.5831 m/s)(0.71475 s)
1
− (9.81 m/s 2 )(0.71475 s)2
2
= 2.87 m
yC � 2.43 m (height of net) � ball clears net �
(b)
At B, y = 0:
1
0 = 2.1 m + (4.5831 m/s)t − (9.81 m/s 2 )t 2
2
Solving
t B = 1.271175 s (the other root is negative)
Then
d = (vx )0 t B = (12.5919 m/s)(1.271175 s)
= 16.01 m
b = (16.01 − 9.00) m = 7.01 m from the net �
The ball lands
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147
148
PROBLEM 11.102
Milk is poured into a glass of height 140 mm and inside
diameter 66 mm. If the initial velocity of the milk is 1.2 m/s
at an angle of 40° with the horizontal, determine the range of
values of the height h for which the milk will enter the glass.
SOLUTION
First note
(vx )0 = (1.2 m/s) cos 40° = 0.91925 m/s
(v y )0 = −(1.2 m/s) sin 40° = −0.77135 m/s
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
Vertical motion. (Uniformly accelerated motion)
y = y0 + ( v y ) 0 t −
1 2
gt
2
Milk enters glass at B
x:
0.08 m = (0.91925 m/s) t or tB = 0.087028 s
y : 0.140 m = hB + (−0.77135 m/s)(0.087028 s)
1
− (9.81 m/s 2 )(0.087028 s)2
2
or
hB = 0.244 m
Milk enters glass at C
x : 0.146 m = (0.91925 m/s) t or tC = 0.158825 s
y : 0.140 m = hC + (−0.77135 m/s)(0.158825 s)
1
− (9.81 m/s 2 )(0.158825 s)2
2
or
hC = 0.386 m
0.244 m � h � 0.386 m �
�
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148
PROBLEM 11.103
A golfer hits a golf ball with an
initial velocity of 48 m/s at an angle
of 25° with the horizontal. Knowing
that the fairway slopes downward at
an average angle of 5°, determine the
distance d between the golfer and
Point B where the ball first lands.
SOLUTION
(vx )0 = (48 m/s) cos 25°
First note
(v y )0 = (48 m/s) sin 25°
xB = d cos 5°
and at B
Now
yB = − d sin 5°
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
d cos 5° = (48 cos 25°)t
At B
or tB =
cos 5°
d
48 cos 25°
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (v y )0 t − gt2
2
(g = 9.81 m/s2)
1 2
gtB
2
At B:
−d sin 5° = (48 sin 25°)tB −
Substituting for t B
cos 5°
1
cos 5°
− d sin 5° = (48 sin 25°)
d− g
48 cos 25°
2 48 cos 25°
or
2
d2
2
(48 cos 25°)2(tan 5° + tan 25°)
9.81 cos 5°
= 214.49 m
d=
d = 215 m
or
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150
PROBLEM 11.104
Water flows from a drain spout with an initial velocity of
0.75 m/s at an angle of 15° with the horizontal. Determine
the range of values of the distance d for which the water
will enter the trough BC.
SOLUTION
First note
(vx )0 = (0.75 m/s) cos 15° = 0.724 m/s
(v y )0 = −(0.75 m/s) sin 15° = −0.194 m/s
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y ) 0 t −
1 2
gt
2
At the top of the trough
−2.64 m = (−0.194 m/s) t −
or
1
(9.81 m/s2)t 2
2
t BC = 0.714 s (the other root is negative)
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
In time t BC
xBC = (0.724 m/s)(0.714 s) = 0.517 m
Thus, the trough must be placed so that
xB
0.517 m xC
0.517 m
Since the trough is 0.6 m wide, it then follows that
0
d
0.52 m
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151
150
PROBLEM 11.105
Sand is discharged at A from a conveyor belt and falls onto
the top of a stockpile at B. Knowing that the conveyor belt
forms an angle α = 20° with the horizontal, determine the
speed v0 of the belt.
SOLUTION
(v0 ) x = v0 cos 20°
= 0.9397 v0
(v0 ) y = v0 sin 20°
= 0.3420 v0
Horizontal motion.
At B:
Vertical motion.
At B:
Using Eq. (1):
x = (v0 ) x t = (0.9397 v0 ) t
9 m = (0.9397 v0 ) t
y = (v0 ) y t −
t = 9.578/v0
(1)
1 2
gt
2
1
−5.4 m = v0 (0.3420) t − 9.81t 2
2
−5.4 = (0.3420)(9.578) − (4.905)
9.578
−8.676 = −(4.905)
v0
9.578
v0
2
2
v02 = 51.864 m2/s2
v0 = 7.202 m/s
v0 = 7.2 m/s
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151
152
PROBLEM 11.106
A basketball player shoots when she is 5 m from the backboard. Knowing that the ball has an initial velocity v0 at an
angle of 30° with the horizontal, determine the value of v0
when d is equal to (a) 0.2 m, (b) 0.38 m.
SOLUTION
First note
(vx )0 = v0 cos 30°
(v y )0 = v0 sin 30°
Horizontal motion. (Uniform)
x = 0 + (vx ) 0 t
At B:
(5 − d) = (v0 cos 30°)t
or tB =
5−d
v0 cos 30°
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
1 2
gt
2
At B:
1 = (v0 sin 30°) tB −
Substituting for tB
1 = (v0 sin 30°)
or
v02 =
( g = 9.81 m/s2)
1 2
gtB
2
5−d
1
5−d
− g
v0 cos 30°
v0 cos 30°
2
2g(5 − d)2
1
3
(5 − d) − 1
3
2
d~m
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152
PROBLEM 11.106 (Continued)
(a)
d = 0.2 m:
v02 =
2(9.81)(5 − 0.2)2
3
1
3
(5 − 0.2) − 1
v0 = 9.2 m/s
or
(b)
d = 0.38 m:
v02 =
2(9.81)(5 − 0.38)2
3
1
3
(5 − 0.38) − 1
v0 = 9.15 m/s
or
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153
154
PROBLEM 11.107
A group of children are throwing
balls through a 0.72-m-inner-diameter
tire hanging from a tree. A child
throws a ball with an initial velocity v0
at an angle of 3° with the horizontal.
Determine the range of values of v0 for
which the ball will go through the tire.
SOLUTION
First note
(vx )0 = v0 cos 3°
(v y )0 = v0 sin 3°
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
When x = 6 m:
6 = (v0 cos 3°)t or t6 =
6
v0 cos 3°
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
1 2
gt
2
( g = 9.81 m/s 2 )
When the ball reaches the tire, t = t6
�
�
6
yB ,C = (v0 sin 3°) �
�
� v0 cos 3° �
�
1 �
6
− g�
�
2 � v0 cos 3° �
2
or
v02 =
18(9.81)
cos 3°(6 tan 3° − yB , C )
or
v02 =
177.065
0.314447 − yB ,C
At B, y = −0.53 m:
v02 =
177.065
0.314447 − (−0.53)
or
2
(v0 ) B = 14.48 m/s
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154
PROBLEM 11.107 (Continued)
At C, y = −1.25 m:
177.06 s
0.314447 − ( −1.25)
(v0 )C = 10.64 m/s
or
�
v02 =
10.64 m/s � v0 � 14.48 m/s �
�
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155
156
PROBLEM 11.108
The nozzle at A discharges cooling water with an initial
velocity v0 at an angle of 6° with the horizontal onto a grinding
wheel 350 mm in diameter. Determine the range of values of
the initial velocity for which the water will land on the
grinding wheel between Points B and C.
SOLUTION
First note
(vx )0 = v0 cos 6°
(v y )0 = −v0 sin 6°
Horizontal motion. (Uniform)
x = x0 + (vx )0 t
Vertical motion. (Uniformly accelerated motion)
y = y0 + (v y )0 t −
At Point B:
1 2
gt
2
( g = 9.81 m/s 2 )
x = (0.175 m) sin 10°
y = (0.175 m) cos 10°
x : 0.175 sin 10° = −0.020 + (v0 cos 6°)t
or
tB =
0.050388
v0 cos 6°
y : 0.175 cos 10° = 0.205 + (−v0 sin 6°)t B −
1 2
gtB
2
Substituting for t B
� 0.050388 � 1
� 0.050388 �
−0.032659 = (−v0 sin 6°) �
� − (9.81) �
�
� v0 cos 6° � 2
� v0 cos 6° �
2
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156
PROBLEM 11.108 (Continued)
v02
or
=
1
2
(9.81)(0.050388)2
cos 2 6°(0.032659 − 0.050388 tan 6°)
(v0 ) B = 0.678 m/s
or
At Point C:
x = (0.175 m) cos 30°
y = (0.175 m) sin 30°
x : 0.175 cos 30° = −0.020 + (v0 cos 6°)t
or
tC =
0.171554
v0 cos 6°
y : 0.175 sin 30° = 0.205 + (−v0 sin 6°)tC −
1 2
gtC
2
Substituting for tC
� 0.171554 � 1
� 0.171554 �
−0.117500 = (−v0 sin 6°) �
� − (9.81) �
�
� v0 cos 6° � 2
� v0 cos 6° �
or
or
v02 =
1
2
2
(9.81)(0.171554) 2
cos 2 6°(0.117500 − 0.171554 tan 6°)
(v0 )C = 1.211 m/s
0.678 m/s � v0 � 1.211 m/s �
�
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157
158
PROBLEM 11.109
While holding one of its ends, a worker lobs a coil of rope over
the lowest limb of a tree. If he throws the rope with an initial
velocity v0 at an angle of 65° with the horizontal, determine the
range of values of v0 for which the rope will go over only the
lowest limb.
SOLUTION
First note
(vx )0 = v0 cos 65°
(v y )0 = v0 sin 65°
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At either B or C, x = 5 m
s = (v0 cos 65°)t B,C
or
t B ,C =
5
(v0 cos 65°)
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
1 2
gt
2
( g = 9.81 m/s 2 )
At the tree limbs, t = tB ,C
�
� 1 �
�
5
5
yB ,C = (v0 sin 65°) �
�− g�
�
� v0 cos 65° � 2 � v0 cos 65° �
2
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158
PROBLEM 11.109 (Continued)
or
v02 =
=
2
1
2
(9.81)(25)
cos 65°(5 tan 65° − yB , C )
686.566
5 tan 65° − yB, C
At Point B:
v02 =
686.566
5 tan 65° − 5
or
(v0 ) B = 10.95 m/s
At Point C:
v02 =
686.566
5 tan 65° − 5.9
or
(v0 )C = 11.93 m/s
10.95 m/s � v0 � 11.93 m/s ��
�
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159
160
PROBLEM 11.110
A ball is dropped onto a step at Point A and rebounds with a
velocity v0 at an angle of 15° with the vertical. Determine the
value of v0, knowing that just before the ball bounces at Point B,
its velocity vB forms an angle of 12° with the vertical.
SOLUTION
First note
(vx )0 = v0 sin 15°
(v y )0 = v0 cos 15°
Horizontal motion. (Uniform)
vx = (vx )0 = v0 sin 15°
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (v0 cos 15°) t − gt 2
2
v y = (v y )0 − gt
y = 0 + (v y ) 0 t −
= v0 cos 15° − gt
At Point B, v y � 0
Then
tan 12° =
(v x ) B
v0 sin 15°
=
|(v y ) B | gt B − v0 cos 15°
� sin 15°
�
+ cos 15° � g = 9.81 m/s 2
�
� tan 12°
�
= 0.22259 v0
v0
g
or
tB =
Noting that
yB = −0.2 m,
We have
1
−0.2 = (v0 cos 15°)(0.22259 v0 ) − (9.81)(0.22259 v0 ) 2
2
v0 = 2.67 m/s ��
or�
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161
160
PROBLEM 11.111
A model rocket is launched from Point A with an initial
velocity v 0 of 75 m/s. If the rocket’s descent parachute
does not deploy and the rocket lands 120 m from A,
determine (a) the angle α that v0 forms with the vertical,
(b) the maximum height above Point A reached by the
rocket, and (c) the duration of the flight.
SOLUTION
Set the origin at Point A.
x0 = 0,
y0 = 0
Horizontal motion:
x = v0 t sin α
Vertical motion:
y = v0t cos α −
cos α =
sin 2 α + cos 2 α =
sin α =
x
v0t
(1)
1 2
gt
2
1
1
y + gt 2
2
v0 t
(2)
1
1
x 2 + y + gt 2
2
(v0 t )2
x 2 + y 2 + gyt 2 +
(
2
=1
1 24
g t = v02t 2
4
)
1 24
g t − v02 − gy t 2 + ( x 2 + y 2 ) = 0
4
At Point B,
(3)
x 2 + y 2 = 120 m, x = 120 cos 30° m
y = −120 sin 30° = −60 m
1
(9.81)2t4 − [752 − (9.81)(−60)]t 2 + 1202 = 0
4
24.06t 4 − 6213.6t 2 − 14400 = 0
t = 16 s and 1.53 s
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162
PROBLEM 11.111 (Continued)
Restrictions on α :
α
0
tan α =
120°
x
y + gt
1
2
2
=
120 cos 30°
= 0.086915
−60 + (4.905)(16)2
α = 4.967°
120 cos 30°
= −2.141953
−60 + (4.905)(1.53)2
α = 115.03°
and
Use α = 4.967° corresponding to the steeper possible trajectory.
(a)
Angle α .
(b)
Maximum height.
α = 5°
v y = 0 at
y = ymax
v y = v0 cos α − gt = 0
t=
v0 cos α
g
ymax = v0 t cos α −
=
(c)
Duration of the flight.
v 2 cos 2 α
1
gt = 0
2
2g
(75)2 cos2 4.967°
(2)(9.81)
ymax = 285 m
(time to reach B)
t = 16 s
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163
162
PROBLEM 11.112
The initial velocity v0 of a hockey
puck is 168 km/h. Determine (a) the
largest value (less than 45°) of the
angle α for which the puck will enter
the net, (b) the corresponding time
required for the puck to reach the net.
SOLUTION
First note
v0 = 168 km/h = 46.7 m/s
(vx )0 = v0 cos α = (46.7 m/s) cos α
and
(v y )0 = v0 sin α = (46.7 m/s) sin α
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (46.7 cos α )t
At the front of the net,
Then
or
x = 4.8 m
4.8 = (46.7 cos α ) t
tenter =
4.8
46.7 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (46.7 sin α ) t − gt 2
2
y = 0 + (v y )0 t −
( g = 9.81 m/s2 )
At the front of the net,
yfront = (46.7 sin α ) tenter −
1 2
gtenter
2
4.8
= (46.7 sin α )
46.7 cos α
= 4.8 tan α −
Now
1
4.8
g
2
46.7 cos α
2
11.52 g
2180.9 cos2 α
1
= sec2 α = 1 + tan 2 α
cos 2 α
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164
PROBLEM 11.112 (Continued)
Then
or
Then
or
or
yfront = 4.8 tan α −
tan2 α −
tan α =
11.52g
(1 + tan2 α )
2180.9
2180.9
2180.9
yfront
tan α + 1 +
11.52g
2.4g
2180.9
2.4g
±
(−
2180.9
tan α =
±
4.8 × 9.81
2180.9
2.4g
)
2
(
=0
)
1/2
2180.9
y
− 4 1 + 11.52
g front
2
2180.9
−
4.8 × 9.81
2
2180.9
− 1+
yfront
11.52 × 9.81
1/2
tan α = 46.3154 ± [(46.3154)2 − (1 + 19.2981 yfront )] 1/2
Now 0 yfront 1.2 m so that the positive root will yield values of α 45° for all values of yfront.
When the negative root is selected, α increases as yfront is increased. Therefore, for α max , set
yfront = yC = 1.2 m
(b)
Then
tan α = 46.3154 − [(46.3154)2 − (1 + 19.2981 × 1.2)]1 / 2
or
α max = 14.66°
We had found
tenter =
α max = 14.7°
4.8
46.7 cos α
4.8
=
46.7 cos 14.66°
tenter = 0.106 s
or
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164
PROBLEM 11.113
The pitcher in a softball game
throws a ball with an initial
velocity v0 of 72 km/h at an angle
α with the horizontal. If the height
of the ball at Point B is 0.68 m,
determine (a) the angle α, (b) the
angle θ that the velocity of the ball
at Point B forms with the horizontal.
SOLUTION
First note
v0 = 72 km/h = 20 m/s
(vx )0 = v0 cos α = (20 m/s) cos α
and
(v y )0 = v0 sin α = (20 m/s) sin α
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (20 cos α ) t
At Point B:
14 = (20 cos α )t or t B =
7
10 cos α
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
At Point B:
1 2
1
gt = (20 sin α )t − gt 2
2
2
0.08 = (20 sin α )t B −
( g = 9.81 m/s 2 )
1 2
gt B
2
Substituting for t B
�
7
0.08 = (20 sin α ) �
α
10
cos
�
or
Now
8 = 1400 tan α −
� 1 �
�
7
�− g�
�
α
2
10
cos
�
�
�
2
1
49
g
2 cos 2 α
1
= sec2 α = 1 + tan 2 α
2
cos α
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166
PROBLEM 11.113 (Continued)
Then
or
Solving
8 = 1400 tan α − 24.5 g (1 + tan 2 α )
240.345 tan 2 α − 1400 tan α + 248.345 = 0
α = 10.3786° and α = 79.949°
Rejecting the second root because it is not physically reasonable, we have
α = 10.38° �
(b)
We have
vx = (vx )0 = 20 cos α
and
v y = (v y )0 − gt = 20 sin α − gt
At Point B:
(v y ) B = 20 sin α − gt B
= 20 sin α −
7g
10 cos α
Noting that at Point B, v y � 0, we have
tan θ =
=
=
|(v y ) B |
vx
7g
10 cos α
− 20 sin α
20 cos α
7
9.81
200 cos 10.3786°
− sin 10.3786°
cos 10.3786°
θ = 9.74° ��
or
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166
PROBLEM 11.114*
A mountain climber plans to jump from A to B over a
crevasse. Determine the smallest value of the climber’s initial
velocity v0 and the corresponding value of angle α so that he
lands at B.
SOLUTION
First note
(vx )0 = v0 cos α
(v y )0 = v0 sin α
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (v0 cos α ) t
At Point B:
or
1.8 = (v0 cos α )t
tB =
1.8
v0 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (v0 sin α ) t − gt 2
2
y = 0 + (v y )0 t −
At Point B:
−1.4 = (v0 sin α ) t B −
( g = 9.81 m/s 2 )
1 2
gt B
2
Substituting for t B
� 1.8 � 1 � 1.8 �
−1.4 = (v0 sin α ) �
�− g�
�
� v0 cos α � 2 � v0 cos α �
or
v02 =
1.62 g
cos α (1.8 tan α + 1.4)
=
1.62 g
0.9 sin 2α + 1.4 cos 2 α
2
2
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PROBLEM 11.114* (Continued)
Now minimize v02 with respect to α.
We have
dv02
−(1.8 cos 2α − 2.8 cos α sin α )
= 1.62 g
=0
dα
(0.9 sin 2α + 1.4 cos 2 α ) 2
1.8 cos 2α − 1.4 sin 2α = 0
or
tan 2α =
or
18
14
α = 26.0625° and α = 206.06°
or
Rejecting the second value because it is not physically possible, we have
α = 26.1° �
Finally,
v02 =
1.62 × 9.81
cos 26.0625°(1.8 tan 26.0625° + 1.4)
2
(v0 )min = 2.94 m/s �
or
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168
PROBLEM 11.115
An oscillating garden sprinkler which
discharges water with an initial velocity
v0 of 8 m/s is used to water a vegetable
garden. Determine the distance d to the
farthest Point B that will be watered and
the corresponding angle α when (a) the
vegetables are just beginning to grow,
(b) the height h of the corn is 1.8 m.
SOLUTION
First note
(vx )0 = v0 cos α = (8 m/s) cos α
(v y )0 = v0 sin α = (8 m/s) sin α
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (8 cos α ) t
At Point B:
or
x = d : d = (8 cos α ) t
d
tB =
8 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (8 sin α ) t − gt 2 ( g = 9.81 m/s 2 )
2
1
0 = (8 sin α ) t B − gt B2
2
y = 0 + (v y )0 t −
At Point B:
Simplifying and substituting for t B
0 = 8 sin α −
d=
or
(a)
1 � d �
g�
�
2 � 8 cos α �
64
sin 2α
g
(1)
When h = 0, the water can follow any physically possible trajectory. It then follows from
Eq. (1) that d is maximum when 2α = 90°
α = 45° �
or
Then
d=
64
sin (2 × 45°)
9.81
d max = 6.52 m �
or
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PROBLEM 11.115 (Continued)
(b)
Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as α increases
in value from 0 to 45° and then d decreases as α is further increased. Thus, d max occurs for the value
of α closest to 45° and for which the water just passes over the first row of corn plants. At this row,
xcom = 1.5 m
tcorn =
so that
1.5
8 cos α
Also, with ycorn = h, we have
h = (8 sin α ) tcorn −
1 2
gtcorn
2
Substituting for tcorn and noting h = 1.8 m,
� 1.5 � 1 � 1.5 �
1.8 = (8 sin α ) �
�− g�
�
� 8 cos α � 2 � 8 cos α �
or
Now
Then
or
1.8 = 1.5 tan α −
2
2.25 g
128 cos 2 α
1
= sec2 α = 1 + tan 2 α
cos 2 α
1.8 = 1.5 tan α −
2.25(9.81)
(1 + tan 2 α )
128
0.172441 tan 2 α − 1.5 tan α + 1.972441 = 0
α = 58.229° and α = 81.965°
Solving
From the above discussion, it follows that d = d max when
α = 58.2° �
Finally, using Eq. (1)
d=
64
sin (2 × 58.229°)
9.81
d max = 5.84 m �
or
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170
PROBLEM 11.116
A worker uses high-pressure water to
clean the inside of a long drainpipe. If
the water is discharged with an initial
velocity v0 of 11.5 m/s, determine
(a) the distance d to the farthest Point B
on the top of the pipe that the worker
can wash from his position at A, (b) the
corresponding angle α.
SOLUTION
First note
(vx )0 = v0 cos α = (11.5 m/s) cos α
(v y )0 = v0 sin α = (11.5 m/s) sin α
By observation, d max occurs when
ymax = 1.1 m.
Vertical motion. (Uniformly accelerated motion)
v y = (v y )0 − gt
= (11.5 sin α ) − gt
y = ymax
When
Then
at
1 2
gt
2
1
= (11.5 sin α ) t − gt 2
2
y = 0 + (v y ) 0 t −
B , (v y ) B = 0
(v y ) B = 0 = (11.5 sin α ) − gt
11.5 sin α
g
( g = 9.81 m/s 2 )
or
tB =
and
yB = (11.5 sin α ) t B −
1 2
gt B
2
Substituting for t B and noting yB = 1.1 m
� 11.5 sin α � 1 � 11.5 sin α �
1.1 = (11.5 sin α ) �
�− g�
�
g
g
�
� 2 �
�
1
=
(11.5) 2 sin 2 α
2g
or
sin 2 α =
2.2 × 9.81
11.52
2
α = 23.8265°
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PROBLEM 11.116 (Continued)
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (11.5 cos α ) t
At Point B:
where
Then
x = d max
tB =
and t = t B
11.5
sin 23.8265° = 0.47356 s
9.81
d max = (11.5)(cos 23.8265°)(0.47356)
d max = 4.98 m �
or
(b)
α = 23.8° ��
From above
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PROBLEM 11.117
As slider block A moves downward at a speed of 0.5 m/s, the velocity
with respect to A of the portion of belt B between idler pulleys C and D
is vCD/A = 2 m/s
θ. Determine the velocity of portion CD of the belt
when (a) θ = 45°, (b) θ = 60°.
SOLUTION
We have
vCD = v A + vCD/A
v A = (0.5 m/s)(− cos 65°i − sin 65° j)
where
= (0.21131 m/s)i − (0.45315 m/s) j
vCD/A = (2 m/s)(cos θ i + sin θ j)
and
vCD = [(−0.21131 + 2 cos θ ) m/s]i + [(−0.45315 + 2 sin θ ) m/s] j
Then
(a)
We have
vCD = (−0.21131 + 2 cos 45°)i + (−0.45315 + 2 sin 45°) j
= (1.20290 m/s)i + (0.96106 m/s) j
vCD = 1.540 m/s
or
(b)
We have
38.6°
vCD = (−0.21131 + 2 cos 60°)i + ( −0.45315 + 2 sin 60°) j
= (0.78869 m/s)i + (1.27890 m/s) j
vCD = 1.503 m/s
or
58.3°
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PROBLEM 11.118
The velocities of skiers A and B are
as shown. Determine the velocity of
A with respect to B.
SOLUTION
We have
vA = vB + vA/B
The graphical representation of this equation is then as shown.
Then
v 2A/B = 102 + 142 − 2(10)(14) cos 15°
or
vA/B = 5.05379 m/s
and
or
10
5.05379
=
sin α
sin 15°
α = 30.8°
vA/B = 5.05 m/s
55.8° �
Alternative solution.
vA/B = vA − vB
= 10 cos 10i − 10 sin 10 j − (14 cos 25i − 14 sin 25 j)
= −2.84i + 4.14 j
= 5.05 m/s
55.8°
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175
174
PROBLEM 11.119
Shore-based radar indicates that a ferry leaves its slip with a
70°, while instruments aboard the
velocity v = 18 km/h
ferry indicate a speed of 19 km/h and a heading of 30° west
of south relative to the river. Determine the velocity of the
river.
SOLUTION
We have
v F = v R + v F/R
or
v F = v F/R + v R
The graphical representation of the second equation is then as shown.
We have
vR2 = 182 + 192 − 2(18)(19) cos 10°
or
vR = 3.375 km/h
and
or
18
3.375
=
sin α sin 10°
α = 67.84°
Noting that
v R = 3.4 km/h
7.8°
Alternatively one could use vector algebra.
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175
176
PROBLEM 11.120
Airplanes A and B are flying at the same altitude and are tracking the
eye of hurricane C. The relative velocity of C with respect to A
75°, and the relative velocity of C with respect
is vC/A = 235 km/h
to B is vC/B = 260 km/h
40°. Determine (a) the relative velocity
of B with respect to A, (b) the velocity of A if ground-based radar
indicates that the hurricane is moving at a speed of 24 mi/h due north,
(c) the change in position of C with respect to B during a 15-min
interval.
SOLUTION
(a)
We have
v C = v A + v C/ A
and
v C = v B + v C/ B
Then
v A + vC/A = v B + v C/B
or
v B − v A = v C/A − vC/B
Now
v B − v A = v B/A
so that
v B/A = vC/A − vC/B
or
vC/A = vC/B + v B/A
The graphical representation of the last equation is then as shown.
We have
vB2 /A = 2352 + 2602 − 2(235)(260) cos 65°
or
vB/A = 266.798 km/h
and
or
260 266.798
=
sin α
sin 65°
α = 62.03°
v B/A = 267 km/h
(b)
We have
v C = v A + v C/ A
or
v A = (24 km/h)j − (235 km/h)(−cos 75°i − sin 75°j)
12.97°
v A = (60.822 km/h)i + (250.99 km/h)j
v A = 258 km/h
or
76.4°
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176
PROBLEM 11.120 (Continued)
(c)
Noting that the velocities of B and C are constant. We have
rB = (rB )0 + v B t
Now
rC = (rC )0 + v C t
rC/B = rC − rB = [(rC )0 − (rB )0 ] + ( vC − v B )t
= [(rC )0 − (rB )0 ] + v C/B t
Then
∆rC/B = (rC/B )t2 − (rC/B )t1 = v C/B (t2 − t1 ) = vC/B ∆t
For ∆t = 15 min:
∆rC/B = (260 km/h)
1
h = 65 km
4
∆rC/B = 65 km
40°
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178
PROBLEM 11.121
The velocities of commuter trains
A and B are as shown. Knowing
that the speed of each train is
constant and that B reaches the
crossing 10 min after A passed
through the same crossing,
determine (a) the relative velocity
of B with respect to A, (b) the
distance between the fronts of the
engines 3 min after A passed
through the crossing.
SOLUTION
(a)
We have
v B = v A + v B/A
The graphical representation of this equation is then as shown.
Then
vB2 /A = 662 + 482 − 2(66)(48) cos 155°
or
vB/A = 111.366 km/h
and
or
48
111.366
=
sin α sin 155°
α = 10.50°
v B/A = 111.4 km/h
(b)
10.50° �
First note that
at t = 3 min, A is (66 km/h) ( 603 ) = 3.3 km west of the crossing.
7
at t = 3 min, B is (48 km/h) ( 60
) = 5.6 km southwest of the crossing.
Now
rB = rA + rB/A
Then at t = 3 min, we have
rB2/ A = 3.32 + 5.62 − 2(3.3)(5.6) cos 25°
rB/A = 2.96 km ��
or
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179
178
PROBLEM 11.122
Knowing that the velocity of block B with respect to block A
70°, determine the velocities of A and B.
is v B/A = 5.6 m/s
SOLUTION
From the diagram
2 x A + 3xB = constant
Then
2v A + 3vB = 0
2
vA
3
or
|vB | =
Now
v B = v A + v B/A
and noting that vA and v B must be parallel to surfaces A and B, respectively, the graphical representation of
this equation is then as shown. Note: Assuming that vA is directed up the incline leads to a velocity diagram
that does not “close.”
First note
Then
or
or
α = 180° − (40° + 30° + θ B )
= 110° − θ B
2
v
vA
5.6
= 3 A =
sin (110° − θ B ) sin 40° sin (30° + θ B )
v A sin 40° =
2
v A sin (110° − θ B )
3
sin (110° − θ B ) = 0.96418
or
θ B = 35.3817°
and
θ B = 4.6183°
For θ B = 35.3817° :
vB =
or
v A = 5.94 m/s vB = 3.96 m/s
2
5.6 sin 40°
vA =
3
sin (30° + 35.3817°)
v A = 5.94 m/s
v B = 3.96 m/s
30° �
35.4° �
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179
180
PROBLEM 11.122 (Continued)
2
5.6 sin 40°
vA =
3
sin (30° + 4.6183°)
For θ B = 4.6183°:
vB =
or
v A = 9.50 m/s
vB = 6.34 m/s
v A = 9.50 m/s
v B = 6.34 m/s
30° �
4.62° ��
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181
180
PROBLEM 11.123
Knowing that at the instant shown block A has a velocity of 8 cm/s
and an acceleration of 6 cm/s2 both directed down the incline,
determine (a) the velocity of block B, (b) the acceleration of block B.
SOLUTION
From the diagram
2 x A + xB/A = constant
Then
2v A + vB/A = 0
| vB/A | = 16 cm/s
or
2a A + aB/A = 0
and
| aB/A | = 12 cm/s2
or
Note that v B/A and a B/A must be parallel to the top surface of block A.
(a)
We have
v B = v A + v B/A
The graphical representation of this equation is then as shown. Note that because A is moving
downward, B must be moving upward relative to A.
We have
vB2 = 82 + 162 − 2(8)(16) cos 15°
or
vB = 8.5278 cm/s
and
or
8
8.5278
=
sin α sin 15°
α = 14.05°
v B = 8.53 cm/s
(b)
54.1°
The same technique that was used to determine v B can be used to determine a B . An alternative method
is as follows.
We have
a B = a A + a B/A
= (6i ) + 12(− cos 15°i + sin 15° j)*
= –(5.5911 cm/s2)i + (3.1058 cm/s2)j
a B = 6.40 cm/s2
or
54.1°
* Note the orientation of the coordinate axes on the sketch of the system.
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182
PROBLEM 11.124
Knowing that at the instant shown assembly A has a velocity of 9 cm/s and
an acceleration of 15 cm/s2 both directed downward, determine (a) the
velocity of block B, (b) the acceleration of block B.
SOLUTION
Length of cable = constant
L = x A + 2 xB/A = constant
v A + 2vB/A = 0
(1)
a A + 2aB/A = 0
(2)
a A = 15 cm/s2 ↓
Data:
v A = 9 cm/s ↓
Eqs. (1) and (2)
a A = −2aB/A
v A = −2vB/A
15 = −2aB/A
9 = −2vB/A
aB/A = −7.5 cm/s2
vB/A = −4.5 cm/s
a B/A = 7.5 cm/s2
(a)
40°
v B/A = −4.5 cm/s
Velocity of B.
v B = v A + v B/A
Law of cosines:
vB2 = (9) 2 + (4.5) 2 − 2(9)(4.5) cos 50°
40°
vB = 7.013 cm/s
Law of sines:
sin β sin 50°
=
4.5
7.013
β = 29.44°
α = 90° − β = 90° − 29.44° = 60.56°
v B = 7 cm/s
60.6°
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182
PROBLEM 11.124 (Continued)
(b)
Acceleration of B. a B may be found by using analysis similar to that used above for vB . An alternate
method is
a B = a A + a B/A
a B = 15 cm/s2 ↓ + 7.5 cm/s2
40°
= −15 j − (7.5 cos 40°)i + (7.5 sin 40°) j
= −15 j − 5.745i + 4.821j
a B = −5.745i − 10.179 j
a B = 11.7 cm/s2
60.6°
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184
PROBLEM 11.125
The assembly of rod A and wedge B starts from rest and moves
to the right with a constant acceleration of 2 mm/s2. Determine
(a) the acceleration of wedge C, (b) the velocity of wedge C
when t = 10 s.
SOLUTION
(a)
We have
a C = a B + a C/ B
The graphical representation of this equation is then as shown.
First note
Then
α = 180° − (20° + 105°)
= 55°
aC
2
=
sin 20° sin 55°
aC = 0.83506 mm/s 2
(b)
aC = 0.835 mm/s 2
75° �
vC = 8.35 mm/s
75° �
For uniformly accelerated motion
vC = 0 + aC t
At t = 10 s:
vC = (0.83506 mm/s 2 )(10 s)
= 8.3506 mm/s
or
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184
PROBLEM 11.126
As the truck shown begins to back up with a constant acceleration
of 1.2 m/s2, the outer section B of its boom starts to retract with a
constant acceleration of 0.5 m/s2 relative to the truck. Determine
(a) the acceleration of section B, (b) the velocity of section B
when t = 2 s.
SOLUTION
(a)
We have
a B = a A + a B/A
The graphical representation of this equation is then as shown.
We have
aB2 = 1.22 + 0.52 − 2(1.2)(0.5) cos 50°
or
aB = 0.95846 m/s 2
and
or
0.5
0.95846
=
sin α
sin 50°
α = 23.6°
a B = 0.958 m/s 2
(b)
23.6° �
For uniformly accelerated motion
vB = 0 + a B t
At t = 2 s:
vB = (0.95846 m/s 2 )(2 s)
= 1.91692 m/s
v B = 1.917 m/s
or
23.6° ��
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186
PROBLEM 11.127
Conveyor belt A, which forms a 20°
angle with the horizontal, moves at
a constant speed of 1.2 m/s and is
used to load an airplane. Knowing
that a worker tosses duffel bag B
with an initial velocity of 0.7 m/s at
an angle of 30° with the horizontal,
determine the velocity of the bag
relative to the belt as it lands on the
belt.
SOLUTION
First determine the velocity of the bag as it lands on the belt. Now
[(vB ) x ]0 = (vB )0 cos 30°
= (0.7 m/s) cos 30°
[(vB ) y ]0 = (vB )0 sin 30°
= (0.7 m/s) sin 30°
Horizontal motion. (Uniform)
x = 0 + [(vB ) x ]0 t
(vB ) x = [(vB ) x ]0
= 0.7 cos 30°
= (0.7 cos 30°)t
Vertical motion. (Uniformly accelerated motion)
y = y0 + [(vB ) y ]0 t −
1 2
gt
2
= 0.5 + (0.7 sin 30°) t −
(vB ) y = [(vB ) y ]0 − gt
1 2
gt
2
= 0.7 sin 30° − gt
The equation of the line collinear with the top surface of the belt is
y = x tan 20°
Thus, when the bag reaches the belt
0.5 + (0.7 sin 30°)t −
1 2
gt = [(0.7 cos 30°)t] tan 20°
2
or
1
(9.81)t 2 + 0.7(cos 30° tan 20° − sin 30°)t − 0.5 = 0
2
or
4.905t 2 − 0.129t − 0.5 = 0
Solving
t = 0.333 s and t = −0.306 s (Reject)
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186
PROBLEM 11.127 (Continued)
The velocity v B of the bag as it lands on the belt is then
v B = (0.7 cos 30°)i + [0.7 sin 30° − 9.81(0.333)]j
= (0.606 m/s)i − (2.917 m/s)j
Finally
or
v B = v A + v B/A
v B/A = (0.606i − 2.917j) − 1.2(cos 20°i + sin 20°j)
= −(0.5216 m/s)i − (3.3274 m/s)j
v B/A = 3.37 m/s
or
81.1°
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187
188
PROBLEM 11.128
Determine the required velocity of the belt B if the relative
velocity with which the sand hits belt B is to be (a) vertical,
(b) as small as possible.
SOLUTION
A grain of sand will undergo projectile motion.
vsx = vsx = constant = −1.5 m/s
0
vs y = 2 gh = (2)(9.81 m/s2)(1 m) = 4.43 m/s ↓
y-direction.
v S/B = v S − v B
Relative velocity.
(a)
(1)
If vS/B is vertical,
−vS/B j = −1.5i − 4.43j − (−vB cos 15i + vB sin 15j)
= −1.5i − 4.43j + vB cos 15i − vB sin 15j
Equate components.
i : 0 = −1.5 + vB cos 15°
vB =
1.5
= 1.55 m/s
cos 15°
v B = 1.55 m/s
(b)
15°
vS/C is as small as possible, so make vS/B ⊥ to vB into (1).
−vS/B sin 15°i − vS/B cos 15° j = −1.5i − 4.43j + vB cos 15°i − vB sin 15°j
Equate components and transpose terms.
(sin 15°) vS/B + (cos 15°) vB = 1.5
(cos 15°) vS/B − (sin 15°) vB = 4.43
Solving,
vS/B = 4.667 m/s
vB = 0.302 m/s
v B = 0.3 m/s
15°
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188
PROBLEM 11.129
As observed from a ship moving due east at 9 km/h, the wind appears to blow from the south. After the ship
has changed course and speed, and as it is moving due north at 6 km/h, the wind appears to blow from the
southwest. Assuming that the wind velocity is constant during the period of observation, determine the
magnitude and direction of the true wind velocity.
SOLUTION
v wind = v ship + v wind/ship
v w = v s + v w/s
Case �
v s = 9 km/h →; v w/s ↑
Case �
v s = 6 km/h ↑ ; v w/s �
15
= 1.6667
9
α = 59.0°
tan α =
vw = 92 + 152 = 17.49 km/h
v w = 17.49 km/h
59.0° ��
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189
190
PROBLEM 11.130
When a small boat travels north at 5 km/h, a flag mounted on its
stern forms an angle θ = 50° with the centerline of the boat as
shown. A short time later, when the boat travels east at 20 km/h,
angle θ is again 50°. Determine the speed and the direction of the
wind.
SOLUTION
We have
vW = v B + vW/B
Using this equation, the two cases are then graphically represented as shown.
With vW now defined, the above diagram is redrawn for the two cases for clarity.
Noting that
θ = 180° − (50° + 90° + α )
= 40° − α
We have
φ = 180° − (50° + α )
= 130° − α
vW
vW
5
20
=
=
sin 50° sin (40° − α ) sin 50° sin (130° − α )
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190
PROBLEM 11.130 (Continued)
Therefore
or
or
5
20
=
sin (40° − α ) sin (130° − α )
sin 130° cos α − cos 130° sin α = 4(sin 40° cos α − cos 40° sin α )
tan α =
sin 130° − 4 sin 40°
cos 130° − 4 cos 40°
or
α = 25.964°
Then
vW =
5 sin 50°
= 15.79 km/h
sin (40° − 25.964°)
vW = 15.79 km/h
26.0° ��
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191
192
PROBLEM 11.131
As part of a department store display, a
model train D runs on a slight incline
between the store’s up and down
escalators. When the train and shoppers
pass Point A, the train appears to a
shopper on the up escalator B to move
downward at an angle of 22° with the
horizontal, and to a shopper on the down
escalator C to move upward at an angle
of 23° with the horizontal and to travel to
the left. Knowing that the speed of the
escalators is 1 m/s, determine the speed
and the direction of the train.
SOLUTION
v D = v B + v D/B
We have
v D = vC + v D/C
The graphical representations of these equations are then as shown.
Then
vD
1
=
sin 8° sin (22° + α )
vD
1
=
sin 7° sin (23° − α )
Equating the expressions for vD
sin 8°
sin 7°
=
sin (22° + α ) sin (23° − α )
or
or
or
Then
sin 8° (sin 23° cos α − cos 23° sin α ) = sin 7° (sin 22° cos α
+ cos 22° sin α )
tan α =
sin 8° sin 23° − sin 7° sin 22°
sin 8° cos 23° + sin 7° cos 22°
α = 2.0728°
vD =
1 sin 8°
= 0.341 m/s
sin (22° + 2.0728°)
v D = 0.34 m/s
2.07°
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193
192
PROBLEM 11.131 (Continued)
Alternate solution using components.
v B = (1 m/s)
30° = (0.866 m/s)i + (0.5 m/s)j
vC = (1 m/s)
30° = (0.866 m/s)i − (0.5 m/s)j
v D/B = u1
22° = −(u1 cos 22°)i − (u1 sin 22°) j
v D/C = u2
23° = −(u2 cos 23°)i + (u2 sin 23°) j
v D = vD
α = −(vD cos α )i + (vD sin α ) j
v D = v B + v D/B = vC + v D/C
0.866i + 0.5j − (u1 cos 22°)i − (u1 sin 22°)j = 0.866i − 0.5j − (u2 cos 23°)i + (u2 sin 23°)j
Separate into components, and transpose and change
u1 cos 22° − u2 cos 23° = 0
u1 sin 22° + u1 sin 23° = 1
Solving for u1 and u2 ,
u1 = 1.302 m/s
u2 = 1.311 m/s
v D = 0.866i + 0.5j − (1.302 cos 22°)i − (1.302 sin 22°)j
= −(0.341 m/s)i + (0.01226 m/s)j
or
v D = 0.866i − 0.5j − (1.311 cos 23°)i + (1.311 sin 23°)j
= −(0.341 m/s)i + (0.01226 m/s)j
vD = 0.34 m/s
2.07°
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194
PROBLEM 11.132
The paths of raindrops during a storm appear to form an angle of 75° with the vertical and to be directed to the
left when observed through a left-side window of an automobile traveling north at a speed of 40 km/h. When
observed through a right-side window of an automobile traveling south at a speed of 30 km/h, the raindrops
appear to form an angle of 60° with the vertical. If the driver of the automobile traveling north were to stop, at
what angle and with what speed would she observe the drops to fall?
SOLUTION
v R = ( v A )1 + ( v R/A )1
We have
v R = ( v A )2 + ( v R/A ) 2
The graphical representations of these equations are then as shown. Note that the line of action of ( v R/A ) 2
must be directed as shown so that the second velocity diagram “closes.”
From the diagram
(vR ) y = [40 + (vR ) x ]tan 15°
and
(vR ) y = [30 − (vR ) x ]tan 30°
Equating the expressions for (vR ) y
[40 + (vR ) x ] tan 15° = [30 − (vR ) x ] tan 30°
or
(vR ) x = 7.8109 km/h
Then
(vR ) y = (40 + 7.8109) tan 15° = 12.8109 km/h
v R = 15.00 km/h
58.6°
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195
194
PROBLEM 11.133
Determine the peripheral speed of the centrifuge test cab A for
which the normal component of the acceleration is 10g.
SOLUTION
an = 10 g = 10(9.81 m/s 2 ) = 98.1 m/s 2
an =
v2
ρ
v 2 = ρ an = (8 m)(98.1 m/s2 ) = 784.8 m 2 /s 2
v = 28.0 m/s �
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195
196
PROBLEM 11.134
To test its performance, an automobile is driven around a circular test track of diameter d. Determine (a) the
value of d if when the speed of the automobile is 72 km/h, the normal component of the acceleration
is 3.2 m/s 2 , (b) the speed of the automobile if d = 180 m and the normal component of the acceleration is
measured to be 0.6 g.
SOLUTION
(a)
First note
Now
or
v = 72 km/h = 20 m/s
an =
v2
ρ
d (20 m/s)2
=
2 3.2 m/s 2
d = 250 m �
or
(b)
v2
ρ
We have
an =
Then
�1
�
v 2 = (0.6 × 9.81 m/s 2 ) � × 180 m �
�2
�
or
v = 23.016 m/s
v = 82.9 km/h ��
or
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196
PROBLEM 11.135
Determine the smallest radius that should be used for a highway
if the normal component of the acceleration of a car traveling at
72 km/h is not to exceed 0.7 m/s2.
SOLUTION
an =
v2
ρ
an = 0.7 m/s2
v = 72 km/h = 20 m/s
0.7 m/s2 =
(20 m/s)2
ρ
ρ = 571 m
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197
198
PROBLEM 11.136
Determine the maximum speed that the cars of the roller-coaster
can reach along the circular portion AB of the track if the normal
component of their acceleration cannot exceed 3 g.
SOLUTION
We have
an =
v2
ρ
Then
(vmax )2AB = (3 × 9.81 m/s2)(24 m)
or
(vmax ) AB = 26.577 m/s
(vmax ) AB = 95.7 km/h
or
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198
PROBLEM 11.137
Pin A, which is attached to link AB, is constrained to move in
the circular slot CD. Knowing that at t = 0 the pin starts from
rest and moves so that its speed increases at a constant rate
of 20 mm/s 2 , determine the magnitude of its total acceleration
when (a) t = 0, (b) t = 2 s.
SOLUTION
(a)
At t = 0, v A = 0, which implies (a A ) n = 0
a A = (a A )t
a A = 20 mm/s 2 ��
or
(b)
We have uniformly accelerated motion
v A = 0 + ( a A )t t
At t = 2 s:
Now
Finally,
v A = (20 mm/s 2 )(2 s) = 40 mm/s
(a A ) n =
v A2 (40 mm/s) 2
=
= 17.7778 mm/s 2
90 mm
ρA
a A2 = ( a A )t2 + (a A ) 2n
= (20) 2 + (17.7778) 2
a A = 26.8 mm/s 2 �
or
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199
200
PROBLEM 11.138
A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the
maximum total acceleration of the train must not exceed 1.5 m/s 2 , determine (a) the shortest distance in
which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at .
SOLUTION
When v = 72 km/h = 20 m/s and ρ = 400 m,
an =
v 2 (20)2
=
= 1.000 m/s 2
ρ
400
a = an2 + at2
But
at = a 2 − an2 = (1.5) 2 − (1.000) 2 = ± 1.11803 m/s 2
Since the train is accelerating, reject the negative value.
(a)
Distance to reach the speed.
v0 = 0
Let
x0 = 0
v12 = v02 + 2at ( x1 − x0 ) = 2at x1
x1 =
(b)
v12
(20) 2
=
2at (2)(1.11803)
x1 = 178.9 m �
Corresponding tangential acceleration.
at = 1.118 m/s 2 �
�
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201
200
PROBLEM 11.139
An outdoor track is 125 m in diameter. A runner increases her speed at a constant
rate from 4 to 7 m/s over a distance of 28 m. Determine the total acceleration of the
runner 2 s after she begins to increase her speed.
SOLUTION
We have uniformly accelerated motion
v22 = v12 + 2at ∆s12
Substituting
or
(7 m/s)2 = (4 m/s)2 + 2at (28 m)
at = 0.59 m/s2
Also
v = v1 + at t
At t = 2 s:
v = 7 m/s + (0.59 m/s2)(2 s) = 8.18 m/s
Now
an =
v2
ρ
At t = 2 s:
an =
(8.18 m/s)2
= 0.535 m/s2
125 m
Finally
a 2 = at2 + an2
At t = 2 s:
a 2 = 0.592 + 0.5352
a = 0.8 m/s2
or
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201
202
PROBLEM 11.140
At a given instant in an airplane race, airplane A is flying
horizontally in a straight line, and its speed is being increased
at the rate of 8 m/s 2 . Airplane B is flying at the same altitude
as airplane A and, as it rounds a pylon, is following a circular
path of 300-m radius. Knowing that at the given instant the
speed of B is being decreased at the rate of 3 m/s 2 , determine,
for the positions shown, (a) the velocity of B relative to A,
(b) the acceleration of B relative to A.
SOLUTION
First note
v A = 450 km/h vB = 540 km/h = 150 m/s
(a)
v B = v A + v B/A
We have
The graphical representation of this equation is then as shown.
We have
vB2 /A = 4502 + 5402 − 2(450)(540) cos 60°
or
vB/A = 501.10 km/h
and
or
540 501.10
=
sin α sin 60°
α = 68.9°
v B/A = 501 km/h
(b)
First note
Now
a A = 8 m/s 2 → (a B )t = 3 m/s 2
( aB ) n =
v 2B
ρB
=
60°
(150 m/s) 2
300 m
(a B ) n = 75 m/s 2
or
�
Then
68.9° �
30° ��
a B = (a B )t + (a B )n
= 3(− cos 60° i + sin 60° j) + 75(−cos 30° i − sin 30° j)
= −(66.452 m/s 2 )i − (34.902 m/s 2 ) j
Finally
or
a B = a A + a B/A
a B/A = ( −66.452i − 34.902 j) − (8i )
= −(74.452 m/s 2 )i − (34.902 m/s 2 ) j
a B/A = 82.2 m/s 2
or
25.1° �
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203
202
PROBLEM 11.141
A motorist traveling along a straight portion of a
highway is decreasing the speed of his automobile at
a constant rate before exiting from the highway onto a
circular exit ramp with a radius of 168-m. He continues
to decelerate at the same constant rate so that 10 s after
entering the ramp, his speed has decreased to 32 km/h,
a speed which he then maintains. Knowing that at this
constant speed the total acceleration of the automobile
is equal to one-quarter of its value prior to entering the
ramp, determine the maximum value of the total
acceleration of the automobile.
SOLUTION
First note
v10 = 32 km/h = 8.9 m/s
While the car is on the straight portion of the highway.
a = astraight = at
and for the circular exit ramp
a = at2 + an2
where
an =
v2
ρ
By observation, amax occurs when v is maximum, which is at t = 0 when the car first enters the ramp.
For uniformly decelerated motion
v = v0 + at t
and at t = 10 s:
v = constant
a=
Then
or
a = an =
2
v10
ρ
1
ast.
4
astraight = at
v 2 (8.9 m/s)2
1
at = 10 =
ρ
4
168 m
at = −1.9 m/s2
(The car is decelerating; hence the minus sign.)
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203
204
PROBLEM 11.141 (Continued)
Then at t = 10 s:
or
Then at t = 0:
8.9 m/s = v0 + (−1.9 m/s2)(10 s)
v0 = 27.9 m/s
amax =
at2
v2
+ 0
ρ
2
(27.9 m/s)2
= (−1.9 m/s ) +
168 m
2
1/2
2 2
amax = 5 m/s2
or
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205
204
PROBLEM 11.142
Racing cars A and B are traveling on circular portions
of a race track. At the instant shown, the speed of A
is decreasing at the rate of 7 m/s 2 , and the speed of B
is increasing at the rate of 2 m/s 2 . For the positions
shown, determine (a) the velocity of B relative to A,
(b) the acceleration of B relative to A.
SOLUTION
First note
v A = 162 km/h = 45 m/s
vB = 144 km/h = 40 m/s
(a)
We have
v B = v A + v B/A
The graphical representation of this equation is then as shown.
We have
vB2 /A = 1622 + 1442 − 2(162)(144)cos165°
or
vB/A = 303.39 km/h
and
or
144
303.39
=
sin α sin165°
α = 7.056°
v B/A = 303 km/h
(b)
First note
(a A )t = 7 m/s 2
60°
(a B )t = 2 m/s 2
45°
52.9° �
v2
ρ
Now
an =
Then
(a A ) n =
or
(a A ) n = 6.75 m/s 2
(45 m/s) 2
300 m
( aB ) n =
30°
(40 m/s) 2
250 m
(a B ) n = 6.40 m/s 2
45°
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205
206
PROBLEM 11.142 (Continued)
Noting that
We have
a = at + a n
a A = 7(cos 60°i − sin 60° j) + 6.75(− cos 30°i − sin 30° j)
= −(2.3457 m/s 2 )i − (9.4372 m/s 2 ) j
and
Finally
or
a B = 2(cos 45°i − sin 45° j) + 6.40(cos 45°i + sin 45° j)
= (5.9397 m/s 2 )i + (3.1113 m/s 2 ) j
a B = a A + a B/A
a B/A = (5.9397i + 3.1113j) − (−2.3457i − 9.4372 j)
= (8.2854 m/s 2 )i + (12.5485 m/s 2 ) j
a B/A = 15.04 m/s 2
or
56.6° ��
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207
206
PROBLEM 11.143
A golfer hits a golfball from Point A with an initial velocity of 50 m/s at
an angle of 25° with the horizontal. Determine the radius of curvature
of the trajectory described by the ball (a) at Point A, (b) at the highest
point of the trajectory.
SOLUTION
(a)
We have
or
(a A ) n =
ρA =
v 2A
ρA
(50 m/s)2
(9.81 m/s 2 ) cos 25°
ρ A = 281 m �
or
(b)
We have
( aB ) n =
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A cos 25°
Then
ρB =
[(50 m/s) cos 25°]2
9.81 m/s 2
ρ B = 209 m �
or
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207
208
PROBLEM 11.144
From a photograph of a homeowner using a snowblower, it
is determined that the radius of curvature of the trajectory
of the snow was 8.5 m as the snow left the discharge chute
at A. Determine (a) the discharge velocity v A of the snow,
(b) the radius of curvature of the trajectory at its maximum
height.
SOLUTION
(a)
We have
or
(a A ) n =
v 2A
ρA
v A2 = (9.81 cos 40°)(8.5 m)
= 63.8766 m 2 /s 2
v A = 7.99 m/s
or
(b)
We have
( aB ) n =
40° �
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A cos 40°
Then
ρB =
(63.8766 m 2 /s 2 ) cos 2 40°
9.81 m/s 2
ρ B = 3.82 m ��
or
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209
208
PROBLEM 11.145
A basketball is bounced on the ground at Point A and rebounds with a velocity v A of
magnitude 2 m/s as shown. Determine the radius of curvature of the trajectory
described by the ball (a) at Point A, (b) at the highest point of the trajectory.
SOLUTION
(a)
(a A ) n =
We have
ρA =
or
v 2A
ρA
(2 m/s)2
(9.81 m/s2) sin 15°
ρ A = 1.6 m
or
(b)
( aB ) n =
We have
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A sin 15°
ρB =
Then
[(2 m/s) sin 15°]2
9.81 m/s2
ρ B = 0.027 m
or
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209
210
PROBLEM 11.146
Coal is discharged from the tailgate A of a dump truck with
an initial velocity vA = 2 m/s
50° . Determine the radius
of curvature of the trajectory described by the coal (a) at
Point A, (b) at the point of the trajectory 1 m below Point A.
SOLUTION
(a)
(b)
(a A ) n =
We have
v 2A
ρA
(2 m/s)2
(9.81 m/s2) cos 50°
or
ρA =
or
ρ A = 0.634 m
Horizontal motion. (Uniform)
(vB ) x = (v A ) x = (2 m/s) cos 50° = 1.286 m/s
Vertical motion. (Uniformly accelerated motion)
We have
(
v 2y = (v A ) 2y − 2 g y − y 0A
)
( g = 9.81 m/s2)
where
(v A ) y = (2 m/s) sin 50° = 1.532 m/s
At Point B, y = −3 ft:
(vB ) 2y = (1.532 m/s)2 − 2(9.81 m/s2)(−1 m)
or
(vB ) y = 4.678 m/s
Then
and
or
Now
or
vB = (vB ) 2x + (vB ) 2y = (1.286)2 + (4.687)2 = 4.86 m/s
tan θ =
vy
vx
=
4.687
1.286
θ = 74.66°
( aB ) n =
ρB =
vB2
ρB
(4.86 m/s)2
(9.81 m/s2) cos 74.66°
ρ B = 9.1 m
or
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211
210
PROBLEM 11.147
A horizontal pipe discharges at Point A a stream of water into a
reservoir. Express the radius of curvature of the stream at Point B
in terms of the magnitudes of the velocities v A and v B .
SOLUTION
vB2
We have
( aB ) n =
where
(aB ) n = aB cosθ = g cosθ
ρB
Noting that the horizontal motion is uniform, we have
( vB ) x = v A
where
(vB ) x = vB cos θ
cos θ =
Then
ρB =
vA
vB
vB2
g
( )
vA
vB
ρB =
or
vB2
�
gv A
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212
PROBLEM 11.148
A child throws a ball from Point A with an initial velocity v A of
20 m/s at an angle of 25° with the horizontal. Determine the
velocity of the ball at the points of the trajectory described by the
ball where the radius of curvature is equal to three-quarters of its
value at A.
SOLUTION
Assume that Point B and C are the points of interest, where yB = yC and vB = vC .
(a A ) n =
Now
v 2A
ρA
or
ρA =
v 2A
g cos 25°
Then
ρB =
v A2
3
3
ρA =
4
4 g cos 25°
vB2
We have
( aB ) n =
where
(aB ) n = g cos θ
so that
or
ρB
v A2
vB2
3
=
4 g cos 25° g cos θ
vB2 =
3 cos θ 2
vA
4 cos 25°
(1)
Noting that the horizontal motion is uniform, we have
( v A ) x = ( vB ) x
where
Then
or
(v A ) x = v A cos 25° (vB ) x = vB cos θ
v A cos 25° = vB cos θ
cos θ =
vA
cos 25°
vB
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No part
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212
PROBLEM 11.148 (Continued)
Substituting for cos θ in Eq. (1), we have
or
vB2 =
� v A2
3 � vA
� cos 25° �
4 � vB
� cos 25°
vB3 =
3 3
vA
4
3
vB = 3 v A = 18.17 m/s
4
4
cos 25°
3
θ = ± 4.04°
cos θ = 3
v B = 18.17 m/s
v B = 18.17 m/s
and�
4.04° �
4.04° �
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213
214
PROBLEM 11.149
A projectile is fired from Point A with an
initial velocity v 0 . (a) Show that the radius
of curvature of the trajectory of the
projectile reaches its minimum value at
the highest Point B of the trajectory.
(b) Denoting by θ the angle formed by
the trajectory and the horizontal at a given
Point C, show that the radius of curvature
of the trajectory at C is ρ = ρ min /cos3 θ .
SOLUTION
For the arbitrary Point C, we have
(aC ) n =
ρC =
or
vC2
ρC
vC2
g cos θ
Noting that the horizontal motion is uniform, we have
(v A ) x = (vC ) x
(v A ) x = v0 cos α
where
v0 cos α = vC cos θ
Then
cos α
v0
cos θ
or
vC =
so that
1
ρC =
g cos θ
(a)
2
� cos α �
v 2 cos 2 α
v0 � = 0
�
g cos3 θ
� cos θ �
In the expression for ρC , v0 , α , and g are constants, so that ρC is minimum where cos θ is maximum.
By observation, this occurs at Point B where θ = 0.
ρ min = ρ B =
(b)
(vC ) x = vC cos θ
ρC =
1
cos3 θ
ρC =
ρ min
cos3 θ
v02 cos 2 α
g
� v02 cos 2 α
��
g
�
Q.E.D.
�
Q.E.D.
�
�
��
�
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214
PROBLEM 11.150
A projectile is fired from Point A with an initial
velocity v 0 which forms an angle α with the
horizontal. Express the radius of curvature of
the trajectory of the projectile at Point C in
terms of x, v0 , α , and g.
SOLUTION
We have
or
(aC ) n =
ρC =
vC2
ρC
vC2
g cos θ
Noting that the horizontal motion is uniform, we have
x = 0 + (v0 ) x t = (v0 cos α )t
(v A ) x = (vC ) x
where (v A ) x = v0 cos α
Then
or
so that
(vC ) x = vC cos θ
v0 cos α = vC cos θ
cos θ =
ρC =
and
(vC ) x = v0 cos α
(1)
v0
cos α
vC
vC3
g v0 cos α
For the uniformly accelerated vertical motion have
(vC ) y = (v0 ) y − gt = v0 sin α − gt
From above
Then
Now
x = (v0 cos α )t or t =
(vC ) y = v0 sin α − g
x
v0 cos α
x
v0 cos α
(2)
vC2 = (vC ) 2x + (vC ) 2y
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215
216
PROBLEM 11.150 (Continued)
Substituting for (vC ) x [Eq. (1)] and (vC ) y [Eq. (2)]
�
x �
vC2 = (v0 cos α )2 + � v0 sin α − g
�
v0 cos α �
�
� 2 gx tan α
g 2 x2 �
= v02 �1 −
+
�
�
v02
v04 cos 2α ��
�
or
vC3
=
v03
� 2 gx tan α
g 2 x2
+
��1 −
v02
v04 cos 2α
�
�
��
�
2
3/ 2
Finally, substituting into the expression for ρC , obtain
v02 � 2 gx tan α
g 2 x2
ρ=
+
��1 −
g cos α �
v02
v04 cos 2 α
�
��
�
3/ 2
�
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217
216
PROBLEM 11.151*
Determine the radius of curvature of the path described by the particle of Problem 11.95 when t = 0.
SOLUTION
We have
v=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R (sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
= R − ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t i
dt
(
)
(
)
+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k
or
a = ωn R [−(2 sin ωn t + ωn t cos ωn t )i
+ (2cos ωn t − ωn t sin ωn t ) k ]
Now
v 2 = R 2 (cos ωn t − ωn t sin ωn t )2 + c 2
+ R 2 (sin ωn t + ωn t cos ωn t )2
(
)
= R 2 1 + ωn2 t 2 + c 2
Then
and
Now
(
1/ 2
R 2ωn2 t
dv
=
1/ 2
dt � 2
R 1 + ωn2 t 2 + c 2 �
�
�
(
)
a 2 = at2 + an2
2
2
� dv � � v
= � � + ��
� dt � � ρ
At t = 0:
)
v = � R 2 1 + ωn2 t 2 + c 2 �
�
�
�
��
�
2
dv
=0
dt
a = ωn R(2 k ) or a = 2ωn R
v2 = R2 + c2
Then, with
we have
or
dv
= 0,
dt
a=
2ωn R =
v2
ρ
R 2 + c2
ρ
ρ=
or
R 2 + c2
�
2ωn R
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217
218
PROBLEM 11.152*
Determine the radius of curvature of the path described by the particle of Problem 11.96 when t = 0, A = 3,
and B = 1.
SOLUTION
With
A = 3,
B =1
We have
r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k
Now
v=
and
�
�
2
dv
a=
= 3(− sin t − sin t − t cos t )i + 3 � t + 1 − t ��
�
dt
t2 + 1
��
)
(
� 3t �
dr
j + (sin t + t cos t )k
= 3(cos t − t sin t )i + � 2
� t + 1 ��
dt
�
�
t
t
2
��
�
+1 �� j
�
��
+ (cos t + cos t − t sin t )k
= − 3(2sin t + t cos t )i + 3
1
j
(t + 1)1/2
2
+ (2 cos t − t sin t )k
v 2 = 9 (cos t − t sin t )2 + 9
Then
t2
+ (sin t + t cos t )2
t2 + 1
Expanding and simplifying yields
v 2 = t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t
v = [t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t ]1/ 2
Then
and
Now
dv 4t 3 + 38t + 8(−2 cos t sin t + 4t 3sin 2 t + 2t 4 sin t cos t ) − 8[(3t 2 + 1)sin 2t + 2(t 3 + t ) cos 2t ]
=
dt
2[t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t ) sin 2t ]1/ 2
2
a =
at2
+
an2
2
2
� dv � � v �
= � � + �� ��
� dt � � ρ �
2
PROPRIETARY
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218
PROBLEM 11.152* (Continued)
At t = 0:
a = 3j + 2k
or
a = 13 m/s2
dv
=0
dt
v 2 = 9 (m/s)2
Then, with
dv
= 0,
dt
we have
a=
or
ρ=
v2
ρ
9 m2/s2
13 m/s2
ρ = 2.50 m
or
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220
PROBLEM 11.153
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite.
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 160 km above the surface of the planet.
Venus: g = 8.53 m/s 2 , R = 6161 km.
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 6161 km
8.53 m/s 2
3600 s
×
3
1h
(6161 + 160) × 10 m
vcirc. = 25.8 × 103 km/h �
or
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220
PROBLEM 11.154
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite.
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 160 km above the surface of the planet.
Mars: g = 3.83 m/s 2 , R = 3332 km.
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 3332 km
3.83 m/s 2
3600 s
×
3
1h
(3332 + 160) × 10 m
vcirc. = 12.56 × 103 km/h �
or
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222
PROBLEM 11.155
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite.
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 160 km above the surface of the planet.
Jupiter: g = 26.0 m/s 2 , R = 69,893 km.
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 69,893
26.0 m/s 2
3600 s
×
3
1h
(69,893 + 160) × 10 m
vcirc. = 153.3 × 103 km/h �
or
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222
PROBLEM 11.156
Knowing that the diameter of the sun is 1382 × 103 km and that the acceleration of gravity at its surface is
270 m/s2, determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is
circular. (See information given in Problems 11.153–11.155.)
Earth: (vmean )orbit = 106500 km/h
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
r=g
R
v
2
where R =
1
d
2
For the given data
2
rearth = (270 m/s )
1
2
× 1382 × 103 km
106500 km/h
2
×
1 km
3600 s
=
1000 m
1h
2
rearth = 147.3 × 106 km
or
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224
PROBLEM 11.157
Knowing that the diameter of the sun is 1382 × 103 km and that the acceleration of gravity at its surface is
270 m/s2, determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is
circular. (See information given in Problems 11.153–11.155.)
Saturn: (vmean )orbit = 34500 km/h
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
r=g
R
v
2
where R =
1
d
2
For the given data
rsaturn = (270 m/s2)
1
2
× 1382 × 103 km
34500 km/h
2
×
1 km
3600 s
×
1000 m
1h
2
rsaturn = 1403.7 × 106 km
or
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224
PROBLEM 11.158
Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space
Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth.
(See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
v=R
g
r
where
r =R+h
The circumference s of the circular orbit is equal to
s = 2π r
Assuming that the speed of the telescope is constant, we have
s = vtorbit
Substituting for s and v
2π r = R
or
torbit =
=
g
torbit
r
2π r 3/2
R g
2π
[(6370 + 590) km]3/2
1h
×
2 1/2
−3
6370 km [9.81 × 10 km/s ]
3600 s
torbit = 1.606 h ��
or
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226
PROBLEM 11.159
A satellite is traveling in a circular orbit around Mars at an altitude of 300 km. After the altitude of the satellite
is adjusted, it is found that the time of one orbit has increased by 10 percent. Knowing that the radius of Mars
is 3310 km, determine the new altitude of the satellite. (See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
v=R
g
r
where
r =R+h
The circumference s of a circular orbit is equal to
s = 2π r
Assuming that the speed of the satellite in each orbit is constant, we have
s = vtorbit
Substituting for s and v
2π r = R
or
torbit =
=
Now
or
or
g
torbit
r
2π r 3/2
R g
2π ( R + h)3/2
R
g
(torbit ) 2 = 1.1(torbit )1
2π ( R + h2 )3/2
2π ( R + h1 )3/2
= 1.1
R
R
g
g
h2 = (1.1)2/3 ( R + h1 ) − R
= (1.1)2/3 (3310 + 300) km − (3310 km)
h2 = 537 km
or
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226
PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around
the earth at altitudes of 190 and 320 km, respectively. If at t = 0 the
satellites are aligned as shown and knowing that the radius of the earth is
R = 6370 km, determine when the satellites will next be radially aligned.
(See information given in Problems 11.153–11.155.)
SOLUTION
R2
r2
an = g
We have
Then
g
and an =
R2 v2
=
r
r2
or
v2
r
v=R
g
r
r =R+h
where
The circumference s of a circular orbit is
s = 2π r
equal to
Assuming that the speeds of the satellites are constant, we have
s = vT
Substituting for s and v
2π r = R
T=
or
Now
hB
g
T
r
2π r 3/ 2 2π ( R + h)3/ 2
=
R g
R
g
hA
(T ) B
(T ) A
Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B
completes N orbits, satellite A must complete ( N + 1) orbits.
Thus,
TC = N (T ) B = ( N + 1)(T ) A
or
N
2π ( R + hB )3/2
2π ( R + hA )3/ 2
= ( N + 1)
R
R
g
g
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PROBLEM 11.160 (Continued)
N=
or
=
Then
( R + hA )3/2
( R + hB )
(
1
6370 + 320
6370 +190
3/ 2
)
− ( R + hA )
3/2
−1
3/2
=
(
1
R + hB
R + hA
)
3/2
−1
= 33.476 orbits
2π ( R + hB )3/2
R
g
TC = N (T ) B = N
2π [(6370 + 320) km]
1h
= 33.476
×
1/2
6370 km 9.81 m/s2 × 1 km
3600s
3/2
(
1000 m
)
TC = 50.7 h
or
Alternative solution
From above, we have (T ) B (T ) A . Thus, when the satellites are next radially aligned, the angles θ A and θ B
swept out by radial lines drawn to the satellites must differ by 2π . That is,
θ A = θ B + 2π
For a circular orbit
s = rθ
From above
s = vt and v = R
Then
θ=
At time TC :
or
R g
( R + hA )
3/2
TC =
TC =
=
g
r
R g
R g
s vt 1
g
= =
R
t = 3/2 t =
t
r r r
r
( R + h)3/2
r
R g
( R + hB )3/ 2
TC + 2π
2π
R g
1
( R + hA )3/ 2
(
−
1
( R + hB )3/ 2
2π
1 km
(6370 km) 9.81 m/s2 × 1000
m
×
×
1
[(6370 + 190) km] 3/2
1
−
)
1/2
1
[(6370 + 320) km] 3/2
1h
3600 s
TC = 50.7 h
or
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228
PROBLEM 11.161
The path of a particle P is a limaçon. The motion of the particle is
defined by the relations r = b(2 + cos π t ) and θ = π t , where t and θ
are expressed in seconds and radians, respectively. Determine (a) the
velocity and the acceleration of the particle when t = 2 s, (b) the values
of θ for which the magnitude of the velocity is maximum.
SOLUTION
We have
r = b(2 + cos π t )
θ = πt
Then
r� = −π b sin π t
θ� = π
and
r�� = −π 2 b cos π t
� = 0
Now
v = r�er + rθ�eθ = −(π b sin π t ) er + π b(2 + cos π t ) eθ
and
a = (��
r − rθ� 2 ) e r + (rθ�� + 2r�θ�) eθ
= [ −π 2 b cos π t − π 2 b(2 + cos π t )]er
+ (0 − 2π 2 b sin π t )eθ
= −2π 2b[(1 + cos π t ) e r + (sin π t ) eθ ]
(a)
At t = 2 s:
v = −(0)er + π b(2 + 1)eθ
v = 3π beθ �
or
a = −2π 2b[(1 + 1) e r + (0) eθ ]
a = −4π 2 ber �
or
(b)
We have
v = π b (− sin π t )2 + (2 + cos π t ) 2
= π b 5 + 4cos π t θ = π t
= π b 5 + 4cos θ
By observation,
v = vmax
when cos θ = 1
θ = 2nπ , n = 0, 1, 2, . . . �
or
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229
230
PROBLEM 11.162
The two-dimensional motion of a particle is defined by the relation r = 2b cos ω t and θ = ω t where b and ω
are constant. Determine (a) the velocity and acceleration of the particle at any instant, (b) the radius of
curvature of its path. What conclusion can you draw regarding the path of the particle?
SOLUTION
r = 2b cos ω t
θ = ωt
r� = −2bω sin ω t
r�� = −2bω 2 cos ω t
(a)
Velocity.
θ� = ω
� = 0
vr = ��
r = −2bω sin ω t vθ = rθ� = 2bω cos ω t
v 2 = vr2 + vθ2 = (2bω )2 [(−sin ω t) 2 + (cos ω t) 2 ] = (2bω )2
Acceleration.
v = 2bω �
ar = ��
r − rθ� 2 = −2bω 2 cos ω t − (2b cos ω t)(ω ) 2
= −4bω 2 cos ω t
aθ = rθ�� + 2r�θ� = (2b cos ω t)(0) + 2( −2bω sin ω t)(ω )
= −4bω 2 sin ω t
a 2 = ar2 + aθ2 = (−4bω 2 ) 2 (cos 2 ω t + sin 2 ω t) = (4bω 2 ) 2
(b)
Since
Thus:
a = 4bω 2 �
v = 2bω = constant, at = 0
an = a = 4bω 2
an =
v2
ρ
; ρ=
v 2 (2bω ) 2
=
=b ρ =b
an
4bω 2
For the path, ρ = constant.
Thus, path is a circle.
�
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230
PROBLEM 11.163
The rotation of rod OA about O is defined by the relation θ = π (4t 2 − 8t ), where θ
and t are expressed in radians and seconds, respectively. Collar B slides along the
rod so that its distance from O is r = 10 + 6sin π t , where r and t are expressed in
meters and seconds, respectively. When t = 1 s, determine (a) the velocity of the
collar, (b) the total acceleration of the collar, (c) the acceleration of the collar
relative to the rod.
SOLUTION
We have
r = 10 + 6sin π t
θ = π (4t 2 − 8t )
Then
r = 6π cos π t
θ = θπ (t − 1)
and
r = −6π 2 sin π t
θ = θπ
At t = 1 s:
r = 10 m
θ = −4π rad
r = −6π m/s
θ =0
r =0
θ = 8π rad/s 2
(a)
We have
v B = re r + rθ eθ
v B = −(6π m/s) er
so that
(b)
We have
a B = ( r − rθ 2 )er + ( rθ + 2rθ )eθ
= (10)(8π )eθ
a B = (80π m/s2) eθ
or
(c)
We have
a B/OA = r
a B/OA = 0
so that
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232
PROBLEM 11.164
The oscillation of rod OA about O is defined by the relation θ = (2/π )(sin πτ),
where θ and t are expressed in radians and seconds, respectively. Collar B slides
along the rod so that its distance from O is r = 25/(t + 4), where r and t are
expressed in meters and seconds, respectively. When t = 1 s, determine (a) the
velocity of the collar, (b) the total acceleration of the collar, (c) the acceleration of
the collar relative to the rod.
SOLUTION
25
t+4
θ=
2
sin π t
We have
r=
Then
r =−
and
r=
At t = 1 s:
r =5m
θ =0
r = −1 m/s
θ = −2 rad/s
r = 0.4 m/s2
θ =0
(a)
We have
25
(t + 4)2
π
θ = 2 cos π t
50
(t + 4) 2
θ = −2π sin π t
v B = re r + rθ eθ = (−1)e r + (5)(−2)eθ
v A = −(1 m/s)er − (10 m/s)eθ
or
(b)
We have
a B = ( r − rθ 2 )er + ( rθ + 2rθ )eθ
= [0.4 − (5)(2) 2 ]e r + [0 + 2(−1)(−2)]eθ
a B = −(19.6 m/s2)er + (4 m/s2)eθ
or
(c)
We have
a B/OA = r
a B/OA = (0.4 m/s2 ) er
so that
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232
PROBLEM 11.165
The path of particle P is the ellipse defined by the relations
r = 2/(2 − cos π t ) and θ = π t , where r is expressed in meters, t is
in seconds, and θ is in radians. Determine the velocity and the
acceleration of the particle when (a) t = 0, (b) t = 0.5 s.
SOLUTION
We have
r=
2
2 − cos π t
θ = πt
Then
r� =
−2π sin π t
(2 − cos π t )2
θ� = π
and
r�� = −2π
π cos π t (2 − cos π t ) − sin π t (2π sin π t )
(2 − cos π t )3
= −2π 2
(a)
At t = 0:
Now
2cos π t − 1 − sin 2 π t
(2 − cos π t )3
r =2m
θ =0
r� = 0
θ� = π rad/s
r�� = −2π 2 m/s 2
� = 0
v = r�er + rθ�eθ = (2)(π )eθ
v = (2π m/s)eθ �
or
and
� = 0
a = (r�� − rθ� 2 )e r + (rθ�� + 2r�θ�)eθ
= [ −2π 2 − (2)(π )2 ]er
a = −(4π 2 m/s 2 )er �
or
(b)
At t = 0.5 s:
θ=
r =1 m
r� =
−2π
π
= − m/s
2
2
(2)
r�� = −2π 2
−1 − 1 π 2
=
m/s 2
3
2
(2)
π
2
rad
θ� = π rad/s
� = 0
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234
PROBLEM 11.165 (Continued)
Now
� π�
v = r�er + rθ�eθ = � − � e r + (1)(π )eθ
� 2�
�π
�
v = − � m/s � er + (π m/s)eθ �
2
�
�
or
and
a = (r�� − rθ� 2 )e r + (rθ�� + 2r�θ�)eθ
�π 2
�
� � π� �
=�
− (1)(π ) 2 � e r + � 2 � − � (π ) � eθ
� � 2� �
� 2
�
�π2
�
a = − ��
m/s 2 �� er − (π 2 m/s 2 )eθ �
� 2
�
or
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PROBLEM 11.166
The two-dimensional motion of a particle is defined by the relations r = 2a cos θ and θ = bt 2 /2, where a and b
are constants. Determine (a) the magnitudes of the velocity and acceleration at any instant, (b) the radius of
curvature of the path. What conclusion can you draw regarding the path of the particle?
SOLUTION
(a)
1
2
We have
r = 2a cos θ
θ = bt 2
Then
r� = −2aθ sin θ
θ� = bt
and
r�� = −2a (θ�� sin θ + θ� 2 + cos θ )
� = b
Substituting for θ� and θ��
r� = −2abt sin θ
��
r = −2ab(sin θ + bt 2 cos θ )
Now
Then
vθ = rθ� = 2abt cos θ
vr = r� = −2abt sin θ
v = vr2 + vθ2 = 2abt[( − sin θ ) 2 + (cos θ ) 2 ]1/2
v = 2abt �
or
Also
ar = ��
r − rθ� 2 = −2ab(sin θ + bt 2 cos θ ) − 2ab 2 t 2 cos θ
= −2ab(sin θ + 2bt 2 cos θ )
and
aθ = rθ�� + 2r�θ� = 2ab cos θ − 4ab 2 t 2 sin θ
= −2ab(cos θ − 2bt 2 sin θ )
Then
a = ar2 + aB2 = 2ab[(sin θ + 2bt 2 cos θ ) 2
+ (cos θ − 2bt 2 sin θ ) 2 ]1/2
a = 2ab 1 + 4b 2 t 4 �
or
(b)
2
at2
an2
2
2
� dv � � v �
= � � + �� ��
� dt � � ρ �
Now
a =
Then
dv d
= (2abt ) = 2ab
dt dt
+
2
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235
236
PROBLEM 11.166 (Continued)
so that
or
(
2ab 1 + 4b 2 t 4
)
2
= (2ab) 2 + an2
4a 2b 2 (1 + 4b 2t 4 ) = 4a 2 b 2 + an2
or
an = 4ab 2 t 2
Finally
an =
v2
ρ
�ρ=
(2abt ) 2
4ab 2 t 2
ρ =a �
or
Since the radius of curvature is a constant, the path is a circle of radius a.
��
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236
PROBLEM 11.167
To study the performance of a race car, a high-speed motionpicture camera is positioned at Point A. The camera is mounted
on a mechanism which permits it to record the motion of the
car as the car travels on straightway BC. Determine the speed
of the car in terms of b, θ , and θ�.
SOLUTION
We have
r=
b
cos θ
Then
r� =
bθ� sin θ
cos 2 θ
We have
v 2 = vr2 + vθ2 = (r�)2 + (rθ�) 2
2
or
� bθ sin θ � � bθ �2
= ��
�� + �
�
2
� cos θ � � cos θ �
� b2θ� 2
b2θ� 2 � sin 2θ
1
=
+
�
�� =
4
cos 2θ �� cos 2θ
� cos θ
bθ�
v=±
cos 2θ
For the position of the car shown, θ is decreasing; thus, the negative root is chosen.
v=
bθ�
�
cos 2θ
v=−
bθ�
�
cos 2θ
Alternative solution.
From the diagram
or
r� = −v sin θ
bθ� sin θ
= −v sin θ
cos 2θ
or
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237
238
PROBLEM 11.168
Determine the magnitude of the acceleration of the race car of
Problem 11.167 in terms of b, θ , θ , and θ .
SOLUTION
We have
r=
b
cos θ
Then
r=
bθ sin θ
cos 2 θ
For rectilinear motion
a=
dv
dt
From the solution to Problem 11.167
v=−
Then
a=
bθ
cos 2θ
d
bθ
−
dt
cos 2θ
= −b
θ cos 2θ − θ (−2θ cos θ sin θ )
cos 4θ
a=−
or
b
(θ + 2θ 2 tan θ )
cos 2θ
Alternative solution
b
cos θ
From above
r=
Then
r =b
=b
r=
bθ sin θ
cos 2 θ
(θ sin θ + θ 2 cos θ )(cos 2θ ) − (θ sin θ )(−2θ cos θ sin θ )
cos 4θ
θ sin θ θ 2 (1 + sin 2 θ )
+
cos 2θ
cos3θ
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238
PROBLEM 11.168 (Continued)
Now
where
a 2 = ar2 + aθ2
� θ�� sin θ θ� 2 (1 + sin 2 θ ) � bθ� 2
ar = ��
r − rθ� 2 = b �
+
�−
2
cos 2θ
� cos θ
� cos θ
2θ� 2 sin 2 θ �
b � ��
θ
θ
sin
=
+
�
�
cos θ ��
cos 2θ ��
ar =
and
b�
bθ� 2 sin θ
aθ = rθ�� + 2r�θ� =
+2
cos θ
cos 2θ
=
Then
b sin θ ��
(θ + 2θ� 2 tan θ )
cos 2θ
b cos θ ��
(θ + 2θ� tan θ )
cos 2θ
a=±
b
(θ�� + 2θ� 2 tan θ )[(sin θ ) 2 + (cos θ )2 ]1/ 2
2
cos θ
For the position of the car shown, � is negative; for a to be positive, the negative root is chosen.
a=−
b
(θ�� + 2θ� 2 tan θ ) �
cos 2 θ
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239
240
PROBLEM 11.169
After taking off, a helicopter climbs in
a straight line at a constant angle β . Its
flight is tracked by radar from Point A.
Determine the speed of the helicopter
in terms of d, β , θ , and θ�.
SOLUTION
From the diagram
r
d
=
sin (180° − β ) sin ( β − θ )
or
d sin β = r (sin β cos θ − cos β sin θ )
tan β
tan β cos θ − sin θ
or
r=d
Then
r� = d tan β
−(− tan β sin θ − cos θ ) �
θ
(tan β cos θ − sin θ )2
= dθ tan β
tan β sin θ + cos θ
(tan β cos θ − sin θ ) 2
From the diagram
vr = v cos ( β − θ )
where
vr = r�
Then
tan β sin θ + cos θ
= v(cos β cos θ + sin β sin θ )
dθ� tan β
(tan β cos θ − sin θ )2
= v cos β (tan β sin θ + cos θ )
v=
or
dθ� tan β sec β
�
(tan β cosθ − sin θ )2
Alternative solution.
We have
v 2 = vr2 + vθ2 = (r�) 2 + ( rθ�) 2
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240
PROBLEM 11.169 (Continued)
Using the expressions for r and r� from above
�
tan β sin θ + cos θ �
v = � dθ� tan β
�
(tan β cos θ − sin θ ) 2 �
�
2
1/ 2
or
� (tan β sin θ + cos θ )2
�
dθ� tan β
v=±
+ 1�
�
2
(tan β cos θ − sin θ ) � (tan β cos θ − sin θ )
�
1/ 2
�
�
dθ� tan β
tan 2 β + 1
=±
�
2�
(tan β cos θ − sin θ ) � (tan β cos θ − sin θ ) �
Note that as θ increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v=
dθ� tan β sec β
�
(tan β cos θ − sin θ )2
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241
242
PROBLEM 11.170*
Pin P is attached to BC and slides freely in the slot of OA.
Determine the rate of change θ� of the angle θ , knowing that
BC moves at a constant speed v0 . Express your answer in
terms of v0 , h, β , and θ .
SOLUTION
From the diagram
r
h
=
sin (90° − β ) sin ( β + θ )
or
r (sin β cos θ + cos β sin θ ) = h cos β
r=
or
Also,
Then
h
tan β cos θ + sin θ
vθ = v0 sin (β + θ ) where vθ = rθ�
hθ�
= v0 (sin β cos θ + cos β sin θ )
tan β cos θ + sin θ
θ� =
or
v0 cos β
(tan β cos θ + sin θ ) 2 �
h
Alternative solution.
h
tan β cos θ + sin θ
From above
r=
Then
r� = h
tan β sin θ − cos θ �
θ
(tan β cos θ + sin θ )2
Now
v02 = vr2 + vθ2 = (r�) 2 + ( rθ�) 2
or
v02
�
tan β sin θ − cos θ �
= � hθ�
2�
� (tan β cos θ + sin θ ) �
�
�
hθ�
+�
�
� tan β cos β + sin θ �
2
2
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242
PROBLEM 11.170* (Continued)
or
v0 = ±
hθ�
tan β cos θ + sin θ
1/2
� (tan β sin θ − cos θ ) 2
�
+�
+
1
�
2
� (tan β cos θ + sin θ )
�
hθ�
=±
tan β cos θ + sin θ
1/2
�
�
tan 2 β + 1
�
2�
� (tan β cos θ + sin θ ) �
Note that as θ increases, member BC moves in the indicated direction. Thus, the positive root is chosen.
θ� =
v0 cos β
(tan β cos θ + sin θ ) 2 �
h
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243
244
PROBLEM 11.171
For the race car of Problem 11.167, it was found that it
took 0.5 s for the car to travel from the position θ = 60° to
the position θ = 35°. Knowing that b = 25 m, determine
the average speed of the car during the 0.5-s interval.
SOLUTION
From the diagram:
∆r12 = 25 tan 60° − 25 tan 35°
= 25.796 m
Now
vave =
=
∆r12
∆t12
25.796 m
0.5 s
= 51.592 m/s
vave = 185.7 km/h �
or
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244
PROBLEM 11.172
For the helicopter of Problem 11.169, it
was found that when the helicopter was at
B, the distance and the angle of elevation
of the helicopter were r = 900 m and
θ = 20°, respectively. Four seconds later,
the radar station sighted the helicopter at
r = 996 m and θ = 23.1°. Determine the
average speed and the angle of climb β of
the helicopter during the 4-s interval.
SOLUTION
We have
θ0 = 20°
θ 4 = 23.1°
r0 = 900 m
r4 = 996 m
From the diagram:
∆r 2 = 9002 + 9962
−2(900)(996) cos (23.1° − 20°)
or
∆r = 108.8 m
Now
vave =
∆r
∆t
108.8 m
=
4s
= 27.2 m/s
vave = 97.9 km/h
or
Also,
or
∆r cos β = r4 cos θ 4 − r0 cos θ0
cos β =
996 cos 23.1° − 900 cos 20°
108.8
β = 49.7°
or
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245
246
PROBLEM 11.173
A particle moves along the spiral shown; determine the magnitude of the
velocity of the particle in terms of b, θ , and θ�.
SOLUTION
Hyperbolic spiral.
r=
b
θ
dr
b dθ
b
=− 2
= − 2 θ�
dt
θ dt
θ
b
b
vr = r� = − 2 θ� vθ = rθ� = θ�
r� =
θ
θ
2
� 1 � �1�
v = vr2 + vθ2 = bθ� � − 2 � + � �
� θ � �θ �
bθ�
= 2 1+θ 2
2
θ
v=
b
θ
2
1 + θ 2 θ� �
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246
PROBLEM 11.174
A particle moves along the spiral shown; determine the magnitude of the
velocity of the particle in terms of b, θ , and θ�.
SOLUTION
Logarithmic spiral.
r = ebθ
r� =
dr
dθ
= bebθ
= bebθ θ�
dt
dt
vr = r� = bebθ θ� vθ = rθ� = ebθ θ�
v = vr2 + vθ2 = ebθ θ� b 2 + 1
v = ebθ 1 + b 2 θ� �
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248
PROBLEM 11.175
A particle moves along the spiral shown. Knowing that θ� is constant and
denoting this constant by ω , determine the magnitude of the acceleration of
the particle in terms of b, θ , and ω.
SOLUTION
b
Hyperbolic spiral.
r=
From Problem 11.173
r� = −
θ
r�� = −
b
θ2
θ�
b �� 2b � 2
θ + 3θ
2
θ
θ
b
b
2b
ar = r�� − rθ� 2 = − 2 θ�� + 3 θ� 2 − θ� 2
θ
θ
θ
b
b
b
� b �
aθ = rθ�� + 2r�θ� = θ�� + 2 � − 2 θ� �θ� = θ�� − 2 2 θ� 2
θ
θ
θ
� θ �
Since θ� = ω = constant, θ�� = 0, and we write:
b 2 bω 2
2
−
ω
ω = 3 (2 − θ 2 )
θ
θ3
θ
2
b
bω
aθ = −2 2 ω 2 = − 3 (2θ )
θ
θ
ar = +
2b
a = ar2 + aθ2 =
bω 2
θ
3
(2 − θ 2 ) 2 + (2θ ) 2 =
bω 2
θ
3
4 − 4θ 2 + θ 4 + 4θ 2
a=
bω 2
θ
3
4 +θ 4 �
PROPRIETARY
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249
248
PROBLEM 11.176
A particle moves along the spiral shown. Knowing that θ� is constant and
denoting this constant by ω , determine the magnitude of the acceleration of
the particle in terms of b, θ , and ω.
SOLUTION
Logarithmic spiral.
r = ebθ
dr
= bebθ θ�
dt
��
r = bebθ θ�� + b 2 ebθ θ� 2 = bebθ (θ�� + bθ� 2 )
r� =
ar = ��
r − rθ� 2 = bebθ (θ�� + bθ� 2 ) − ebθ θ� 2
a = rθ�� + 2r�θ� = ebθ θ�� + 2(bebθ θ�)θ�
θ
Since θ� = ω = constant, θ�� = θ , and we write
ar = bebθ (bω 2 ) − ebθ ω 2 = ebθ (b 2 − 1)ω 2
aθ = 2bebθ w2
a = ar2 + aθ2 = ebθ ω 2 (b2 − 1)2 + (2b) 2
= ebθ ω 2 b4 − 2b 2 + 1 + 4b 2 = ebθ ω 2 b 4 + 2b 2 + 1
= ebθ ω 2 (b2 + 1) 2 = ebθ ω 2 (b2 + 1)
a = (1 + b 2 )ω 2 ebθ �
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249
250
PROBLEM 11.177
Show that r� = hφ� sin θ knowing that at the instant shown,
step AB of the step exerciser is rotating counterclockwise at
a constant rate φ�.
SOLUTION
From the diagram
r 2 = d 2 + h 2 − 2dh cos φ
Then
Now
or
2rr� = 2dhφ� sin φ
r
d
=
sin φ sin θ
r=
d sin φ
sin θ
Substituting for r in the expression for r�
� d sin φ �
�
� sin θ � r� = dhφ sin φ
�
�
or
r� = hφ� sin θ
Q.E.D.
�
Alternative solution.
First note
α = 180° − (φ + θ )
Now
v = v r + vθ = r�e r + rθ�eθ
With B as the origin
vP = dφ�
( d = constant � d� = 0)
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250
PROBLEM 11.177 (Continued)
With O as the origin
(vP )r = r�
where
(vP )r = vP sin α
Then
Now
or
r� = dφ� sin α
h
d
=
sin α sin θ
d sin α = h sin θ
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251
252
PROBLEM 11.178
The motion of a particle on the surface of a right circular
cylinder is defined by the relations R = A, θ = 2π t , and
z = At 2 /4, where A is a constant. Determine the magnitudes
of the velocity and acceleration of the particle at any time t.
SOLUTION
We have
R=A
θ = 2π t
z=
1 2
At
4
Then
R� = 0
θ� = 2π
z� =
1
At
2
and
�� = 0
R
� = 0
��
z=
1
A
2
Now
v 2 = vR2 + vθ2 + vz2
0
= (�
R� ) 2 + ( Rθ�) 2 + ( z� ) 2
�1 �
= 0 + ( A + 2π ) 2 + � At �
�2 �
1 �
�
= A2 � 4π 2 + t 2 �
4 �
�
2
v=
or
and
1
A 16π 2 + t 2 �
2
a 2 = aR2 + aθ2 + az2
0
0
�� 0 − Rθ� 2 ) 2 + ( R�
�
= (R
� + 2�
R� θ ) 2 + ( ��
z )2
�1 �
= [ − A(2π ) 2 ]2 + 0 + � A �
�2 �
1�
�
= A2 �16π 4 + �
4�
�
2
a=
or
1
A 64π 4 + 1 �
2
PROPRIETARY
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253
252
PROBLEM 11.179
The three-dimensional motion of a particle is defined by the cylindrical coordinates (see Figure 11.26)
R = A/(t + 1), θ = Bt , and z = Ct/(t + 1). Determine the magnitudes of the velocity and acceleration when
(a) t = 0, (b) t = ∞.
SOLUTION
A
,
t +1
We have
R=
Then
R� = −
�� =
R
and
θ = Bt ,
A
,
(t + 1) 2
2A
(t + 1)3
z=
θ� = B
� = 0
Ct
t +1
z� = C
(t + 1) − t
(t + 1)2
=
C
(t + 1) 2
��
z=−
zC
(t + 1)3
Now
v 2 = (vR ) 2 + (vθ )2 + (vz ) 2
= ( R� ) 2 + ( Rθ�) 2 + ( z� ) 2
and
a 2 = ( aR )2 + (aθ ) 2 + (az ) 2
0
�� + 2 R�θ�)2 + ( ��
�� − Rθ� 2 ) 2 + ( Rθ�
z )2
= (R
(a)
At t = 0:
R= A
R = −A
θ� = B
�� = 2 A
R
Then
z� = C
��
z = −2C
v 2 = (− A) 2 + ( AB) 2 + (C ) 2
v = (1 + B 2 ) A2 + C 2 �
or
and
a 2 = (2 A − AB 2 ) 2 + [ z (− A)( B)]2 + (−2C ) 2
�� 1 � 2
C2 �
= 4 A2 ��1 − B 2 � + B 2 + 2 �
A ��
��� 2 �
�� 1 �
�
= 4 � � 1 + B 4 � A2 + C 2 �
�� 4 �
�
� 1 �
a = 2 � 1 + B 4 � A2 + C 2 �
� 4 �
or
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253
254
PROBLEM 11.179 (Continued)
(b)
As t → ∞ :
R=0
R� = 0
�� = 0
R
θ� = B
z� = 0
��
z=0
v=0 �
a=0 �
and
PROPRIETARY
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255
254
PROBLEM 11.180*
For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis.
SOLUTION
First note that the vectors v and a lie in the osculating plane.
Now
r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k
Then
v=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
dt
(
)
= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i
(
)
+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k
= ωn R[−(2sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t )k ]
It then follows that the vector ( v × a) is perpendicular to the osculating plane.
i
j
k
( v × a) = ωn R R(cos ωn t − ωn t sin ωn t ) c R(sin ωn t + ωn t cos ωn t )
−(2sin ωn t + ωn t cos ωn t ) 0 (2cos ωn t − ωn t sin ωn t )
= ωn R{c(2 cos ωn t − ωn t sin ωn t )i + R[−(sin ωn t + ωn t cos ωn t )(2sin ωn t + ωn t cos ωn t )
− (cos ωn t − ωn t sin ωn t )(2 cos ωn t − ωn t sin ωn t )] j + c(2sin ωn t + ωn t cos ωn t )k
(
)
= ωn R �c(2 cos ωn t − ωn t sin ωn t )i − R 2 + ωn2 t 2 j + c(2sin ωn t + ωn t cos ωn t )k �
�
�
PROPRIETARY MATERIAL.
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255
256
PROBLEM 11.180* (Continued)
The angle α formed by the vector ( v × a) and the y axis is found from
cos α =
Where
( v × a) ⋅ j
| ( v × a) || j |
|j| =1
(
( v × a) ⋅ j = −ωn R 2 2 + ωn2 t 2
)
(
|( v × a) | = ωn R ��c 2 (2 cos ωn t − ωn t sin ωn t ) 2 + R 2 2 + ωn2 t 2
�
)
2
1/ 2
+ c 2 (2sin ωn t + ωn t cos ωn t )2 ��
(
)
2 �1/2
(
) ��
= ωn R ��c 2 4 + ωn2 t 2 + R 2 2 + ωn2t 2
�
Then
(
)
ω R ��c ( 4 + ω t ) + R ( 2 + ω t ) ��
�
�
−R ( 2 + ω t )
=
�c 4 + ω t + R 2 + ω t �
) (
) ��
�� (
−ωn R 2 2 + ωn2t 2
cos α =
2
n
2 2
n
2
2 2
n
2
1/2
2 2
n
2
2 2
n
2
2 2
n
2 1/2
The angle β that the osculating plane forms with y axis (see the above diagram) is equal to
β = α − 90°
Then
cos α = cos ( β + 90°) = −sin β
− sin β =
Then
tan β =
(
)
) + R ( 2 + ω t ) ���
− R 2 + ωn2t 2
(
�c 2 4 + ω 2t 2
n
��
(
R 2 + ωn2 t 2
2
2 2
n
2
1/ 2
)
c 4 + ωn2 t 2
(
� R 2 + ωn2t 2
β = tan �
� c 4 + ω 2t 2
n
�
−1
or
) �� �
�
�
PROPRIETARY
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257
256
PROBLEM 11.181*
Determine the direction of the binormal of the path described by the particle of Problem 11.96 when
(a) t = 0, (b) t = π /2 s.
SOLUTION
Given:
)
(
r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k
r − m, t − 5; A = 3, B − 1
First note that eb is given by
eb =
v×a
|v ×a |
)
(
Now
r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k
Then
v=
dr
dt
3t
= 3(cos t − t sin t )i +
j + (sin t + t cos t )k
(
t
)
2
dv
t 2 +1
j
a=
= 3( − sin t − sin t − t cos t )i + 3 t + 12− t
dt
t +1
+ (cos t + cos t − t sin t )k
3
j + (2 cos t − t sin t )k
= −3(2sin t + t cos t )i + 2
(t + 1)3/2
and
(a)
t2 +1
At t = 0:
v = (3 m/s)i
a = (3 m/s2)j + (2 m/s2)k
Then
and
Then
v × a = 3i × (3j + 2k )
= 3(−2 j + 3k )
| v × a | = 3 (−2) 2 + (3) 2 = 3 13
eb =
3( −2 j + 3k )
3 13
cos θ x = 0
or
=
cos θ y = −
θ x = 90°
1
13
2
(−2 j + 3k )
13
θ y = 123.7°
cos θ z =
3
13
θ z = 33.7°
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258
PROBLEM 11.181* (Continued)
(b)
At t =
π
2
s:
v=−
3π
m/s i +
2
a = −(6 m/s2) i +
Then
i
3π
v×a = −
2
π2 +4
m/s j + (1 m/s)k
π
24
m/s2 j −
m/s2 k
3/2
2
(π + 4)
2
j
3π
(π 2 + 4)1/ 2
24
2
(π + 4)3/ 2
−6
=−
3π
k
1
−
π
2
3π 2
24
3π 2
i
j
+
−
6
+
4
2(π 2 + 4)1/2 (π 2 + 4)3/ 2
+ −
36π
18π
k
+ 2
3/ 2
(π + 4)
(π + 4)1/ 2
2
= −4.43984i − 13.40220 + 12.99459k
and
Then
or
| v × a | = [(−4.43984) 2 + ( −13.40220)2 + (12.99459)2 ]1/ 2
= 19.18829
1
(−4.43984i − 13.40220 j + 12.99459k )
19.1829
4.43984
13.40220
12.99459
cos θ x = −
cos θ y = −
cos θ z =
19.18829
19.18829
19.18829
eb =
θ x = 103.4°
θ y = 134.3°
θ z = 47.4°
PROPRIETARY
PROPRIETARY MATERIAL.
MATERIAL. ©
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258
PROBLEM 11.182
The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x and t are expressed in
meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total
distance traveled when the acceleration is zero.
SOLUTION
x = 2t 3 − 15t 2 + 24t + 4
dx
= 6t 2 − 30t + 24
dt
dv
= 12t − 30
a=
dt
v=
so
(a)
0 = 6t 2 − 30t + 24 = 6 (t 2 − 5t + 4)
Times when v = 0.
(t − 4)(t − 1) = 0
(b)
t = 1.00 s, t = 4.00 s
�
Position and distance traveled when a = 0.
a = 12t − 30 = 0
t = 2.5 s
x2 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4
so
x = 1.50 m �
Final position
For
0 � t � 1 s, v � 0.
For
1 s � t � 2.5 s, v � 0.
At
t = 0,
At
t = 1 s, x1 = (2)(1)3 − (15)(1) 2 + (24)(1) + 4 = 15 m
x0 = 4 m.
Distance traveled over interval: x1 − x0 = 11 m
For
1 s � t � 2.5 s,
v�0
Distance traveled over interval
| x2 − x1 | = |1.5 − 15 | = 13.5 m
Total distance:
d = 11 + 13.5
d = 24.5 m �
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259
260
PROBLEM 11.183
The acceleration of a particle is defined by the relation a = −60 x −1.5, where a and x are expressed in m/s 2
and meters, respectively. Knowing that the particle starts with no initial velocity at x = 4 m, determine the
velocity of the particle when (a) x = 2 m, (b) x = 1 m, (c) x = 100 mm.
SOLUTION
v
We have
When x = 4 m, v = 0:
0
vdv =
�
x
4
(−60 x −1.5 )dx
1 2
v = 120[ x −0.5 ]4x
2
or
� 1
1�
v 2 = 240 �
− �
� x 2�
or
(a)
�
v
dv
= a = −60 x −1.5
dx
When x = 2 m:
� 1
1�
v 2 = 240 �
− �
� 2 2�
v = −7.05 m/s �
or
(b)
When x = 1 m:
� 1�
v 2 = 240 �1 − �
� 2�
v = −10.95 m/s �
or
(c)
When x = 0.1 m:
� 1
1�
v 2 = 240 �
− �
� 0.1 2 �
v = −25.3 m/s �
or
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260
PROBLEM 11.184
A projectile enters a resisting medium at x = 0 with an initial velocity
v0 = 270 m/s and travels 100 mm before coming to rest. Assuming that the
velocity of the projectile is defined by the relation v = v0 − kx, where v is
expressed in m/s and x is in meters, determine (a) the initial acceleration
of the projectile, (b) the time required for the projectile to penetrate 97.5
mm into the resisting medium.
SOLUTION
First note
When x = 0.1 m, v = 0:
0 = (270 m/s) − k (0.1 m)
or
k = 2700
(a)
1
s
We have
v = v0 − kx
Then
a=
or
a = −k (v0 − kx)
At t = 0:
1
a = 2700 (270 m/s − 0)
s
dv d
= (v0 − kx) = −kv
dt dt
a0 = −729 × 103 m/s2
or
(b)
dx
= v = v0 − kx
dt
We have
At t = 0, x = 0:
or
x
0
dx
=
v0 − kx
t
0
dt
1
− [ln(v0 − kx)]0x = t
k
or
t=
v0
1
1
1
= ln
ln
k
v0 − kx
k
1 − vk x
0
When x = 0.0975 m:
t=
1
ln
1−
2700 1s
2700 1/s
270 m/s
1
(0.0975 m)
t = 1.366 × 10−3 s
or
PROPRIETARY MATERIAL.
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261
262
PROBLEM 11.185
A freight elevator moving upward with a constant velocity of 1.8 m/s passes a passenger elevator which is
stopped. Four seconds later, the passenger elevator starts upward with a constant acceleration of 0.72 m/s2.
Determine (a) when and where the elevators will be at the same height, (b) the speed of the passenger elevator
at that time.
SOLUTION
(a)
For t
t
0:
y F = 0 + vF t
4 s:
yP = 0 + 0(t − 4) +
1
aP (t − 4) 2
2
yF = yP
When
(1.8 m/s) t =
1
(0.72 m/s2)(t − 4)2
2
Expanding and simplifying
t 2 − 13t + 16 = 0
t = 1.3765 s and t = 11.6235 s
Solving
Most require t
At t = 11.6235 s:
t = 11.62 s
4s
yF = (1.8 m/s)(11.6235 s)
yF = yP = 20.9 m
or
(b)
For t
4 s:
At t = 11.6235 s:
vP = 0 + aP (t − 4)
vP = (0.72 m/s2)(11.6235 − 4)s
vP = 5.5 m/s
or
PROPRIETARY
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263
262
PROBLEM 11.186
Block C starts from rest at t = 0 and moves upward with a constant acceleration
of 25 mm/s2. Knowing that block A moves downward with a constant velocity
of 75 mm/s, determine (a) the time for which the velocity of block B is zero,
(b) the corresponding position of block B.
SOLUTION
The cable lengths are constant.
L1 = 2 yC + 2 yD + constant
L2 = y A + yB + ( yB − yD ) + constant
Eliminate yD .
L1 + 2 L2 = 2 yC + 2 yD + 2 y A + 2 yB + 2( yB − yD ) + constant
2(yC + y A + 2yB ) = constant
Differentiate to obtain relationships for velocities and accelerations, positive downward.
vC + v A + 2vB = 0
(1)
aC + a A + 2aB = 0
(2)
Initial velocities.
(vC )0 = 0, (v A )0 = 75 mm/s
From (1),
1
(vB )0 = − [(vC )0 + (vA )0 ] = −37.5 mm/s
2
Accelerations.
aC = −25 mm/s a A = 0
From (2),
1
aB = − ( aC + a A ) = 12.5 mm/s
2
Motion of block B.
vB = (vB )0 + aB t = −37.5 + 12.5t
∆ y B = ( vB ) 0 t +
(a)
Time at vB = 0.
(b)
Corresponding position.
1
aB t 2 = −37.5t + 6.25t 2
2
−37.5 + 12.5t = 0
t = 3.00 s �
∆yB = (−37.5)(3.00) + (6.25)(3.00)2
∆yB = −56.25 mm �
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263
264
PROBLEM 11.187
The three blocks shown move with constant velocities. Find the velocity of each
block, knowing that the relative velocity of A with respect to C is 300 mm/s upward
and that the relative velocity of B with respect to A is 200 mm/s downward.
SOLUTION
From the diagram
Cable 1:
y A + yD = constant
Then
v A + vD = 0
Cable 2:
(1)
( yB − yD ) + ( yC − yD ) = constant
vB + vC − 2vD = 0
Then
(2)
Combining Eqs. (1) and (2) to eliminate vD
2v A + vB + vC = 0
(3)
Now
v A/C = v A − vC = −300 mm/s
(4)
and
vB/A = vB − v A = 200 mm/s
(5)
Then
(3) + (4) − (5) �
(2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200)
v A = 125 mm/s ↑ �
or
vB − (−125) = 200
and using Eq. (5)
v B = 75 mm/s ↓ �
or
−125 − vC = −300
Eq. (4)
vC = 175 mm/s ↓ �
or
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264
PROBLEM 11.188
An oscillating water sprinkler at Point A rests on an incline
which forms an angle α with the horizontal. The sprinkler
discharges water with an initial velocity v 0 at an angle φ with
the vertical which varies from −φ0 to +φ0 . Knowing that
v0 = 9 m/s, φ0 = 40°, and α = 10°, determine the horizontal
distance between the sprinkler and Points B and C which
define the watered area.
SOLUTION
First note
(v0 ) x = v0 sin φ = (9 m/s) sin φ
(v0 ) y = v0 cos φ = (9 m/s) cos φ
Also, along incline Cab
y = x tan10°
Horizontal motion. (Uniform)
x = 0 + (v0 ) x t = (9 sin φ ) t or t =
x
9 sin φ
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v0 ) y t −
1 2
1
gt = (9 cos φ )t− gt 2
2
2
Substituting for t
x
1
x
− g
y = (9 cos φ )
9 sin φ
2 9 sin φ
=
At B:
φ = 40°, x = d B :
d B tan10° =
g
x2
x
−
tan φ 162 sin 2 θ
( g = 9.81 m/s2)
dB
9.81 d B2
−
tan 40° 162 sin 2 40°
dB = 7 m
or
At C:
2
φ = −40°, x = − dC :
− dC tan10° =
− dC
9.81 (− dC ) 2
−
tan (−40°) 162 sin 2 (−40°)
dC = 9.5 m
or
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265
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PROBLEM 11.189
As the driver of an automobile travels north at 25 km/h in a parking lot, he observes a truck approaching from
the northwest. After he reduces his speed to 15 km/h and turns so that he is traveling in a northwest direction,
the truck appears to be approaching from the west. Assuming that the velocity of the truck is constant during
the period of observation, determine the magnitude and the direction of the velocity of the truck.
SOLUTION
We have
vT = v A + vT/A
Using this equation, the two cases are then graphically represented as shown.
From the diagram
(vT ) x = 25 − 15sin 45°
= 14.3934 km/h
(vT ) y = 15sin 45°
= 10.6066 km/h
vT = 17.88 km/h
36.4° �
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266
PROBLEM 11.190
The driver of an automobile decreases her speed at a constant rate from 72 to 48 km/h over a distance of
225 m along a curve of 450-m radius. Determine the magnitude of the total acceleration of the automobile
after the automobile has traveled 150 m along the curve.
SOLUTION
First note
v1 = 72 km/h = 20 m/s
v2 = 48 km/h = 13.3 m/s
We have uniformly decelerated motion
v 2 = v12 + 2at ( s − s1 )
When v = v2 :
(13.3 m/s)2 = (20 m/s)2 + 2at(225 m)
or
at = −0.496 m/s2
Then when ∆s = 500 ft:
v 2 = (20 m/s)2 + 2(−0.496 m/s2)(150 m)
= 251.2 m2/s2
Now
an =
v2
ρ
251.2 m2/s2
450 m
= 0.558 m/s2
=
Finally
a 2 = at2 + an2
= (−0.496 m/s2)2 + (0.558 m/s2)2
a = 0.75 m/s2
or
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267
268
PROBLEM 11.191
A homeowner uses a snowblower to clear
his driveway. Knowing that the snow is
discharged at an average angle of 40° with
the horizontal, determine the initial velocity
v0 of the snow.
SOLUTION
First note
(vx )0 = v0 cos 40°
(v y )0 = v0 sin 40°
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At B:
4 = (v0 cos 40°) t or t B =
4
v0 cos 40°
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
At B:
1 2
gt
2
0.4 = (v0 sin 40°) tB −
( g = 9.81 m/s2)
1 2
gtB
2
Substituting for t B
4
1
4
0.4 = (v0 sin 40°)
− g
2
v0 cos 40°
v0 cos 40°
or
v02 =
1
2
2
(9.81)(16)/ cos 2 40°
−0.4 + 4 tan 40°
v0 = 6.7 m/s
or
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PROBLEM 11.192
From measurements of a photograph, it has been
found that as the stream of water shown left the nozzle
at A, it had a radius of curvature of 25 m. Determine
(a) the initial velocity vA of the stream, (b) the radius
of curvature of the stream as it reaches its maximum
height at B.
SOLUTION
(a)
We have
(a A ) n =
v A2
ρA
or
�4
�
v A2 = � (9.81 m/s 2 ) � (25 m)
5
�
�
or
v A = 14.0071 m/s
vA = 14.01 m/s
(b)
We have
Where
Then
( aB ) n =
vB2
ρB
vB = ( v A ) x =
ρB
36.9° �
4
vA
5
( 4 × 14.0071 m/s )
= 5
2
9.81 m/s 2
ρ B = 12.80 m �
or
PROPRIETARY MATERIAL.
MATERIAL. ©
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269
270
PROBLEM 11.193
At the bottom of a loop in the vertical plane, an
airplane has a horizontal velocity of 150 m/s and
is speeding up at a rate of 25 m/s2. The radius of
curvature of the loop is 2000 m. The plane is
being tracked by radar at O. What are the recorded
r , θ� and θ�� for this instant?
values of r�, ��
SOLUTION
Geometry. The polar coordinates are
� 600 �
v = (800) 2 + (600) 2 = 1000 m θ = tan −1 �
� = 36.87°
� 800 �
Velocity Analysis.
v = 150 m/s →
vr = 150 cos θ = 120 m/s
vθ = −150sin θ = −90 m/s
r� = 120 m/s �
vr = r�
v
90
vθ = rθ� θ� = θ = −
1000
r
θ� = −0.0900 rad/s �
Acceleration analysis.
at = 25 m/s 2
an =
v2
ρ
=
(150) 2
= 11.25 m/s 2
2000
a = 25 m/s 2 → +11.25 m/s 2 ↑ = 27.41 m/s 2
24.23°
β = 24.23°
θ − β = 12.64°
PROPRIETARY
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PROPRIETARY MATERIAL.
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No part
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Manual may
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or distributed
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form or
or by
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you
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271
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PROBLEM 11.193 (Continued)
ar = a cos (θ − β ) = 27.41 cos 12.64° = 26.74 m/s 2
aθ = − a sin (θ − β ) = −27.41 sin 12.64° = −6.00 m/s 2
ar = ��
r − rθ� 2 ��
r = ar + rθ� 2
��
r = 26.74 + (1000)(0.0900) 2
aθ = rθ�� + 2r�θ�
a
2r�θ�
θ�� = θ −
r
r
−6.00 (2)(120)(−0.0900)
=
−
1000
1000
r�� = 34.8 m/s 2 �
θ�� = −0.0156 rad/s 2 �
PROPRIETARY MATERIAL.
MATERIAL. ©
© 2010
2009 The
TheMcGraw-Hill
McGraw-Hill Companies,
Companies, Inc.
Inc. All
All rights
rights reserved.
reserved. No
No part
part of
of this
this Manual
Manual may
may be
be displayed,
displayed,
PROPRIETARY
reproduced or
or distributed
distributed in
in any
any form
form or
or by
by any
anymeans,
means, without
without the
the prior
prior written
written permission
permission of
of the
the publisher,
publisher,or
orused
usedbeyond
beyond the
the limited
limited
reproduced
youare
areaastudent
studentusing
usingthis
thisManual,
Manual,
distributionto
toteachers
teachersand
andeducators
educatorspermitted
permittedby
byMcGraw-Hill
McGraw-Hillfor
fortheir
theirindividual
individual course
course preparation.
preparation. IfIf you
distribution
youare
areusing
using itit without
without permission.
permission.
you
271
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