CHAPTER
CHAPTER11
3
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
v=
= 6t 3 - 60t + 5
dt
dv
a=
= 18t 2 - 60
dt
Evaluate expressions at t = 4 s.
x = 1.5( 4) 4 - 30( 4)2 + 5( 4) + 10 = -66 m v = 6( 4)3 - 60( 4) + 5 = 149 m/s
a = 18( 4)2 - 60 = 228 m/s2 x = -66.0 m b
v = 149.0 m/s b
a = 228.0 m/s2 b
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3
PROBLEM 11.2
The motion of a particle is defined by the relation x = 12t 3 - 18t 2 + 2t + 5, 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
v=
= 36t 2 - 36t + 2
dt
dv
a=
= 72t - 36
dt
Find the time for a = 0.
72t - 36 = 0 fi t = 0.5 s
Substitute into above expressions.
x = 12(0.5)3 - 18(0.5)2 + 2(0.5) + 5 = 3 v = 36(0.5)2 - 36(0.5) + 2
= -7 m/s
x = 3.00 m b
v = -7.00 m/s b
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4
PROBLEM 11.3
5 3 5 2
t - t - 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
Then
and
When v = 0:
or
At t = 3 s:
5
5
x = t 3 - t 2 - 30t + 8
3
2
dx
v=
= 5t 2 - 5t - 30
dt
dv
a=
= 10t - 5
dt
5t 2 - 5t - 30 = 5(t 2 - t - 6) = 0
t = 3 s and t = -2 s ( Reject) t = 3.00 s b
5
5
x3 = (3)3 - (3)2 - 30(3) + 8 3
2
or
a3 = 10(3) - 5 or a3 = 25.0 m/s2 b
x3 = -59.5 m b
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5
PROBLEM 11.4
The motion of a particle is defined by the relation x = 6t 2 - 8 + 40 cos p 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 p t
Then
v=
dx
= 12t - 40p sin p t
dt
and
a=
dv
= 12 - 40p 2 cos p t
dt
At t = 6 s:
x6 = 6(6)2 - 8 + 40 cos 6p
or
x6 = 248 m b
v6 = 12(6) - 40p sin 6p or
v6 = 72.0 m/s b
a6 = 12 - 40p 2 cos 6p or a6 = -383 m/s2 b
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6
PROBLEM 11.5
The motion of a particle is defined by the relation x = 6t 4 - 2t 3 - 12t 2 + 3t + 3, where x and t are expressed in
meters and seconds, respectively. Determine the time, the position, and the velocity when a = 0.
SOLUTION
We have
x = 6t 4 - 2t 3 - 12t 2 + 3t + 3
Then
v=
dx
= 24t 3 - 6t 2 - 24t + 3
dt
and
a=
dv
= 72t 2 - 12t - 24
dt
When a = 0:
72t 2 - 12t - 24 = 12(6t 2 - t - 2) = 0
(3t - 2)(2t + 1) = 0
or
or
2
At t = s:
3
t=
2
1
s and t = - s ( Reject ) 3
2
4
x2/3
3
2
Ê 2ˆ
Ê 2ˆ
Ê 2ˆ
Ê 2ˆ
= 6 Á ˜ - 2 Á ˜ - 12 Á ˜ + 3 Á ˜ + 3
Ë 3¯
Ë 3¯
Ë 3¯
Ë 3¯
3
t = 0.667 s b
or
x2/3 = 0.259 m b
or
v2/3 = -8.56 m/s b
2
Ê 2ˆ
Ê 2ˆ
Ê 2ˆ
v2/3 = 24 Á ˜ - 6 Á ˜ - 24 Á ˜ + 3
Ë 3¯
Ë 3¯
Ë 3¯
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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
v=
= 6t 2 - 30t + 24
dt
dv
a=
= 12t - 30
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
b
t = 2.5 s
x2.5 = 2(2.5)3 - 15(2.5)2 + 24(2.5) + 4
x2.5 = + 1.500 m b
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 Æ
x2.5 - x1 = 1.5 - 15 = 13.5 m ¨
Total distance traveled = 11 m + 13.5 m = 24.5 m b
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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)
When v = 0:
3t 2 - 12t - 36 = 3(t 2 - 4t - 12) = 0
(t + 2)(t - 6) = 0
or
t = -2 s ( Reject ) and t = 6 s or
(b)
When x = 0:
Factoring
Now observe that
t = 6.00 s b
t 3 - 6t 2 - 36t - 40 = 0
(t - 10)(t + 2)(t + 2) = 0 or t = 10 s
0 £ t < 6 s:
v<0
6 s < t £ 10 s:
v>0
and at t = 0:
x0 = -40 m
x6 = (6)3 - 6(6)2 - 36(6) - 40
t = 6 s:
= -256 m
t = 10 s:
Then
v10 = 3(10)2 - 12(10) - 36 or v10 = 144.0 m/s b
a10 = 6(10) - 12 or
a0 = 48.0 m/s2 b
| x6 - x0 | = | - 256 - ( -40) | = 216 m
x10 - x6 = 0 - ( -256) = 256 m
Total distance traveled = (216 + 256) m = 472 m b
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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
(b)
or
When a = 0:
t = 2.00 s and t = 4.00 s b
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:
or
x3 = 10.00 m b
v>0
2 s < t £ 3 s:
v<0
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 b
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10
PROBLEM 11.9
The acceleration of a particle is defined by the relation a = -8 m/s2. 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/s2
dt
We have
Ú dv = Ú -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:
Ú20 dx = Ú 4(-8t + C ) dt
or
x - 20 = [-4t 2 + Ct ]t4
x = -4t 2 + C (t - 4) + 84 ( m)
or
When v = 16 m/s, x = 4 m:
16 = -8t + C fi 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 v11 = - (8)(11) + 32 At t = 0:
x0 = - 44 m
t = 4 s:
x4 = 20 m
t = 11 s:
x11 = - 4(11)2 + 32(11) - 44 = -176 m
or t = 4.00 s b
v11 = -56.0 m/s b
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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 b
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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:
Ú18 dv = Ú6 kt
or
1
v - 18 = k (t 3 - 216)
3
2
dt
1
v = 18 + k (t 3 - 216)( m/s)
3
or
1
dx
= v = 18 + k (t 3 - 216)
dt
3
Also
tÈ
x
1
˘
- 216)˙ dt
˚
At t = 0, x = 24 m:
Ú 24 dx = Ú 0 ÍÎ18 + 3 k (t
or
1 Ê1
ˆ
x - 24 = 18t + k Á t 4 - 216t ˜
¯
3 Ë4
3
Now
At t = 6 s, x = 96 m:
or
Then
or
and
or
1 È1
˘
96 - 24 = 18(6) + k Í (6)4 - 216(6)˙
3 Î4
˚
1
k = m/s 4
9
1 Ê 1ˆ Ê 1
ˆ
x - 24 = 18t + Á ˜ Á t 4 - 216t ˜
¯
Ë
¯
Ë
3 9 4
x (t ) =
1 4
t + 10t + 24 108
b
1 Ê 1ˆ
v = 18 + Á ˜ (t 3 - 216)
3 Ë 9¯
v (t ) =
1 3
t + 10 27
b
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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 = 400 mm/s. Knowing that v = 375 mm/s and that x = 500 mm when t = 1 s, determine the velocity, the
position, and the total distance traveled when t = 7 s.
SOLUTION
a = kt k = constant
dv
= a = kt
dt
We have
Now
At t = 0, v = 400 mm/s :
or
or
At t = 1 s, v = 375 mm/s:
v
v - 400 =
1 2
kt
2
1
v = 400 + kt 2 (mm /s)
2
1
375 mm/s = 400 mm/s + k(1 s)2
2
k = -50 mm/s3
or
and v = 400 - 25t 2
dx
= v = 400 - 25t 2 = 25(16 - t 2 )
dt
Also
At t = 1 s, x = 500 mm:
t
Ú16 dv = Ú 0 kt dt
x
t
Ú500 dx = Ú1 25(16 - t
2
) dt
t
or
or
Then
At t = 7 s:
1 ˘
È
x - 500 = 25 Í16t - t 3 ˙
3 ˚1
Î
È -t 3
˘
25
x = 25 Í
+ 16t ˙ + 500 - 400 +
3
3
Î
˚
325
= 25( -t 3 / 3 + 16t ) +
3
{
}
v7 = 25 16 - (7)2 13 ¸
Ï 1
x7 = 25 Ì- (7)3 + 16(7) + ˝ 3˛
Ó 3
When v = 0:
or v7 = -825 mm/s b
or
x7 = 50.0 mm b
25(16 - t 2 ) = 0 or t = 4 s
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14
PROBLEM 11.11 (Continued)
At t = 0:
x0 =
325
3
13 ˆ
Ê 1
x4 = 25 Á - ( 4)3 + 16( 4) + ˜ = 1175 mm
Ë 3
3¯
t = 4 s:
Now observe that
Then
0 £ t < 4 s:
v>0
4 s < t £ 7 s:
v<0
13 ˆ
Ê
x4 - x0 = 25 Á 47 - ˜ = 1067 mm
Ë
3¯
| x7 - x4 | = |50 - 1775 | = 1125 mm
Total distance traveled = (1067 + 1125) mm = 2192 mm 2190 mm b
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15
PROBLEM 11.12
The acceleration of a particle is defined by the relation a = kt 2. (a) Knowing that v = -10 m /s when t = 0 and
that v = +10 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 = -10 m/s and t = 4 s, v = + 10 m/s
4
10
Ú-10 dv = Ú 0 kt
(a)
2
dt
1
10 - ( -10) = k ( 4)3 3
(b)
k = 0.9375 m/s 4 b
Substituting k = 0.9375 m/s 4 into (1)
dv
= a = 0.9375t 2 dt
v
t
Ú-10 dv = Ú 0 0.9375t
t = 0, v = -10 m/s:
2
a = 0.9375t 2 b
dt
1
v - ( -10) = 0.9375(t )3 3
dx
= v = 0.3125t 3 - 10
dt
t = 4 s, x = 0:
v = (0.3125t 3 - 10) b
t4
dx
=
(
0
.
3125
t
10
)
dt
;
x
=
0
.
3125
- 10t
Ú0
Ú4
4
x
t
t
3
4
˘
È 0.3125t 4
˘ È 0.3125( 4) 4
- 10( 4)˙
- 10t ˙ - Í
x=Í
4
4
˚
Î
˚ Î
Ê 0.3125t 4
ˆ
- 10t ˜ - 19.968 + 40
x=Á
4
Ë
¯
x = 0.078125t 4 - 10t + 20.032 b
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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
2
)dt
v = At - 2t 3 ( m/s)
30 = A(1) - 2(1)3
At t = 1 s, v = 30 m/s:
A = 32 m/s2
or
and v = 32t - 2t 3
dx
= v = 32t - 2t 3
dt
Also
At t = 0, x = 8 m:
x
t
Ú8 dx = Ú 0 (32t - 2t
3
)dt
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)
t
Ú 0 dv = Ú 0( A - 6t
or
(a)
A = constant
At t = 4 s:
t = 5 s:
b
1
x4 = 8 + 16( 4)2 - ( 4) 4 = 136 m
2
1
x5 = 8 + 16(5)2 - (5) 4 = 95.5 m
2
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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
b
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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
t
v - ( -15) = -
or
k
Ú-15 dv = Ú 2 t 3 dt
v=
or
At t = 10 s, v = 0.36 m/s:
0.36 =
k2 È 1
1 ˘
Í 2 - 2˙
2 Ît
( 2) ˚
kÊ1 1ˆ
Á - ˜ - 15 ( m/s)
2 Ë 4 t2 ¯
kÊ1
1 ˆ
ÁË - 2 ˜¯ - 15
2 4 10
or
k = 128 m ◊ s
and
v =1-
(a)
t2
( m/s)
64
dx
= v =1- 2
dt
t
We have
64 ˆ
Ê
Ú dx = Ú ÁË1 - t 2 ˜¯ dt + C
Then
x=t+
or
Now x2 = 2 x10 :
64
2+
C = constant
64
+ C ( m)
t
64
64
ˆ
Ê
+ C = 2 Á10 +
+ C˜
¯
Ë
2
10
or
C = 1.2 m
and
x=t+
64
+ 1.2 ( m)
t
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19
PROBLEM 11.14 (Continued)
64
+ 1.2 2
64
x10 = 10 +
+ 1.2 10
At t = 2 s:
x2 = 2 +
t = 10 s:
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-
At t = 8 s:
64
t2
x2 = 35.2 m b
or
x10 = 17.60 m b
4
m
15
11
13
m, x10 = -5 m
15
15
= 0 or 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 b
Note: The total distance traveled is the same for both cases.
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20
PROBLEM 11.15
The acceleration of a particle is defined by the relation a = - k /x. It has been experimentally determined that
v = 5 m /s when x = 0.2 m and that v = 3 m /s when x = 0.4 m. Determine (a) the velocity of the particle when
x = 0.5 m, (b) the position of the particle at which its velocity is zero.
SOLUTION
a=
vdv - k
=
dx
x
Separate and integrate using x = 0.2 m, v = 5 m/s.
v
x
Ú5 vdv = -k Ú0.2
dx
x
v
1 2
= - k ln x
v
2 0.2
x
0.2
1 2 1 2
Ê x ˆ
v - (5) = - k ln Á
Ë 0.2 ˜¯
2
2
(1)
When v = 3 m/s, x = 0.4 m
1 2 1 2
Ê 0.4 ˆ
(3) - (5) = - k ln Á
Ë 0.2 ˜¯
2
2
Solve for k.
k = 11.5416 m2 /s2
(a)
Substitute
k = 11.5416 m2 /s2
and
x = 0.5 m into (1).
1 2 1 2
Ê 0.5 ˆ
v - (5) = -11.5416 ln Á
Ë 0.2 ˜¯
2
2
(b)
For v = 0,
1
Ê x ˆ
0 - (5)2 = -11.5416 ln Á
Ë 0.2 ˜¯
2
Ê x ˆ
ln Á
= 1.083
Ë 0.2 ˜¯
v = 1.962 m/s b
x = 0.591 m b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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21
PROBLEM 11.16
A particle starting from rest at x = 0.3 m is accelerated so that its velocity doubles in magnitude between
x = 0.6 m and x = 2.4 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 8.7 m /s when x = 4.8 m.
SOLUTION
v
We have
When x = 0.3 m, v = 0:
dv
Aˆ
Ê
= a = kÁx - ˜
Ë
dx
x¯
v
x
Ê
Aˆ
Ú0 vdv = Ú0.3 k ÁË x - x ˜¯ dx
x
1 2
È1
˘
v = k Í x 2 - A ln x ˙
2
2
Î
˚ 0.3
or
x ˆ
Ê1
= k Á x 2 - 0.045 - A ln
˜
Ë2
0.3¯
At x = 0.6 m:
1 2
0.6
È1
˘
v0.6 = k Í (0.6)2 - A ln
- 0.045˙ = k (0.135 - A ln 2)
2
0.3
Î2
˚
x = 2.4 m:
1 2
2.4
È1
˘
v2.4 = k Í (2.4)2 - A ln
- 0.045˙ = k (2.835 - A ln 8)
.
0
3
2
2
Î
˚
Now
1 2
2 v2.4
1 2
2 v0.6
v2.4
= 2:
v0.6
= (2)2 =
k (2.835 - A ln 8)
k (0.135 - A ln 2)
or
0.54 - 4 A ln 2 = (2.835 - A ln 8)
or
Ê 1ˆ
2.295 = A(ln 8 - 4 ln 2) = A(ln 8 - ln 24 ) = A ln Á ˜
Ë 2¯
A = -3.31 m2 b
or
When x = 4.8 m, v = 8.7 m/s:
˘
È1
1
2.295 Ê 4.8 ˆ
˙
(8.7)2 = k Í ( 4.8)2 ln
.
0
045
˜
Á
Ë 0.3 ¯
2
ln 12
˙˚
ÍÎ 2
Noting that
Ê 1ˆ
Ê 4.8 ˆ
= ln(16) = 4 ln 2 and ln Á ˜ = - ln(2)
ln Á
˜
Ë 2¯
Ë 0.3 ¯
We have
()
37.845 = k [11.52 + 9.18 - 0.045]
k = 1.832 s -2 b
or
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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)
We have
When x = 40 mm, v = 0:
v
v
dv
= a = k (100 - x )
dx
x
Ú0 vdv = Ú40 k (100 - x)dx
x
or
1 2
1 ˘
È
v = k Í100 x - x 2 ˙
2
2 ˚ 40
Î
or
1 2
1
ˆ
Ê
v = k Á100 x - x 2 - 3200˜
¯
Ë
2
2
When x = 100 mm, v = 18 mm/s:
1
1
È
˘
(18)2 = k Í100(100) - (100)2 - 3200 ˙
2
2
Î
˚
k = 0.0900 s -2 b
or
(b)
When x = 120 mm:
1 2
1
È
˘
v = 0.09 Í100(120) - (120)2 - 3200 ˙ = 144
2
2
Î
˚
v = ± 16.97 mm/s b
or
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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
dv
k
=a=
dx
( x + 4)2
x
k
Ú0 vdv = Ú0 ( x + 4)2 dx
1 2
1ˆ
Ê 1
v = -k Á
Ë x + 4 4 ˜¯
2
1 2
1ˆ
Ê 1
( 4) = - k Á
- ˜
Ë 8 + 4 4¯
2
or
(b)
When v = 4.5 m/s:
1
1ˆ
Ê 1
( 4.5)2 = -48 Á
- ˜
Ë x + 4 4¯
2
x = 21.6 m b
or
(c)
k = 48 m3 /s2 b
Note that when v = vmax, a = 0.
Now a Æ 0 as x Æ • so that
1 2
1 ˆ
Ê 1ˆ
-1 Ê 1
= 48 Á ˜
vmax = 48ln
˜
xÆ • Á Ë 4¯
Ë 4 x + 4¯
2
vmax = 4.90 m/s b
or
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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
Úv
0
xf
vdv = - Ú k x dx
0
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/s2
a = 800 m/s2 ≠ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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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
We have
When x = 0, v = 1 m/s:
v
v
dv
x ˆ
Ê
= a = - Á 0.1 + sin
˜
Ë
0.8 ¯
dx
x
x ˆ
Ê
Ú1 vdv = Ú0 - ÁË 0.1 + sin 0.8 ˜¯ dx
x
x ˘
1 2
È
( v - 1) = - Í0.1x - 0.8 cos
.8 ˙˚ 0
2
0
Î
or
x
1 2
v = -0.1x + 0.8 cos
- 0.3
2
0.8
or
(a)
When x = -1 m:
1 2
-1
v = -0.1( -1) + 0.8 cos
- 0.3
0.8
2
v = ± 0.323 m/s b
or
(b)
When v = vmax, a = 0:
or
(c)
x ˆ
Ê
- Á 0.1 + sin
˜ =0
Ë
0.8 ¯
x = -0.080 134 m x = -0.0801 m b
When x = -0.080 134 m:
-0.080 134
1 2
vmax = -0.1( -0.080 134) + 0.8 cos
- 0.3
0.8
2
= 0.504 m2 /s2
vmax = 1.004 m/s b
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
x
When x = 0, v = 0:
Ú0
or
È v 2 + 49 ˘ = 0.8 x
ÎÍ
˚˙0
or
v 2 + 49 - 7 = 0.8 x
v 2 + 49
= Ú 0.8 dx
0
v
(a)
When v = 24 m/s:
242 + 49 - 7 = 0.8 x
x = 22.5 m b
or
(b)
When x = 40 m:
v 2 + 49 - 7 = 0.8( 40)
v = 38.4 m/s b
or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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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
x
When x = 0, v = 81 m/s:
Ú81
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 = Ú - k dx
0
2
(363/ 2 - 729) = -k (18)
3
k = 19 m/s2
or
Finally
When x = 20 m:
2 3/ 2
( v - 729) = -19(20)
3
v 3/ 2 = 159 or
(b)
We have
dv
= a = -19 v
dt
At t = 0, v = 81 m/s:
Ú81
v
t
dv
= Ú -19dt
0
v
or
v
2[ v ]81
= -19t
or
2( v - 9) = -19t
When v = 0:
v = 29.3 m/s b
2( -9) = -19t
t = 0.947 s b
or
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28
PROBLEM 11.23
The acceleration of a particle is defined by the relation a = -0.8v, where a is expressed in m /s2 and v in m /s.
Knowing that at t = 0 the velocity is 1 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 = 1 m/s when x = 0.
v
x
Ú1 dv = -0.8Ú0 dx
v - 1 = -0.8 x
Distance traveled.
For v = 0,
(b)
Separate.
x=
-1
fi x = 1.25 m -0.8
a=
dv
= -0.8v
dt
v dv
Ú1
b
x
= -Ú 0.8dt
0
v
ln v - ln 1 = -0.8t
ln v = -0.8t
Ê 1ˆ
t = 1.25 ln Á ˜
Ë v¯
For v = 0, we get t = •.
(c)
t=• b
For v = 0.5 m /s,
Ê 1 ˆ
= 0.866 s t = 1.25 ln Á
Ë 0.5 ˜¯
b
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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 = 3 - 0.1 v 2 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 = 3 - 0.1v 2 = k (c 2 - v 2 )
where
k = 0.1 m -1 and c 2 =
3
m2
= 30 2
0.1
s
c = 5.4772 m/s
Since v0 > c, write
a=v
dv
= - k ( v 2 - c2 )
dx
vdv
v - c2
2
= - k dx
v
Integrating,
1
ln( v 2 - c 2 ) = - k ( x - x0 )
2
v0
ln
v 2 - c2
= -2k ( x - x0 )
v02 - c 2
v 2 - c2
= e -2 k ( x - x0 )
2
2
v0 - c
(
)
v 2 = c 2 + v02 - c 2 e -2 k ( x - x0 )
= 30 + [(8) - 30] e -2( 0.1)(10 )
2
= 34.6014 m2 /s2
v = 5.88 m/s b
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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
v
dv
t
Ú0 1 - kv = Ú0 0.4 dt
At t = 0, v = 0:
1
- [ln(1 - kv )]0v = 0.4t
k
or
or
ln(1 - kv ) = -0.4 kt 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 b
or
(b)
v
We have
v
vdv
x
Ú0 1 - kv = Ú4 0.4 dx
When x = 4 m, v = 0:
1/k
v
1
=- +
1 - kv
k 1 - kv
Now
Then
dv
= a = 0.4(1 - kv )
dx
vÈ
1
1
˘
x
Ú0 ÍÎ- k + k (1 - kv) ˙˚ dv = Ú4 0.4 dx
v
or
È v 1
˘
x
ÍÎ- k - k 2 ln(1 - kv )˙˚ = 0.4[ x ]4
0
or
Èv 1
˘
- Í + 2 ln(1 - kv )˙ = 0.4( x - 4)
Îk k
˚
When v = 6 m/s:
È
˘
6
1
ln(
.
)
-Í
+
¥
1
0
145
703
6
˙ = 0.4( x - 4)
2
Î 0.145 703 (0.145 703)
˚
or
0.4( x - 4) = 56.4778
x = 145.2 m b
or
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31
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 b
or
An alternative solution is to begin with Eq. (1).
Then
ln(1 - kv ) = -0.4 kt
1
v = (1 - k -0.4 kt )
k
Thus, vmax is attained as t Æ •
vmax =
1
k
as above.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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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 = - 0.6v 3/ 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:
dv
= a = -0.6v 3/ 2
dx
v - (3/2)
Ú9 v
x
dv = Ú -0.6 dx
0
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 b
or
(b)
dv
= a = -0.6v 3/ 2
dt
We have
v - (3/2)
t
When t = 0, v = 9 m/s:
Ú9 v
dv = Ú -0.6dt
or
-2[v - (1/2) ]9v = -0.6t
0
or
1 1
- = 0.3t
v 3
When v = 1 m/s:
1 1
- = 0.3t
1 3
t = 2.22 s b
or
(c)
We have
(1)
1 1
- = 0.3t
v 3
2
or
Now
9
Ê 3 ˆ
=
v=Á
˜
Ë 1 + 0.9t ¯
(1 + 0.9t )2
dx
9
=v=
dt
(1 + 0.9t )2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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33
PROBLEM 11.26 (Continued)
x
9
t
Ú 0 dx = Ú0 (1 + 0.9t )2 dt
At t = 0, x = 0:
t
1 ˘
È 1
x = 9 ÍÎ 0.9 1 + 0.9t ˙˚ 0
or
1 ˆ
Ê
= 10 Á1 Ë 1 + 0.9t ˜¯
When x = 6 m:
=
9t
1 + 0.9t
6=
9t
1 + 0.9t
t = 1.667 s b
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
dx
t
Ú0 (3 - 0.3x)2 = Ú0 dt
x
or
t=
x
1 È 1 ˘
=
0.3 ÍÎ 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 = 12(1 - 0.06 x )0.3 , where v and x are expressed in km / h and km, 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 9 km.
SOLUTION
(a)
dx
= v = 12(1 - 0.06 x )0.3
dt
We have
or
dx
x
t
Ú0 (1 - 0.06 x)0.3 = Ú 012dt
At t = 0, x = 0:
1 Ê
1 ˆ
0.7 x
˜ [(1 - 0.06 x ) ]0 = 12t
Á0.7 Ë 0.06 ¯
1 - (1 - 0.06 x )0.7 = 0.504t or
or
x=
1
[1 - (1 - 0.504t )1/ 0.7 ]
0.06
At t = 1 h:
x=
1
{1 - [1 - 0.504(1)]1/ 0.7 }
0.06
x = 10.55 km b
or
(b)
We have
(1)
dv
a=v
dx
= 12(1 - 0.06 x )0.3
d
[12(1 - 0.06 x )0.3 ]
dx
= 122 (1 - 0.06 x )0.3 [(0.3)( -0.06)(1 - 0.06 x ) -0.7 ]
= -2.592(1 - 0.06 x )-0.4
At t = 0, x = 0:
a0 =
-2.592 km
h2
¥
1000 m Ê 1 h ˆ
¥Á
Ë 3600 ˜¯
1 km
a0 = -2 ¥ 10 -4 m/s2 b
or
(c)
2
From Eq. (1)
t=
1
[1 - (1 - 0.06 x )0.7 ]
0.504
When x = 9 km:
t=
1
{1 - [1 - 0.06(9)]0.7 }
0.504
= 0.832 hrs
t = 49.9 min b
or
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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
a=v
We have
=
0.18v0 d Ê 0.18v0 ˆ
Á
˜
x dx Ë x ¯
0.0324v02
=When x = 2 m:
x3
0.0324(3.6)2
(2)3
a=-
a = -0.0525 m/s2 b
or
(b)
0.18v0
dx
=v=
dt
x
We have
From x = 1 m to x = 3 m:
3
t3
Ú1 xdx = Út
1
0.18v0 dt
3
or
È1 2 ˘
ÍÎ 2 x ˙˚ = 0.18v0 (t3 - t1 )
1
or
(t3 - t1 ) =
1
2 (9 - 1)
0.18(3.6)
t3 - t1 = 6.17 s b
or
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36
PROBLEM 11.29
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
-9.81
a=
[1 + ( y /6.37 ¥ 106 )]2
where a and y are expressed in m /s2 and m, 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) 11,180 m /s.
SOLUTION
v
We have
- 9.81
dv
=a=2
dy
y
Ê
ˆ
1
+
ÁË
˜
6.37 ¥ 106 ¯
y = 0,
When
v = v0
y = ymax, v = 0
0
- 9.81
ymax
Úv v dv = Ú0
Then
0
y
Ê
ˆ
ÁË1 +
6˜
6.37 ¥ 10 ¯
2
dy
y
or
È
˘ max
Í
˙
1
1
- v02 = -9.81 Í- 6.37 ¥ 106
˙
y
2
Í
˙
1+
ÍÎ
6.37 ¥ 106 ˙˚0
or
ˆ
Ê
˜
Á
1
v02 = 124.9794 ¥ 106 Á1 ymax ˜
˜
Á 1+
Ë
6.37 ¥ 106 ¯
È
˘
v02
ymax = 6.37 ¥ 106 Í
6
2˙
ÍÎ124.9794 ¥ 10 - v0 ˙˚
or
(a)
v0 = 540 m/s:
È
˘
5402
ymax = 6.37 ¥ 106 ¥ Í
6
2˙
Î124.9794 ¥ 10 - 540 ˚
ymax = 14.90 ¥ 103 m b
or
(b)
v0 = 900 m/s:
È
˘
9002
ymax = 6.37 ¥ 106 Í
6
2˙
Î124.9794 ¥ 10 - 900 ˚
or
ymax = 41.6 ¥ 103 m b
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37
PROBLEM 11.29 (Continued)
(c)
v0 = 11,180 m/s :
È
˘
(11,180)2
ymax = 6.37 ¥ 106 Í
6
2 ˙
Î (124.9794 ¥ 10 - 11,180 ) ˚
ymax = -6.12 ¥ 1010 m b
or
The velocity 11,180 m /s is approximately the escape velocity vR from the earth. For vR
ymax Æ • b
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38
PROBLEM 11.30
The acceleration due to gravity of a particle falling toward the earth is a = - gR2 /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
v
We have
dv
gR2
=a=- 2
dr
r
r = R, v = ve
When
r = •, v = 0
Then
0
Úv
e
•
gR2
R
r2
vdv = Ú -
dr
•
or
1
È1 ˘
- ve2 = gR2 Í ˙
2
Î r ˚R
or
ve = 2 gR
1000 m ˆ
Ê
= Á 2 ¥ 9.81 m/s2 ¥ 6370 km ¥
˜
Ë
1 km ¯
1/ 2
ve = 11,180 m/s b
or
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39
PROBLEM 11.31
The velocity of a particle is v = v0 [1 - sin (p 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
È
Ê pt ˆ ˘
= v = v0 Í1 - sin Á ˜ ˙
Ë T ¯˚
dt
Î
We have
At t = 0, x = 0 :
x
t
Ê pt ˆ ˘
È
Ú0 dx = Ú0 v0 ÍÎ1 - sin ÁË T ˜¯ ˙˚ dt
t
or
È T
Ê pt ˆ ˘
x = v0 Ít + cos Á ˜ ˙
Ë T ¯ ˚0
Î p
È T
Ê pt ˆ T ˘
= v0 Ít + cos Á ˜ - ˙ Ë T ¯ p˚
Î p
At t = 3T :
or
T
È
Ê p ¥ 3T ˆ T ˘
x BT = v0 Í3T + cos Á
Ë T ˜¯ p ˙˚
p
Î
2T ˆ
Ê
= v0 Á 3T - ˜
Ë
p ¯
a=
Also
At t = 3T :
x3T = 2.36 v0T b
dv d Ï È
p
pt
Ê p t ˆ ˘¸
= Ìv0 Í1 - sin Á ˜ ˙ ˝ = - v0 cos
Ë T ¯ ˚˛
dt dt Ó Î
T
T
a3T = - v0
p
p ¥ 3T
cos
T
T
a3T =
or
(b)
(1)
p v0
b
T
Using Eq. (1)
At t = 0:
T˘
È T
x0 = v0 Í0 + cos( 0) - ˙ = 0
p
p˚
Î
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40
PROBLEM 11.31 (Continued)
At t = T :
T
È
Ê pT ˆ T ˘
xT = v0 ÍT + cos Á
Ë T ˜¯ p ˙˚
p
Î
2T ˆ
Ê
= v0 Á T - ˜
Ë
p ¯
= 0.363v0T
Now
vave =
xT - x0 0.363v0T - 0
=
Dt
T -0
vave = 0.363v0 b
or
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41
PROBLEM 11.32
The velocity of a slider is defined by the relation v = v ¢ sin (w nt + f ). 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 + x02w n2 2 x0w n , (b) the maximum value of the velocity occurs when
2
x = x0 [3 - ( v0 /x0w n ) ]/ 2.
SOLUTION
(a)
At t = 0, v = v0 :
Then
v0 = v ¢ sin (0 + f ) = v ¢ sin f
2
cosf = v ¢ - v02 v ¢
dx
= v = v ¢ sin (w nt + f )
dt
Now
x
t
At t = 0, x = x0 :
Ú x dx = Ú0 v ¢ sin (w nt + f )dt
or
È 1
˘
x - x0 = v ¢ Ícos (w nt + f )˙
Î wn
˚0
0
t
x = x0 +
or
v¢
[cos f - cos (w nt + f )]
wn
Now observe that xmax occurs when cos (w nt + f ) = -1. Then
xmax = 2 x0 = x0 +
x0 =
Substituting for cosf
or
v¢
[cos f - ( -1)]
wn
2
2
ˆ
v ¢ Ê v ¢ - v0
+
1
Á
˜
w n ÁË
˜¯
v1
x0w n - v1 = v ¢ 2 - v02
Squaring both sides of this equation
x02w n2 - 2 x0w n + v ¢ 2 = v ¢ 2 - v02
or
v¢ =
v02 + x02w n2
2 x0w n
Q. E. D.
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42
PROBLEM 11.32 (Continued)
(b)
First observe that vmax occurs when w nt + f = p2 . The corresponding value of x is
xvmax = x0 +
= x0 +
v¢
wn
È
Êp ˆ˘
Ícos f - cos ÁË 2 ˜¯ ˙
Î
˚
v¢
cos f
wn
Substituting first for cosf and then for v ¢
xvmax = x0 +
v¢
wn
2
v ¢ - v02
v¢
1/ 2
˘
ÈÊ v 2 + x 2w 2 ˆ 2
2˙
0 n
ÍÁ 0
v
0
ÍË 2 x0w n ˜¯
˙
˚
Î
1
= x0 +
v04 + 2v02 x02w n2 + x04w n4 - 4 x02w n2 v02
2 x0w n2
1
= x0 +
wn
(
= x0 +
= x0 +
x
= 0
2
)
1/ 2
1/ 2
1 È 2 2
2 2˘
w
x
v
0 n
0 ˙
˚
2 x0w n2 ÎÍ
(
)
x02w n2 - v02
2 x0w n2
È Ê v ˆ2˘
Í3 - Á 0 ˜ ˙ ÍÎ Ë x0w n ¯ ˙˚
Q. E. D.
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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/s2 b
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 b
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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)
1
x = x0 + v0 t + at 2
2
Initial velocity.
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 = 25.9 m/s b
vf = 19.00 m/s b
Distance traveled during first 1.5 s.
1
x = x0 + v0 t + at 2
2
1
= 0 + (25.0)(1.5) + ( - 0.6)(1.5)2 2
x = 36.8 m b
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45
PROBLEM 11.35
Assuming a uniform acceleration of 3.3 m /s2 and knowing that
the speed of a car as it passes A is 45 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.
v A = 45 km/h = 12.5 m/s, x A = 0, x B = 50 m, a = 3.3 m/s2
1
x B = x A + v At + at 2
2
1
50 = 0 + 12.5t + (3.3)t 2
2
3.3t 2 + 25t - 100 = 0
-25 ± (25)2 - 4(3.3)( -100)
2(3.3)
t = 2.8943 s or - 10.47 sec
t=
Rejecting the negative root.
(b)
t = 2.8943 s t = 2.89 s b
Speed at B.
v B = v A + at = 12.5 + (3.3)(2.8943) = 22.0512 m/s
v B = 79.4 km/h b
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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 27 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) We have
1
y = y1 + v1t + at 2
2
At tland ,
y=0
Then
0 = 27 m + v1 (16 s)
1
+ ( -9.81 m/s2 )(16 s)2
2
or
v1 = 76.7925 m/s (b) We have
v 2 = v12 + 2a( y - y1 )
v1 = 76.8 m/s b
At
y = ymax ,
v=0
Then
0 = (76.7925 m/s)2 + 2( -9.81 m/s2 )( ymax - 27) m
ymax = 327.57 m or
ymax = 328 m b
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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
Given:
0 £ ¥ £ 35 m, a = constant
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
(b)
for
0 x 35 m
a = 2.40 m/s2 b
First note that v = vmax for 35 m £ x £ 100 m.
Now
(c)
1
x = 0 + 0t + at 2
2
1
35 m = a(5.4 s)2
2
a = 2.4005 m/s2 v 2 = 0 + 2 a( x - 0 )
for
0 £ x £ 35 m
When x = 35 m:
2
vmax
= 2(2.4005 m/s2 )(35 m)
or
vmax = 12.9628 m/s
We have
When x = 100 m:
x = x1 + v0 (t - t1 )
vmax = 12.96 m/s b
for
35 m < x £ 100 m
100 m = 35 m + (12.9628 m/s)(t2 - 5.4) s
t2 = 10.41 s b
or
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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 /s2 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
AÆ B
and
CÆD
v 2 = v02 + 2a( x - x0 )
we have
Then,
2
v BC
= 0 + 2( 4.8 m/s2 )(3 - 0) m
at B
= 28.8 m2 /s2
2
vD2 = v BC
+ 2aCD ( xD - xC )
and at D
d = 2.40 m b
or
For
AÆ B
and
C Æ D,
v = v0 + at
we have
Then
d = xd - xC
(7.2 m/s)2 = (28.8 m2 /s2 ) + 2( 4.8 m/s2 )d
or
(b)
( v BC = 5.3666 m/s)
AÆ B
5.3666 m/s = 0 + (4.8 m/s2 )t AB
t AB = 1.11804 s
or
and
or
CÆD
7.2 m/s = 5.3666 m/s + ( 4.8 m/s2 )tCD
tCD = 0.38196 s
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49
PROBLEM 11.38 (Continued)
Now,
for
we have
B Æ C,
xC = x B + v BC 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 b
or
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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 ) 42 = 90 km/h = 25 m/s
Patrol car:
For 18 s < t £ 26 s:
At t = 26 s:
or
Also,
At t = 26 s:
For 26 s < t £ 42 s :
At t = 42 s:
(b)
( vP )26 = 90 km/h = 25 m/s
vP = 0 + aP (t - 18)
25 m/s = aP (26 - 18) s
aP = 3.125 m/s2
1
xP = 0 + 0(t - 18) - aP (t - 18)2
2
1
(xP )26 = (3.125 m/s2 )(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
For the motorist’s car:
xM = 0 + vM t
At t = 42 s, x M = xP :
500 m = v M ( 42 s)
or
( xP ) 42 = 0.5 km b
v M = 11.9048 m/s
v M = 42.9 km/h b
or
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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:
1
x A = 0 + ( v A ) 0 t + a At 2
2
1
20 m = (12.9 m/s)(1.82 s) + aA (1.82 s)2
2
aA = -2.10 m/s2 b
or
Also
At t = 1.82 s:
v A = ( v A ) 0 + a At
(v A )1.82 = (12.9 m/s) + ( -2.10 m/s2 )(1.82 s)
= 9.078 m/s
For runner B:
v B2 = 0 + 2aB [ x B - 0 ]
When
x B = 20 m, v B = v A : (9.078 m/s)2 + 2aB (20 m)
or
aB = 2.0603 m/s2
aB = 2.06 m/s2 b
(b)
For runner B:
v B = 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/s2 )(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.
b
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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 0.54 and that B has
a constant deceleration of 0.36 determine (a) when and where
A will overtake B, (b) the speed of each automobile at that time.
SOLUTION
aA = +0.54 m/s2
aB = -0.36 m/s2
| v A |0 = 36 km/h = 10 m/s
| v B |0 = 54 km/h = 15 m/s
A overtakes B
Motion of auto A:
v A = ( v A )0 + a At = 10 + 0.54t (1)
1
1
x A = ( x A )0 + ( v A )0 t + aAt 2 = 0 + 10t + (0.54)t 2 2
2
(2)
v B = ( v B )0 + aB t = 15 - 0.36t (3)
1
1
x B = ( x B )0 + ( v B )0 t + aB t 2 = 22.5 + 15t + ( -0.36)t 2 2
2
(4)
Motion of auto B:
(a)
A overtakes B at t = t1 .
x A = x B : 10t1 + 0.27t12 = 22.5 + 15t1 - 0.18t12
0.45t12 - 5t1 - 22.5 = 0
t1 = -3.437 s and t1 = 14.548 s Eq. (2):
x A = 10(14.548) + 0.27(14.548)2 t1 = 14.548 s b
x A = .203 m b
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53
PROBLEM 11.41 (Continued)
(b)
Velocities when
Eq. (1):
t1 = 14.548 s
v A = 10 + (0.54)(14.548)
v A = 17.856 m/s Eq. (3):
v A = 64.3 km/h Æ b
v B = 15 - 0.36(14.548)
v B = 9.7627 m/s
v B = 35.1 km/h Æ b
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54
PROBLEM 11.42
In a boat race, boat A is leading boat B by 40 m
and both boats are traveling at a constant speed
of 160 km / h. At t = 0, the boats accelerate at
constant rates. Knowing that when B passes A,
t = 8 s and v A = 200 km/h, determine (a) the
acceleration of A, (b) the acceleration of B.
SOLUTION
(a)
We have
v A = ( v A ) 0 + a At
400
m/s
9
500
v A = 200 km/h =
m/s
9
500
400
m/s =
m/s + aA (8 s)
9
9
25
aA =
m/s2 18
1
x A = ( x A ) 0 + ( v A ) 0 t + a At 2
2
1
x B = 0 + ( v B ) 0 t + aB t 2
2
x A = xB
( v A )0 = 160 km/h =
At t = 8 s:
Then
or
(b)
We have
and
At t = 8 s:
aA = 1.39 m/s2 b
( x A )0 = 40 m
( v B )0 =
400
m/s
9
1 Ê 25 m ˆ
ˆ
Ê 400
2
40m + Á
m/s˜ (8 s) + Á
˜ (8 s)
¯
Ë 9
2 Ë 18 s2 ¯
1
Ê 400 m ˆ
=Á
(8 s) + aB (8 s)2
˜
Ë 9 s¯
2
or
aB = 2.64 m/s2 b
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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 t S = 1 s be the time when the boxes are 10 m apart.
aA = aB = a; ( x A )0 = ( x B )0 = 0; ( v A )0 = ( v B )0 = 0.
Let
(a)
1 2
at
2
1
t > t R , x B = a( t - t R ) 2
2
t > 0, x A =
For
For
At
t = tR , xA = 6 m
At
t = t R + tS ,
6=
1 2
at R 2
(1)
x A - x B = 10 m
1
1
a( t R + t S ) 2 - a( t R + t S - t R ) 2
2
2
1
1
1
= at R2 + at R t S + at S2 - at S2 = 18 + at R tS
2
2
2
10 =
at R =
10 - 6 4
= = 4 m/s tS
1
(2)
Dividing Equation (1) by Eq. (2),
2
1
2 at R
at R
(b)
1
6
= tR = 2
4
t R = 3.00 s b
Solving Eq. (2) for a,
a=
4
= 4 m/s2 1
a = 4 m/s2 b
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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 vA = 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)
1
x A = 0 + ( v A ) 0 t + a At 2
2
We have
At t = 40 s:
( v A )0 = 108 km/h = 30 m/s
1
1000 m = (30 m/s)(40 s) + aA ( 40 s)2
2
aA = -0.250 m/s2 b
or
1
x B = 0 + ( v B ) 0 t + aB t 2
2
Also,
At t = 42 s:
1
1000 m = (17.5 m/s)(42 s) + aB ( 42 s)2
2
aB = 0.30045 m/s2 or
(b)
( v B )0 = 63 km/h = 17.5 m/s
aB = 0.300 m/s2 b
When the cars pass each other
x A + x B = 1000 m
Then
or
Solving
1
2
(30 m/s)t AB + ( -0.250 m/s)t AB
+ (17.5 m/s)t AB
2
1
2
= 1000 m
+ (0.30045 m/s2 )t AB
2
2
2
0.05045t AB
+ 95t AB
- 2000 = 0
t = 20.822 s and t = -1904 s
t > 0 fi t AB = 20.8 s b
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57
PROBLEM 11.44 (Continued)
(c)
We have
v B = ( v B ) 0 + aB t
At t = t AB :
v B = 17.5 m/s + (0.30045 m/s2 )(20.822 s)
= 23.756 m/s
or
v B = 85.5 km/h b
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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 90 km / h. At t = 0, A starts and
accelerates at a constant rate aA, while
at t = 5 s, B begins to slow down with
a constant deceleration of magnitude
aA /6. Knowing that when the cars
pass each other x = 90 m and vA = vB,
determine (a) the acceleration aA,
(b) when the vehicles pass each other,
(c) the distance d between the vehicles
at t = 0.
SOLUTION
For t ≥ 0:
v A = 0 + a At
1
x A = 0 + 0 + a At 2
2
0 £ t < 5 s:
x B = 0 + ( v B )0 t ( v B )0 = 90 km/h = 25 m/s
At t = 5 s:
x B = (25m/s)(5 s) = 125 m
For t ≥ 5 s:
1
v B = ( v B )0 + aB (t - 5) aB = - aA
6
1
x B = ( x B )5 + ( v B )0 (t - 5) + aB (t - 5)2
2
Assume t > 5 s when the cars pass each other.
At that time (t AB ),
v A = vB :
aAt AB = (25 m /s) -
x A = 90 m:
90 m =
aA
(t AB - 5)
6
1
2
aAt AB
2
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59
PROBLEM 11.45 (Continued)
(
aA 73 t AB - 53
Then
2
t AB
Solving
With t AB > 5 s,
or
(b)
90
2
25t AB
- 210t AB + 150 = 0
or
(a)
) = 25
t AB = 0.788 s and t AB = 7.6117 s
90 m =
1
aA (7.6117)2
2
aA = 3.10677 m/s2 aA = 3.11 m/s2 b
t AB = 7.61 s b
From above
Note: An acceptable solution cannot be found if it is assumed that t AB £ 5 s.
(c)
We have
d = x + ( x B )t AB
1 Ê -3.10677 ˆ
È
2˘
= 90 m + Í125 m + 25 m/ s (7.6117 - 5) + Á
˜¯ (7.6117 - 5) ˙
Ë
2
6
Î
˚
d = 279 m b
or
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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 7 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 2A = ( v A )20 + 2a A [ x A - 0]
When
x A = ( x A ) max ,
Then
or
vA = 0
0 = (8 m/s)2 + 2aA (7 m )
aA =
-32
m/s2
7
or
Now
a A = 4.57 m/s2
b
a B = 1.082 m/s2
b
1
x A = 0 + ( v A ) 0 t + a At 2
2
1
x B = 0 + 0t + a B t 2
2
At t = 1 s, the blocks pass each other.
and
( x A )1 + ( x B )1 = d
At t = 3.4 s, x B = d :
Thus
( x A )1 + ( x B )1 = ( x B )3.4
or
È
Í(8 m /s)(1 s) +
Î
1 Ê -32 m ˆ
˘
(1 s)2 ˙
ÁË
2˜
¯
2 7 s
˚
È1
˘ 1
+ Í aB (1 s)2 ˙ = aB (3.4 s)2
Î2
˚ 2
or
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61
PROBLEM 11.46 (Continued)
(b)
At t = 3.4 s, x B = d :
1
d = (1.08225 m/s2 )(3.4 s)2
2
d = 6.26 m b
or
(c)
We have
v A = ( v A ) 0 + a At
At t = 1 s:
Ê -32 m ˆ
(1 s)
v A = 8 m/s + Á
Ë 7 s ˜¯
v A = 3.43 m/s b
or
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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 y B = constant
Then
v A + 3v B = 0
(1)
and
aA + 3aB = 0 (2)
(a)
Substituting into Eq. (1) 6 m/s + 3v B = 0
v B = 2 m/s ≠ b
or
(b)
From the diagram
y B + yD = constant
Then
v B + vD = 0
v D = 2 m/s Ø b
(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 ≠ b
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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 y B = constant
Then
v A + 3v B = 0
(1)
and
aA + 3aB = 0 (2)
(a)
Eq. (2): aA + 3aB = 0 and a B is constant and
positive fi a A is constant and negative
Also, Eq. (1) and ( v B )0 = 0 fi ( v A )0 = 0
Then
v 2A = 0 + 2aA [ x A - ( x A )0 ]
When |D x A | = 0.4 m:
(4 m/s)2 = 2aA (0.4 m)
a A = 20 m/s2 Æ b
or
Then, substituting into Eq. (2):
-20 m/s2 + 3aB = 0
aB =
or
20
m/s2 3
a B = 6.67 m/s2 Ø b
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64
PROBLEM 11.48 (Continued)
(b)
We have
v B = 0 + aB t
At t = 2 s:
ˆ
Ê 20
m/s2 ˜ (2 s)
vB = Á
¯
Ë 3
vB = 13.33 m/s Ø b
or
Also
At t = 2 s:
1
y B = ( y B ) 0 + 0 + aB t 2
2
1 Ê 20
ˆ
y B - ( y B )0 = Á
m/s2 ˜ (2 s)2
¯
Ë
2 3
yB - ( yB )0 = 13.33 m Ø b
or
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65
PROBLEM 11.49
The elevator shown in the figure moves downward with a constant velocity
of 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 = 5 m/s
But,
or
(b)
vC = -2vE = -10 m/s
vC = 10.00 m/s ≠ b
Velocity of counterweight W.
yW + yE = constant
vW + vE = 0 vW = - vE = -5 m/s (c)
vW = 5.00 m/s ≠ b
Relative velocity of C with respect to E.
vC /E = vC - vE = ( -10 m/s) - ( +5 m/s) = -15 m/s
vC /E = 15.00 m/s ≠ b
(d )
Relative velocity of W with respect to E.
vW /E = vW - vE = ( -5 m/s) - (5 m/s) = -10 m/s
vW /E = 10.00 m/s ≠ b
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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 = 0.8 m/s2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW = 0.800 m/s2 Ø
yW + yE = constant vW + vE = 0, and aW + aE = 0
aE = - aW = -(0.8 m/s2 ), a E = 0.800 m/s2 ≠ b
yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0
aC = -2aE = -2( -0.8 m/s2 ) = 1.6 m/s2 , aC = 1.600 m/s2 Ø b
Velocity of elevator after 5 s.
vE = ( vE )0 + aE t = 0 + ( -0.8 m/s2 )(5 s) = - 4 m/s ( vE )5 = 4.00 m/s ≠ b
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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 0.6 m /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 + y B + ( y B - y A ) = constant
Then
v A + 2v B = 0 (1)
and
aA + 2aB = 0
(2)
(a)
Eq. (1) and ( vA )0 = 0 = ( v B )0
Also, Eq. (2) and a A is constant and negative fi a B is constant
and positive
v A = 0 + a At
Then
v B = 0 + aB t
v B /A = v B - v A = ( aB - aA )t
Now
From Eq. (2)
1
aB = - a A
2
3
v B /A = - a A t
2
So that
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68
PROBLEM 11.51 (Continued)
At t = 8 s:
3
0.6 m/s = - a A (8 s)
2
1
m/s2 20
or
aA = -
and then
1Ê 1
ˆ 1
aB = - Á - m/s2 ˜ =
m/s2
¯ 40
2 Ë 20
a B = 0.250 m/s2 Ø b
or
(b)
At t = 6 s:
ˆ
Ê 1
v B = Á m/s2 ˜ (6 s) = 0.15 m/s
¯
Ë 40
vB = 0.1500 m/s Ø b
or
Now
At t = 6 s:
a A = 0.0500 m/s2 ≠ b
1
y B = ( y B ) 0 + 0 + aB t 2
2
1 Ê 1 mˆ
2
y B - ( y B )0 = Á
˜ (6 s) = 0.45 m
2 Ë 40 s2 ¯
yB - ( yB )0 = 0.450 m Ø b
or
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69
PROBLEM 11.52
In the position shown, collar B moves downward with a velocity of 0.3 m / 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 + y B = ( y B - y A ) = constant
Then
v A + 2v B = 0 (1)
and
aA + 2aB = 0
(2)
(a) Substituting into Eq. (1) v A + 2(0.3 m/s) = 0
vA = 0.600 m/s ≠ b
or
(b) From the diagram
2 y A + yC = constant
2v A + vC = 0
Then
Substituting
2( -0.6 m/s) + vC = 0
vC = 1.200 m/s Ø b
or
(c)
vC /B = vC - v B
We have
= (1.2 m/s) - (0.3 m/s)
vC /B = 0.900 m/s Ø b
or
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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
x B + ( x B - x A ) - 2 x A = constant
Then
2v B - 3v A = 0 (1)
and
2aB - 3aA = 0 (2)
Also, we have
vD + v A = 0 Then
(a)
- x B - x A = constant
Substituting into Eq. (1)
2(300 mm/s) - 3v A = 0
v A = 200 mm/s Æ 
or
(b)
From the diagram
Then
Substituting
(3)
x B + ( x B - xC ) = constant
2v B - vC = 0
2(300 mm/s) - vC = 0
vC = 600 mm/s Æ 
or
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71
PROBLEM 11.53 (Continued)
(c)
From the diagram
Then
Substituting
( xC - x A ) + ( x B - x A ) = constant
vC - 2v A + vD = 0
600 mm/s - 2(200 mm/s) + vD = 0
v D = 200 mm/s ¨ 
or
(d)
vC /A = vC - v A
We have
= 600 mm/s - 200 mm/s
vC /A = 400 mm/s Æ 
or
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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
x B + ( x B - x A ) - 2 x A = constant
From the diagram
Then
2v B - 3v A = 0 (1)
and
2aB - 3aA = 0 (2)
(a)
First observe that if block A moves to the right, v A Æ and Eq. (1) fi v B Æ . 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 fi aA = constant
v 2A = ( v A )20 + 2a A [ x A - ( x A )0 ]
Then
When x A - ( x A )0 = 240 mm:
or
(60 mm/s)2 = (100 mm/s)2 + 2aA (240 mm)
40
aA = mm/s2
3
a A = 13.33 mm/s2 ¨ 
or
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73
PROBLEM 11.54 (Continued)
Then, substituting into Eq. (2)
ˆ
Ê 40
mm/s2 ˜ = 0
2 aB - 3 Á ¯
Ë 3
aB = -20 mm/s2 or
(b)
a B = 20.0 mm/s2 ¨ 
From the solution to Problem 11.53
vD + v A = 0
Then
Substituting
aD + aA = 0
ˆ
Ê 40
mm/s2 ˜ = 0
aD + Á ¯
Ë 3
a D = 13.33 mm/s2 Æ 
or
(c)
We have
v B = ( v B ) 0 + aB t
At t = 4 s:
v B = 150 mm/s + ( -20.0 mm/s2 )( 4 s)
v B = 70.0 mm/s Æ 
or
Also
At t = 4 s:
1
y B = ( y B ) 0 + ( v B ) 0 t + aB t 2
2
y B - ( y B )0 = (150 mm/s)(4 s)
1
+ ( -20.0 mm/s2 )( 4 s)2
2
y B - ( y B )0 = 440 mm Æ 
or
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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 y B + xC = constant
Then
3v A + 4 v B + vC = 0 (1)
and
3aA + 4aB + aC = 0 (2)
v B = 20 mm/s Ø ;
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)
We have
At t = 3 s:
Now
or
1
xC = ( xC )0 + ( vC )0 t + aC t 2
2
1
57 mm = (10 mm/s)(3 s) + aC (3 s)2
2
2
aC = 6 mm/s
or ( vC )0 = 10 mm/s Æ 
aC = 6 mm/s2 Æ 
v B = constant Æ aB = 0
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75
PROBLEM 11.55 (Continued)
Then, substituting into Eq. (2)
3aA + 4(0) + (6 mm/s2 ) = 0
aA = -2 mm/s2
(c)
We have
At t = 5 s:
or a A = 2 mm/s2 ≠ 
1
y A = ( y A ) 0 + ( v A ) 0 t + a At 2
2
1
y A - ( y A )0 = ( -30 mm/s)(5 s) + ( -2 mm/s2 )(5 s)2
2
= -175 mm
y A - ( y A )0 = 175 mm ≠ 
or
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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/s2 . 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 y B + xC = constant
Then
3v A + 4 v B + vC = 0 (1)
and
3aA + 4aB + aC = 0 (2)
( v B ) = 0,
Given:
aA = constant
(aC ) = 75 mm/s2 Æ
v B = 480 mm/s Ø
vC = 280 mm/s Æ
At t = 2 s,
(a)
Eq. (2) and aA = constant and aC = constant fi aB = constant
v B = 0 + aB t
Then
At t = 2 s:
480 mm/s = aB (2 s)
aB = 240 mm/s2
or a B = 240 mm/s2 Ø 
3aA + 4(240 mm/s2 ) + (75 mm/s2 ) = 0
aA = -345 mm/s
or
a A = 345 mm/s2 ≠ 
Substituting into Eq. (2)
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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 Æ 
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:
1
xC = ( xC )0 + ( vC )0 t + aC t 2
2
1
xC - ( xC )0 = (130 mm/s)(3 s) + (75 mm/s2 )(3 s)2
2
= -728 mm
or
xC - ( xC )0 = 728 mm Æ 
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78
PROBLEM 11.57
Collar A starts from rest at t = 0 and moves downward with a constant
acceleration of 175 mm /s2. Collar B moves upward with a constant acceleration,
and its initial velocity is 200 mm/s. Knowing that collar B moves through
500 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 - y B ) = constant
Then
-2v A - v B + 4vC = 0 (1)
and
-2a A - aB + 4aC = 0 (2)
( v A )0 = 0
Given:
(a A ) = 175 mm/s2 Ø
( v B )0 = 200 mm/s ≠
a B = constant ≠
At t = 2 s
y - (y B )0 = 500 mm ≠
(a) We have
At t = 2 s:
1
y B = ( y B ) 0 + ( v B ) 0 t + aB t 2
2
1
-500 mm = ( -200 mm/s)(2 s) + aB (2 s)2
2
aB = -50 mm/s2
or a B = 50.0 mm/s2 ≠ 
Then, substituting into Eq. (2)
-2(175 mm/s2 ) - ( -50 mm/s2 ) + 4aC = 0
aC = 75 mm/s2
or aC = 75.0 mm/s2 Ø 
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79
PROBLEM 11.57 (Continued)
(b)
Substituting into Eq. (1) at t = 0
-2(0) - ( -200 mm/s) + 4( vC )0 = 0 or (vC )0 = -50 mm/s
vC = ( vC )0 + aC t
Now
(c)
When vC = 0:
0 = ( -50 mm/s) + (75 mm/s2 )t
or
t=
2
3
t = 0.667 s 
1
yC = ( yC )0 + ( vC )0 t + aC t 2
2
We have
At t =
2
s
3
s:
Ê2
yC - ( yC )0 = ( -50 mm/s) Á
Ë3
=-
50
mm = -16.667 mm
3
Ê2 ˆ
ˆ 1
s˜ + (75 mm/s2 ) Á s˜
Ë3 ¯
¯ 2
or 2
yC - ( yC )0 = 16.67 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 75t 2 mm/s2 . Knowing that collar B moves downward with a constant
acceleration and that its velocity is 200 mm /s after moving 800 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 - y B ) = constant
Then
-2v A - v B + 4vC = 0 (1)
and
-2aA - aB + 4aC = 0 (2)
( v A )0 = 0
Given:
( v B )0 = 0
a A = 75t 2 mm/s2 ≠
a B = constant Ø
y B - ( y B )0 = 800 mm Ø,
When
v B = 200 mm/s
(a)
We have
v B2 = 0 + 2aB [ y B - ( y B )0 ]
When y B - ( y B )0 = 800 mm:
(200 mm/s)2 = 2aB (800 mm)
aB = 25 mm/s2
or
Then, substituting into Eq. (2)
-2( -75t 2 mm/s2 ) - (25 mm/s2 ) + 4aC = 0
aC =
or
(b)
1
(25 - 150t 2 ) mm/s2 
4
Substituting into Eq. (1) at t = 0
-2(0) - (0) + 4( vC )0 = 0 or (vC )0 = 0
dvC
25
= aC = (1 - 6t 2 )
dt
4
Now
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81
PROBLEM 11.58 (Continued)
At t = 0, vC = 0 :
vC
Ú0
dvC = Ú
t 25
0
vC =
Thus,
vC = 0
25
t (1 - 2t 2 ) = 0
4
t = 0,
or
(1 - 6t 2 )dt
25
(t - 2t 3 )
4
or
At
4
t=
1
s
2
Therefore, block C initially moves downward ( vC > 0) and then moves upward ( vC < 0).
Now
At t = 0, yC = ( yC )0 :
or
At t =
1
2
s:
At t = 3 s:
Total distance traveled
dyC
25
= vC = (t - 2t 3 )
dt
4
yC
t 25
3
Ú( yC )0 dyC = Ú0 4 (t - 2t )dt
25
yC - ( yC )0 = (t 2 - t 4 )
8
yC - ( yC )0 =
2
4
25 ÈÊ 1 ˆ
Ê 1 ˆ ˘ 25
mm
ÍÁ
˜ ˙=
ÁË
8 ÍÎË 2 ˜¯
2 ¯ ˙˚ 32
yC - ( yC )0 =
25
[(3)2 - (3) 4 ] = -225 mm
8
25
Ê 25 ˆ
= 226.5625 mm
= Á ˜ + -225 Ë 32 ¯
32
= 227 mm 
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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/s2 upward and the relative acceleration of block D with respect to block
A is 110 mm/s2 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
Cable 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2v B + vC = 0 (1)
and
2aA + 2aB + aC = 0 (2)
Cable 2:
( yD - y A ) + ( yD - y B ) = constant
Then
- v A - v B + 2v D = 0 (3)
and
- aA - aB + 2aD = 0 (4)
Given: At t = 0, v = 0; all accelerations constant;
aC /B = 60 mm/s2 ≠, aD /A = 110 mm/s2 Ø
(a)
We have aC /B = aC - aB = -60 or aB = aC + 60
aD /A = aD - aA = 110 or aA = aD - 110
and
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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83
PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and aD
aC = 40 mm/s2
aD = -10 mm/s2
Now
vC = 0 + aC t
At t = 3 s:
vC = ( 40 mm/s2 )(3 s)
vC = 120 mm/s Ø 
or
(b)
We have
At t = 5 s:
1
yD = ( yD )0 + (0)t + aD t 2
2
1
yD - ( yD )0 = ( -10 mm/s2 )(5 s)2
2
y D - ( y D )0 = 125 mm ≠ 
or
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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/s2 , (b) the change in position of block D
when the velocity of block C is 600 mm /s upward.
SOLUTION
From the diagram
Cable 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2v B + vC = 0 (1)
and
2aA + 2aB + aC = 0 (2)
( yD - y A ) + ( yD - y B ) = constant
Cable 2:
Then
- v A - v B - 2v D = 0 (3)
and
- aA - aB + 2aD = 0 (4)
t=0
v=0
( y A )0 = ( y B )0 = ( yC )0
Given: At
All accelerations constant at t = 2 s
yC /A = 280 mm ≠
v B /A = 80 mm/s Ø
When
y A - ( y A )0 = 160 mm ≠
y B - ( y B )0 = 320 mm Ø
aB > 10 mm/s2
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85
PROBLEM 11.60* (Continued)
(a)
1
y A = ( y A )0 + (0)t + aAt 2
2
1
yC = ( yC )0 + (0)t + aC t 2
2
1
yC /A = yC - y A = ( aC - aA )t 2
2
We have
and
Then
At t = 2 s, yC /A = -280 mm:
or
1
-280 mm = ( aC - aA )(2 s)2
2
aC = a A - 140 (5)
Substituting into Eq. (2)
2aA + 2aB + ( aA - 140) = 0
or
1
aA = (140 - 2aB ) 3
Now
v B = 0 + aB t
(6)
v A = 0 + a At
v B /A = v B - v A = ( aB - aA )t
Also
When
1
y B = ( y B )0 + (0)t + aB t 2
2
v B /A = 80 mm/s Ø : 80 = ( aB - aA )t (7)
1
a At 2
2
1
Dy B = 320 mm Ø : 320 = aB t 2
2
1
160 = ( aB - aA )t 2
2
Dy A = 160 mm Ø : 160 =
Then
Using Eq. (7)
Then
and
320 = (80)t
or t = 4 s
1
a A ( 4)2
2
1
320 = aB ( 4)2
2
160 =
or a A = 20 mm/s2Ø 
or a B = 40 mm/s2 Ø 
Note that Eq. (6) is not used; thus, the problem is over-determined.
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86
PROBLEM 11.60* (Continued)
Alternative solution:
We have
v 2A = (0) + 2aA [ y A - ( y A )0 ]
v B2 = (0) + 2aB [ y B - ( y B )0 ]
Then
v B /A = v B - v A = 2 a B [ y B - ( y B ) 0 ] - 2 a A [ y A - ( y A ) 0 ]
When
v B /A = 80 mm/s Ø :
80 mm/s = 2 ÈÎ aB (320 mm) - aA (160 mm) ˘˚
20 = 2
or
(
)
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/s2
and into Eq. (4)
-(20 mm/s2 ) - ( 40 mm/s2 ) + 2aB = 0
or
aD = 30 mm/s2
Now
vC = 0 + aC t
When vC = -600 mm/s:
or
Also
At t = 5 s:
-600 mm/s = ( -120 mm/s2 )t
t =5s
1
yD = ( yD )0 + (0)t + aD t 2
2
1
2
yD - ( yD )0 = (30 mm/s )(5 s)2
2
y D - ( y D )0 = 375 mm Ø 
or
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87
PROBLEM 11.61
A subway car leaves station A; it gains speed at the rate of 4 m /s2
for 6 s and then at the rate of 6 m /s2 until it has reached the speed of
36 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:
0 < t < 6:
Change in v = area under a - t curve
v6 - 0 = (6 s)( 4 m/s2 ) = 24 m / s
6 < t < t2:
Since the velocity increases from 24 m /s to 36 m /s,
Change in v = area under a - t curve
36 m/s - 24 m/s = (t2 - 6)(6 m/s2 ) t2 = 8 s
t2 < t < 34:Since the velocity is constant the acceleration is
zero.
34 < t < 40:
Change in v = area under a - t curve
0 - 36 m/s = (6 s)a4
a4 = - 6 m/s2
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.
Change in x = area under v - t curve
1
x6 - 0 = (6)(24) = 72 m
0 < t < 6:
2
1
x10 - x6 = ( 4)(24 + 36) = 120 m
6 < t < 8:
2
8 < t < 34 :
x34 - x10 = ( 42)(36) = 1512 m
34 < t < 40 :
1
x40 - x34 = (6)(36) = 108 m
2
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88
PROBLEM 11.61 (Continued)
Adding the changes in x, we obtain the distance from A to B:
d = x40 - 0 = 1812 m
d = 1812 m 
Position-Time Curve. The points determined above should
be joined by three arcs of parabola and one straight-line
segment.
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89
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)
Initial conditions:
t = 0, v0 = - 6 m/s, x0 = 0
Change in v equals area under a - t curve :
v0 = -6 m/s
2
v4 - v0 = (1 m/s )( 4 s) = 4 m/s
v4 = -2 m/s
4 s t 10 s :
v10 - v4 = (2 m/s2 )(6 s) = +12 m/s
v10 = +10 m/s
10 s t 12 s :
v12 - v10 = ( -2 m/s2 )(2 s) = -4 m/s
v12 = +6 m/s 12 s t 20 s :
v20 - v12 = ( -2 m/s2 )(8 s) = -16 m/s
v20 = -10 m/s
0 t 4 s:

Change in x equals area under v - t curve :
x0 = 0
0 t 4 s:
4 s t 5 s:
5 s t 10 s :
10 s t 12 s :
1
x4 - x0 = ( -6 - 2)( 4) = -16 m
2
1
x5 - x4 = ( -2)(1) = -1 m
2
1
x10 - x5 = ( +10)(5) = +25 m
2
1
x12 - x10 = ( +10 + 6)(2) = +16 m
2
x4 = -16 m
x5 = -17 m
x10 = +8 m
x12 = +24 m 
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90
PROBLEM 11.62 (Continued)
12 s < t < 15 s :
1
x16 - x12 = ( +6)(3) = +9 m
2
x16 = +33 m
15 s < t < 20 s :
1
x20 - x15 = ( -10)(5) = -25 m
2
x20 = +8 m
From v - t and x - t curves, we read
(a)
At t = 10 s: vmax = +10.00 m/s 
(b)
At t = 15 s: xmax = +33.0 m 
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91
PROBLEM 11.63
A particle moves in a straight line with the velocity
shown in the figure. Knowing that x = -180 m 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 fi a = 0
-10 - 20 -30
=
= -1.875 m/s2
26 - 10
16
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant fi a = 0
t = 41 s to t = 46 s:
a=
t > 46 s:
-4 - ( -10)
= 1.2 m/s2
46 - 41
v = constant fi a = 0
x2 = x1 + (area under v - t curve from t1 to t2 )
At t = 10 s:
x10 = -180 + 10(20) = +20 m
Next, find time at which v = 0. Using similar triangles
t v = 0 - 10
20
At
t=
62
s:
3
=
26 - 10
30
or
tv = 0 =
62
s
3
1 Ê 62
380
ˆ
x 62 = 20 + Á - 10˜ (20) =
m
¯
2Ë 3
3
3
380 1 Ê
62 ˆ
- Á 26 - ˜ (10) = 124 m
3 2Ë
3¯
t = 26 s :
x26 =
t = 41 s :
x41 = 124 - 15(10) = -26 m
t = 46 s :
Ê 10 + 4 ˆ
x46 = -26 - 5 Á
= -61 m
Ë 2 ˜¯
t = 50 s :
x50 = -61 - 4( 4) = -77 m
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92
PROBLEM 11.63 (Continued)
(b)
From t = 0 to t =
62
380
s: Distance traveled =
- ( -180)
3
3
920
=
m
3
t = 22 s to t = 50 s: Distance traveled = - 77 =
380
3
611
m
3
1531
Ê 920 611ˆ
m
Total distance traveled = Á
+
˜ m=
Ë 3
3 ¯
3
= 510.33 m (c)
510 m 
Using similar triangles
Between 0 and 10 s :
(t x= 0 )1 - 0 10
=
180
200
(t x= 0 )1 = 9.00 s 
Between 26 s and 41 s:
constant velocity
124 m = 10( Dt )
Dt = 12.4
t = (26 + 12.4) s = 38.4 s
or
(t x= 0 )2 = 38.4 s 
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93
PROBLEM 11.64
A particle moves in a straight line with the velocity
shown in the figure. Knowing that x = -180 m 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 = 30 m.
SOLUTION
(a)
at = slope of v - t curve at time t
From t = 0 to t = 10 s:
v = constant fi a = 0
-10 - 20 -30
=
= -1.875 m/s2
26 - 10
16
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant fi a = 0
t = 41 s to t = 46 s:
a=
t > 46 s:
- 4 - ( -10)
= 1.2 m/s2
46 - 41
v = constant fi a = 0
x2 = x1 + (area under v - t curve from t1 to t2 )
At t = 10 s:
x10 = -180 + 10(20) = +20 m
Next, find time at which v = 0. Using similar triangles
t v = 0 - 10 26 - 10
62
or
tv = 0 =
s
=
20
30
3
At
t=
62
s:
3
1 Ê 62
380
ˆ
x 62 = 20 + Á - 10˜ (20) =
m
¯
2Ë 3
3
3
380 1 Ê
62 ˆ
- Á 26 - ˜ (10) = 124 m
3
2Ë
3¯
t = 26 s :
x26 =
t = 41 s :
x41 = 124 - 15(10) = -26 m
t = 46 s :
Ê 10 + 4 ˆ
x46 = -26 - 5 Á
= -61 m
Ë 2 ˜¯
t = 50 s :
x50 = -61 - 4( 4) = -77 m
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94
PROBLEM 11.64 (Continued)
(b)
Reading from the x - t curve:
(c)
Between 10 s and 62/3 s
30 m =
380
m = 126.667 m 3
xmax = 126.7 m 
380
62
m - (area under v - t curve from t , to
s) m
3
3
or
or
30 =
380 1 Ê 62
ˆ
- Á - t1 ˜ ( v1 )
¯
Ë
3
2 3
580
ˆ
Ê 62
ÁË - t1 ˜¯ ( v1 ) =
3
3
Using similar triangles
v1
20
=
Ê 62
ˆ 32 / 3
ÁË - t1 ˜¯
3
Then
or
v1 =
15 Ê 62
ˆ
Á - t1 ˜¯
8Ë 3
ˆ ˘ 580
ˆ ÈÊ 15 ˆ Ê 62
Ê 62
ÁË - t1 ˜¯ ÍÁË ˜¯ ÁË - t1 ˜¯ ˙ =
3
3
3
Î 8
˚
t1 = 10.512 s and t1 = 30.82 s
or
We have
10 s t1 22 s fi t1 = 10.52 s 
Between 26 s and 41 s:
Using similar triangles
41 - t2 15
=
56
150
t2 = 35.4 s 
or
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95
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 D x = area under v - t curve for given time interval
Then
or
or
or
Ê 55.555 + 13.888 ˆ
(586 - 600) m = -t1 Á
˜¯ m/s
Ë
2
t1 = 0.4032 s
(30 - 586) m = -t2 (13.888 m/s)
t2 = 40.0346 s
1
(0 - 30) m = - (t3 )(13.888 m/s)
2
t3 = 4.3203 s
t total = (0.4032 + 40.0346 + 4.3203) s
t total = 44.8 s 
or
(b)
We have
ainitial
D vinitial
=
t1
=
[-13.888 - ( -55.555)] m/s
0.4032 s
= 103.3 m /s2
a initial = 103.3 m/s2 ≠ 
or
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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 painting
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
v paint = consant, 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
Ê v - 80 ˆ
a1 Á = max
˜¯ is maximum.
t1
Ë
Thus,
or
and
( vmax - 80) mm/s
= 60 mm/s
t1(s)
( v - 80)
t1 = max
60
Ê 80 + vmax ˆ
t1 Á
˜¯ + (20 - t1 )( vmax ) = 4000
Ë
2
Substituting for t1
( vmax - 80) Ê 80 + vmax ˆ Ê
vmax - 80 ˆ
˜ vmax = 4000
ÁË
˜¯ + ÁË 20 60 ¯
60
2
2
vmax
- 2560 vmax + 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
PROBLEM 11.67
A temperature sensor is attached to slider AB, which moves back
and forth through 1500 mm. The maximum velocities of the
slider are 300 mm /s to the right and 750 mm /s to the left. When
the slider is moving to the right, it accelerates and decelerates
at a constant rate of 150 mm /s2; when moving to the left, the
slider accelerates and decelerates at a constant rate of 500 mm /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
=
300 mm/s
150 mm/s2
=2s
vleft 750 mm/s
=
aleft 500 mm/s2
= 1.5 s
td =
A1 = 1500 mm
Now
or
[(t1 - 2) s](300 mm/s) = 1500 mm
t1 = 7 s
or
A2 = 1500 mm
and
or
{[(t2 - 7) - 1.5] s}(750 mm/s) = 1500 mm
t2 = 10.5 s
or
tcycle = t2 Now
tcycle = 10.5 s 
We have xii = xi + (area under v - t curve from ti to tii )
1
At
t = 2 s:
x2 = (2)(300) = 300 mm
2
t = 5 s:
x5 = 300 + (5 - 2)(300)
= 1200 mm
t = 7 s:
t = 8.5 s :
t = 9 s:
t = 10.5 s :
x7 = 1500 mm
1
x8.5 = 1500 - (1.5)(750)
2
= 937.5 mm
x9 = 937.5 mm - (0.5)(750)
x10.5
= 562.5 mm
=0
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98
PROBLEM 11.68
A commuter train traveling at 60 km / h is
4.5 km from a station. The train then decelerates so that its speed is 30 km / h when
it is 0.75 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
3.75 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 = 60 km/h x = 0; when x = 3.75 km, v = 30 km/h;
at t = 7.5 min, x = 4.5 km; constant decelerations
The v - t curve is first drawn is shown.
(a)
A1 = 3.75 km
We have
or
1h
Ê 60 + 30 ˆ
km/h ¥
= 3.75 km
(t1 min) Á
˜
Ë 2 ¯
60 min
t1 = 5 min 
or
(b)
A2 = 0.75 km
We have
or
1h
Ê 30 + v2 ˆ
km/h ¥
= 0.75 km
(7.5 - 5) min ¥ Á
Ë 2 ˜¯
60 min
v2 = 6 km/h 
or
(c)
We have
afinal = a12
=
(6 - 30) km/h 1000 m 1 min
1h
¥
¥
¥
(7.5 - 5) min
km
60 s 3600 s
afinal = -0.0444 m/s2 
or
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99
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 = v B = 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.
5
Note:
8 min 20 s =
h
36
At
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) - (t a )(100) - (tb )(30) = 8
2
2
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100
PROBLEM 11.69 (Continued)
Substituting
1 Ê 100 ˆ
1 Ê 30 ˆ
(100) - Á ˜ (30) = 8
100t1 - Á
˜
2Ë a ¯
2Ë a ¯
t1 = 0.08 +
or
Also A2 = 4 km:
Substituting
or
Then
or
1
ˆ
Ê 5
ÁË - t1 ˜¯ (70) - (tc )(70) = 4
2
36
54.5
a
1 Ê 70 ˆ
ˆ
Ê 5
ÁË - t1 ˜¯ (70) - ÁË ˜¯ (70) = 4
36
2 a
t1 =
0.08 =
103 35
1260 a
54.5 103 35
=
a
1260 a
1000 m Ê 1 h ˆ
a = 51.259 km/h ¥
¥Á
km
Ë 3600 s ˜¯
2
2
a = 3.96 m/s2 
or
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101
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/s2 at t = 0 to -2 m /s2 at
t = t1 and then remains equal to -2 m/s2 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/s2 ;
at
t = 0 a = -12 m/s2 ; at t = t1
a = -2 m/s2 , v = 1.8 m/s2
The a - t and v - t curves are first drawn as shown.
(a)
vt1 = v0 + A1
We have
or
Ê 12 + 2 ˆ
1.8 m/s = 6 m/s - (t1 s) Á
m/s2
Ë 2 ˜¯
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/s2
v1.4 = 0.20 m/s 
or
Now x1.4 = A3 + A4 , where A3 is most easily determined using integrating.
Thus,
50
-2 - ( -12)
for 0 £ t £ t1:
a=
t - 12 = t - 12
0.6
3
dv
Now
=a
dt
50
= t - 12
3
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102
PROBLEM 11.70 (Continued)
At t = 0, v = 6 m/s:
or
We have
Then
t Ê 50
v
Ú6 dv = Ú0 ÁË
3
ˆ
t - 12˜ dt
¯
25 2
t - 12t
3
dx
25
= v = 6 - 12t + t 2
dt
3
v =6+
A3 = Ú
xt1
0
dx = Ú
0.6
0
(6 - 12t +
25 2
t )dt
3
0.6
25 ˘
È
= Í6t - 6t 2 + t 3 ˙ = 2.04 m
9 ˚0
Î
Also
Then
Ê 1.8 + 0.2 ˆ
A4 = (1.4 - 0.6) Á
˜ = 0.8 m
Ë
2 ¯
x1.4 = (2.4 + 0.8) m
x1.4 = 2.84 m 
or
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103
PROBLEM 11.71
A car and a truck are both traveling at the constant
speed of 50 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
50 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 75 km/h ? Draw the
v - t curve.
SOLUTION
Relative to truck, car must move a distance:
Allowable increase in speed:
D x = 4.8 + 12 + 15 + 12 = 43.8 m
125
Dvm = 75 - 50 = 25 km/h =
m/s
18
Acceleration Phase:
t1 =
125/18 125
=
s = 4.6296 s
1.5
27
1 Ê 125 ˆ Ê 125 ˆ
A1 = Á
˜ = 16.075 m
˜Á
2 Ë 18 ¯ Ë 27 ¯
Deceleration Phase:
t3 =
125 /18 125
=
s = 1.1574 s
6
108
1 Ê 125 ˆ Ê 125 ˆ
A3 = Á
˜ = 4.0188 m
˜Á
2 Ë 18 ¯ Ë 100 ¯
But: D x = A1 + A2 + A3 :
43.8 = 16.075 + (125 /18)t2 + 4.0188
t total = t1 + t2 + t3 = 4.6296 s + 3.414 s + 1.15745 s = 9.201 s t2 = 3.414 s
t F = 9.20 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 50 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:
D x = 4.8 + 12 + 15 + 12 = 43.8 m
Dvm = 1.5t1 = 6t2 ;
D x = A1 + A2 :
t2 =
1
t1
4
1
43.8 m = ( Dvm )(t1 + t2 )
2
t ˆ
1
Ê
43.8 m = (1.5t1 ) Á t1 + 1 ˜
Ë
2
4¯
t12 = 46.72 t1 = 6.835 s t2 =
1
t1 = 1.709
4
t total = t1 + t2 = 6.835 + 1.709 t F = 8.54 s 
Dvm = 1.5t1 = 1.5(6.835) = 10.2525 m/s = 36.909 m/s
Speed vtruck = 50 km/h, vm = 50 km/h + 36.91 km/h vm = 86.9 km/h 
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105
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/s2 ≠; For vE = 7.8 m/s, aE = 0;
At t = 2 s, v B = 20 m/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/s2 , while the slope of the curve for the ball is - g ( -9.81 m/s2 ).
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/s2 )t1
or
t1 = 6.5 s
The time t top is the time at which the ball reaches the top of its trajectory.
Thus,
or
or
v B = ( v B )0 - g (t - 2)
0 = 20 m/s - (9.81 m/s2 )(t top - 2) s
t top = 4.0387 s
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106
PROBLEM 11.73 (Continued)
Using the coordinate system shown, we have
0 £ t £ t1:
At t = t top:
and
At
and at t = t1 ,
ˆ
Ê1
yE = -12 m + Á aE t 2 ˜ m
¯
Ë2
1
y B = ( 4.0387 - 2) s ¥ (20 m /s)
2
= 20.387 m
1
yE = -12 m + (1.2 m/s2 )( 4.0387 s)2
2
= -2.213 m
t = [2 + 2( 4.0387 - 2)] s = 6.0774 s, y B = 0
1
yE = -12 m + (6.5 s)(7.8 m/s) = 13.35 m
2
The ball hits the elevator ( y B = yE ) when t top < t < t1 .
For t ³ t top:
È1
˘
y B = 20.387 m - Í g (t - t top )2 ˙ m
Î2
˚
Then,
when
y B = yE
1
20.387 m - (9.81 m /s2 )(t - 4.0387)2
2
1
= -12 m + (1.2 m /s2 )(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
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:
= D 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 
Dt = 8 s
D x = (60 m /s)(14 s - 6 s) - (shaded area)
= (60 m /s)(8 s) - 6 m = 480 m - 6 m = 474 m
vaverage =
D x 474 m
=
8s
Dt
vaverage = 59.25 m /s 
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108
PROBLEM 11.75
Car A is traveling on a highway at a constant
speed ( v A )0 = 90 km / h and is 120 m from
the entrance of an access ramp when car B
enters the acceleration lane at that point at
a speed ( v B )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 90 km/h, which it then maintains.
Determine the final distance between the two
cars.
SOLUTION
Given:
( v A )0 = 90 km / h, (v B )0 = 25 km / h; at t = 0,
( x A )0 = -120 m, ( x B )0 = 0; at t = 5 s,
x B = 60 m; for 25 km / h £ v B < 90 km/ h,
aB = constant; for v B = 90 km / h,
aB = 0
First note
90 km / h = 25 m /s
25 km / h = 125 /18 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, x B = 60 m:
È (125/18 + v B )5 ˘
(5 s) Í
˙ m /s = 60 m
2
Î
˚
( v B )5 =
or
307
m/s = 17.0556 m/s
18
Then, using similar triangles, we have
(25 - 125 /18) m/s (307 /18 - 125 /18) m/s
=
( = aB )
t1
5s
or
t1 = 8.9286 s
Finally, at t = t1
È
˘
Ê 25 + 125/18 ˆ
x B /A = x B - x A = Í(8.9286 s) Á
m/s ˙
˜
Ë
¯
2
Î
˚
- [-120 m + (8.9286 s) (25 m/s)]
or
x B /A = 39.4 m 
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109
PROBLEM 11.76
Car A is traveling at 60 km / h when it
enters a 40 km / h speed zone. The driver
of car A decelerates at a rate of 5 m /s2 until
reaching a speed of 40 km / h, which she
then maintains. When car B, which was
initially 20 m behind car A and traveling
at a constant speed of 70 km / h, enters
the speed zone, its driver decelerates at
a rate of 6 m /s2 until reaching a speed of
35 km / h. Knowing that the driver of car B
maintains a speed of 35 km / h, determine
(a) the closest that car B comes to car A,
(b) the time at which car A is 25 m in
front of car B.
SOLUTION
( v A )0 = 60 km/h; For 40 km/h < v A £ 60 km/h;
Given:
aA = -5 m/s2 ; For v A = 40 km/h, aA = 0;
( x A/B )0 = 20 m; ( v B )0 = 70 km/h;
when x B = 0, aB = - 6 m/s2 ; for v B = 35 km/h,
aB = 0
100
m/s
9
175
35 km/h =
m/s
18
First note 60 km/h = 50 / 3 m/s, 40 km/h =
70 km/h =
At t = 0
175
m/s
9
The v - t curves of the two cars are as shown.
At
t = 0:
car A enters the speed zone
t = (tB)1:
car B enters the speed zone
t = tA:
car A reaches its final speed
t = t min:
vA = vB
t = (tB) 2: car B reaches its final speed
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110
PROBLEM 11.76 (Continued)
(a)
aA =
We have
( v A )final - ( v A )0
tA
Ê 100 50 ˆ
- ˜ m/s
ÁË
9
3¯
-5 m/s2 =
tA
or
t A = 1.1111 s
or
Also
20 m = (t B )1 ( v B )0
or
ˆ
Ê 175
20 m = (t B )1 Á
m/s˜
¯
Ë 9
aB =
and
or
(t B )1 = 1.0286 s
( v B )final - ( v B )0
(t B )2 - (t B )1
Ê 175 175 ˆ
ÁË
˜ m/s
18
9 ¯
2
- 6 m/s =
[(t B )2 - 1.0286] s
or
Car B will continue to overtake car A while v B > v A . Therefore, ( x A/B ) min will occur when v A = v B ,
which occurs for
(t B )1 < t min < (t B )2
For this time interval
100
m/s
9
v B = ( v B )0 + aB [t - (t B )1 ]
vA =
Then at t = t min:
100 175
=
+ ( - 6 m/s2 )(t min - 1.0286) s
9
9
or
t min = 2.4175 s
Finally
( x A/ B ) min = ( x A )tmin - ( x B )tmin
¸
Ï È ( v ) + ( v A )final ˘
= Ìt A Í A 0
˙ + (t min - t A )( v A )final ˝
2
˚
˛
Ó Î
Ï
È ( v ) + ( v A )final ˘ ¸
- Ì( x B )0 + (t B )1 ( v B )0 + [t min - (t B )1 ] Í B 0
˙˝
2
Î
˚˛
Ó
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111
PROBLEM 11.76 (Continued)
È
˘
Ê 50 100 ˆ
100
+
˙
m/s
2
4175
1
1111
s
m/s
+
(
.
.
)
¥
= Í(1.1111 s) Á 3
˜
9
Í
˙
9
ÁË
˜¯
2
Î
˚
È
˘
Ê 175 100 ˆ
ˆ
Ê 175
+
m/s˜ + (2.4175 - 1.0286) s ¥ Á 9
- Í-20 m + (1.0286 s) Á
˜ m/s ˙
9
¯
Ë 9
Í
˙
ÁË
˜¯
2
Î
˚
= (15.432 + 14.516) m - ( -20 + 20.00 + 21.219) m
= 8.729 m
(b)
or
(xA/B)min = 8.73 m 
Since ( x A/B ) £ 20 m for t £ t min , it follows that x A/B = 25 m for t > (t B )2
[Note (t B )2 t min ]. Then, for t > (t B )2
x A/B = ( x A/B ) min + [(t - t min )( v A )final ]
+ ( v B )final ˘
¸
Ï
È (v )
+ [t - (t B )2 ]( v B )final ˝
- Ì[(t B )2 - (t min )] Í A final
˙
2
˚
Î
˛
Ó
100
È
˘
or 25 m = 8.729 m + Í(t - 2.4175) s ¥
m/s ˙
9
Î
˚
È
˘
Ê 100 175 ˆ
Ê 175 ˆ
+
m/s ˙
- Í(2.6489 - 2.4175) s ¥ Á 9
18 ˜ m/s + (t - 2.6489) s ¥ ÁË
˜
Í
˙
ÁË
˜¯
18 ¯
2
Î
˚
t = 14.25 s 
or
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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:
At
t = 0, x = 0, v = 54 km / h;
for t = t1,
v = 54 km/h
First note
(a)
We have
Then
54 km/h = 15 m/s
vb = va + (area under a-t curve from t a to tb )
at t = 2 s:
t = 4.5 s:
t = t1:
v = 15 - (1)(6) = 9 m/s
1
v = 9 - (2.5)(6) = 1.5 m/s
2
1
15 = 1.5 + (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
PROBLEM 11.77 (Continued)
Now
x at t = 18 s
x18 = 0 + S (area under the v - t curve from t = 0 to t = 18 s)
Ê 15 + 9 ˆ
m/s
= (1 s)(15 m/s) + (1 s) Á
Ë 2 ˜¯
1
È
˘
+ Í(2.5 s)(1.5 m/s) + (2.5 s)(7.5 m/s)˙
3
Î
˚
2
È
˘
+ Í(13.5 s)(1.5 m/s) + (13.5 s)(13.5 m/s) ˙
3
Î
˚
= [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 =
Dx (178.75 - 15) m
= 9.6324 m/s
=
Dt
(18 - 1) s
vave = 34.7 km/h 
or
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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: At
(a)
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.
Now at t = 2 s : x = 0 +
t = 4 s: x = 4 -
(2)(8)
=4m
3 +1
(2)(8)
=0
3 +1
The x-t curve is then drawn as shown.
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115
PROBLEM 11.78 (Continued)
(b)
We have
At t = 3 s: a = -3(3 - 2)2 = - 3 m/s2
1
v = 0 - (1)(3) = -1 m/s
3
(1)(1)
x =43 +1
x3 = 3.75 m 
or
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116
PROBLEM 11.79
During a manufacturing process, a conveyor belt starts from rest and travels a total of 400 mm before temporarily
coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ± 1.5 m/s2 per second,
determine (a) the shortest time required for the belt to move 400 mm, (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 = 400 mm = 0.4 m
when
Ê da ˆ
= 1.5 m /s3
x = xmax , v = 0; Á ˜
Ë dt ¯ max
Observing that vmax must occur at t = 12 t min , the a - t curve must have the shape shown. Note that the
magnitude of the slope of each portion of the curve is 1.5 m/s2 / s.
We have
at t = Dt :
t = 2Dt :
1
1
v = 0 + ( Dt )( amax ) = amax Dt
2
2
vmax =
1
1
amax Dt + ( Dt )( amax ) = amax Dt
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
(2Dt )( vmax ) = xmax
vmax = amax Dt fi 2amax Dt 2 = xmax
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117
PROBLEM 11.79 (Continued)
Now
amax
= 1.5 m/s2 /s so that
Dt
2(1.5Dt m/s3 ) Dt 2 = 0.4 m
or Dt = 0.5109 s
Then
t min = 4Dt
t min = 2.04 s 
or
(b)
We have
vmax = amax D t
= (1.5 m/s2 /s ¥ D t ) D t
= 1.5 m/s2 /s ¥ (0.5109s)2
vmax = 0.392 m/s 
or
Also
vave =
Dx
0.4 m
=
Dt total 2.04 s
vave = 0.196 m/s 
or
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118
PROBLEM 11.80
An airport shuttle train travels between two terminals that are 2.5 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
± 0.24 m / s2 per second. If the shuttle has a maximum speed of 30 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
Given:
xmax = 2.5 km; | amax| = 1.2 m /s2
Ê da ˆ
= 0.24 m/s2 /s; vmax = 30 km/h
ÁË ˜¯
dt max
First note
(a)
25
m/s
3
2.5 km = 2500 m
30 km/h =
To obtain tmin, the train must accelerate and decelerate at the maximum rate to maximize the
time for which v = vmax . The time Dt required for the train to have an acceleration of 1.2 m / s2 is found
from
a
Ê da ˆ
= max
ÁË ˜¯
Dt
dt max
1.2 m/s2
or
Dt =
or
Dt = 5 s
0.24 m/s2 /s
Now,
da
1
ˆ
Ê
v5 = ( Dt )( amax ) Á since
= constant˜
¯
Ë
dt
2
1
or
v5 = (5 s)(1.2 m/s2 ) = 3 m/s
2
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 0.24 m /s2/s.
after 5 s, the speed of the train is
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119
PROBLEM 11.80 (Continued)
Now
at t = (10 + Dt1 ) s, v = vmax :
25
È1
˘
2 Í (5 s)(1.2 m/s2 )˙ + ( Dt1 )(1.2 m/s2 ) =
m/s
2
3
Î
˚
or
Dt1 = 1.9444 s
Then
at t = 5 s :
1
v = 0 + (5)(1.2) = 3 m/s
2
t = 6.9444 s : v = 3 + (1.9444)(1.2) = 5.33328 m/s
1
t = 11.9444 s : v = 5.33328 + (5)(1.2) = 8.33328 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
È
˘
Ê 3 + 5.33328 ˆ
2 Í(1.9444 s) Á
˜¯ m/s ˙
Ë
2
Î
˚
+ (10 + Dt2 ) s ¥ (8.33328 m/s) = 2500 m
or
Dt2 = 288.06 s
Then
t min = 4(5 s) + 2(1.9444 s) + 288.06 s
= 311.95 s
t min = 5 min11.95 s 
or
(b)
We have
vave =
Dx
2500 m
=
= 8.014 m/s = 28.850 km/h
Dt 311.95 s
vave = 28.9 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
t = 0, x = 0, v = 0
1.The a-t curve is first approximated with a series of rectangles, each of width Dt = 2 s. The area ( Dt )( aave )
of each rectangle is approximately equal to the change in velocity Dv for the specified interval of time.
Thus,
Dv aave Dt
where the values of aave and Dv are given in columns 1 and 2, respectively, of the following table.
2.
Noting that v0 = 0 and that
v2 = v1 + Dv12
where Dv12 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 Dt = 2 s. The area
( Dt )( vave ) of each rectangle is approximately equal to the change in position Dx for the specified interval
of time.
Thus,
D x vave Dt
where vave and D x are given in columns 4 and 5, respectively, of the table.
4.
With x0 = 0 and noting that
x2 = x1 + D x12
where D 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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121
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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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
Given:
t = 25, v = 0
a-t curve; at
1.The a-t curve is first approximated with a series of rectangles, each of width Dt = 0.25 s. The area
(Dt)(aave) of each rectangle is approximately equal to the change in velocity Dv for the specified interval
of time. Thus,
Dv aave D t
where the values of aave and Dv are given in columns 1 and 2, respectively, of the following table.
2.
2
v(2) = v0 + Ú a dt = 0
Now
0
2
and approximating the area Ú a dt under the a-t curve by Saave Dt ª SDv, the initial velocity is then
0
equal to
v0 = -SDv
Finally, using
v2 = v1 + Dv12
where Dv12 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 ( Dt )( vave )
of each rectangle is approximately equal to the change in position D x for the specified interval of time.
Thus
D x ª vave D t
where vave and D x are given in columns 4 and 5, respectively, of the table.
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123
PROBLEM 11.82 (Continued)
4.
With x0 = 0 and noting that
x2 = x1 + D x12
where D 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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124
PROBLEM 11.83
A training airplane has a velocity of 38 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 = 38 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
Dt =
a
Dx =
or
For the five regions shown above, we have
Region
v1, m /s
v2, m/s
a, m /s2
D x, m
Dt, s
1
38
36
-4
18.5
0.5
2
36
30
-10
19.8
0.6
3
30
24
-13.5
12
0.4444
4
24
12
-16
13.5
0.75
5
12
0
-17
4.235
0.7059
68.035
3.0003
S
(a)
From the table, when v = 0 t = 3.00 s 
(b)
From the table and assuming x0 = 0, when v = 0 x = 68.0 m 
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125
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 = 250 mm, (b) when v = 2000 mm/s.
SOLUTION
Given: v - x curve
First note that the slope of the above curve is
a=v
(a)
dv
.
dx
Now
dv
dx
When
x = 250 mm, v = 1375 mm/s
Then
Ê 1000 m/s ˆ
a = (1375 mm/s ) Á
Ë 337.5 mm ˜¯
a = 4070 mm/s2 
or
(b)
When v = 2000 mm/s, we have
Ê 1000 mm/s ˆ
a = 2000 mm/s Á
Ë 700 mm ˜¯
a = 2860 mm/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 = 50 cm, 1 cm = 50 cm /s). In the above
solution, Dv and D 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 + v0 t + 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 + v0 t + (area under a - t curve) (t-t )
Ê 1 ˆ
= x0 + v0 t + (t ¥ a) Á t - t ˜
Ë 2 ¯
1
= x0 + v0 t + at 2 2
Q.E.D.
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127
PROBLEM 11.86
Using the method of Section 11.8, determine the position of
the particle of Problem 11.61 when t = 4 s.
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 = - 6 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 = - 6 m/s
When t = 14 s:
x = x0 + v0 t + SA(t1 - t )
= 0 - (6 m/s)(14 s) + [(1 m/s2 )( 4 s)](12 s) + [(2 m/s2 )(6 s)](7 s)
+ [( -2 m/s)( 4 s)](2 s)
x14 = -84 m + 48 m + 84 m - 16 m = 32 m x14 = + 32 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
2.5 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
t a 0.75
=
20
25
or
t a = 0.60 s
Now
t
v = v0 + Ú a dt
0
where the integral is equal to the area under the a-t curve. Then, with v0 = 2.5 m/s, vt1 = 0
(b)
We have
1
È1
˘
0 = 2.5 m/s + Í (0.6 s)(20 m/s2 ) - (0.15 s)(5 m/s2 ) - (t1 - 0.75) s ¥ (5 m/s2 ) ˙
2
Î2
˚
or
t1 = 2.375 s t1 = 2.38 s 
From the discussion following Eq. (11.13) and assuming x = 0, we have
x = 0 + v0 t1 + SA(t1 - t )
where A is the area of a region and t is the distance to its centroid. Then for t1 = 2.375 s
ÏÈ 1
˘
x = (2.5 m/s)(2.375 s) + Ì Í (0.6 s)(20 m/s2 )˙ (2.375 - 0.2) s
2
˚
ÓÎ
1
È
˘
- Í (0.15 s)(5 m/s2 )˙ (2.375 - 0.70) s
Î2
˚
1
È
ˆ˘ ¸
Ê
- [(1.625 s)(5 m /s2 )] Í2.375 - Á 0.75 + ¥ 1.625˜ ˙ s ˝
¯˚ ˛
Ë
2
Î
= [5.9375 + (13.05 - 0.6281 - 6.6016)] m
x = 11.76 m 
or
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129
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 = -180 m 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
We have
a=
dv
dt
is the slope of the v - t curve. Then
where
t=0
from
to t = 10 s:
dv
dt
v = constant fi a = 0
-10 - 20 -15
=
m/s2
26 - 10
8
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant fi a = 0
t = 41 s to t = 46 s:
a=
t > 46 s:
v = constant fi a = 0
- 4 - ( -10)
= 1.2 m/s2
46 - 41
The a-t curve is then drawn as shown.
(a)
From the discussion following Eq. (11.13),
we have
x = x0 + v0 t1 + SA(t1 - t )
where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s
x = -180 m + (20 m/s)(52 s) + {-[(16 s)(15 / 8 m/s2 )](52 - 18) s
+ [(5 s)(1.2 m/s2 )](52 - 43.5) s}
= [-180 m + 1040 m + ( -1020 m + 51 m)]
x = -109 m 
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 10 s < t
dt
< 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 = -180 + 20t1
È
˘
Ê 15 ˆ ˘ È 1
- Í(t1 - 10) Á ˜ ˙ Í (t1 - 10) ˙ m
¯
Ë
8 ˚ Î2
˚
Î
15
= -180 + 20tt1 - (t1 - 10)2
16
When x = xmax :
or
Then
dx
-30
(t1 - 10) = 0
= 20
16
dt
62
(t1 ) xmax =
s
3
ˆ
Ê 62 ˆ -15 Ê 62
- 10˜
xmax = -180 + 20 Á ˜
¯
Ë 3 ¯ 16 ÁË 3
2
xmax = 126.7 m 
or
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131
PROBLEM 11.89
The motion of a particle is defined by the equations x = 4t3 - 5t2 + 5t and y = 5t2 - 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
dv
ax = x = 24t - 10
dt
(a)
y = 5t 2 - 15t
dy
vy =
= 10t - 15
dt
dv y
= 10
ay =
dt
When t = 15, v x = 7 mm/s, v y = -5 mm/s
v = 72 + ( -5)2 = 8.60 f = tan -15 / 7 = 35.5∞
ax = 14 mm/s2
v = 8.60 mm/s
35.5° 
a = 17.20 mm/s2
35.5° 
v = 33.4 mm/s
8.62° 
a y = 10 mm/s2
a = 142 + 102 = 17.20 mm/s2
Ê 10 ˆ
f = tan -1 Á ˜ = 35.5∞
Ë 14 ¯
(b)
When t = 25, v x = 33 mm/s, v y = 5 mm/s
ax = 38 mm/s2
a y = 10 mm/s2
a = 39.3 mm/s2
14.74° 
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132
PROBLEM 11.90
The motion of a particle is defined by the equations x = 2 cos p t
and y = 1 - 4 cos 2 p 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 p t
Then
y = 1 - 4(2 cos2 p t - 1)
Ê xˆ
= 5 - 8Á ˜
Ë 2¯
y = 1 - 4 cos 2p t
2
y = 5 - 2 x 2 Q.E.D. or
Now
vx =
dx
= -2p sin p t
dt
vy =
and
ax =
dv x
= -2p 2 cos p t
dt
ay =
(a)
At t = 0 : v x = 0
dy
= 8p sin 2p t
dt
dv y
dt
= 16p 2 cos 2p t
vy = 0 ax = -2p 2 m/s2
v=0 
a y = 16p 2 m/s2
a = 159.1 m/s2
or
(b)

82.9° 
At t = 1.53: v x = -2p sin(1.5p ) = 2p m/s
v y = 8p sin(2p ¥ 1.5) = 0
or
v = 6.28 m/s Æ 
ax = -2p 2 cos(1.5p ) = 0
a y = 16p 2 cos( 2p ¥ 1.5) = -16p 2
a = 157.9 m/s2 Ø 
or
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133
PROBLEM 11.91
The motion of a particle is defined by the equations x = t2 - 8t + 7 and y = 0.5t2 + 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
v x = dx /dt = 2t - 8
y = 0.5t 2 + 2t - 4
vy = t + 2
a y = dv y /ddt = 1
ax = dv x /dt = 2
2
v =
v x2
+
v 2y
2
= (2t - 8) + (t + 2)2
When v is minimum, v2 is also minimum; thus, we write
d (v2 )
= 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 v x = 2(2.8) - 8 v x = - 2.40 m/s ¸
v y = 2.8 + 2
v y = + 4.80 m/s ˝Ô˛
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 = 100t - 50 sin t and y = 100 - 50 cos t, where x and
y are expressed in mm 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
(a)
t, s
x, mm
y, mm
0
0
50
p
2
107.08
100
p
314.25
150
3p
2
521.24
100
2p
628.32
50
We have
y = 4 - 2 cos t
x = 100t - 50 sin t = 25( 4t - 2 sin t )
dx
= 25 ( 4 - 2 cos t )
dt
y = 100 - 50 cos t = 25 ( 4 - 2 cos t )
vy =
dy
= 50 sin t
dt
Then
vx =
Now
v 2 = v x2 + v 2y = 625( 4 - 2 cos t )2 + 2500 sin 2 t
= 12500 - 10000 t
By observation,
for vmin, cos t = 1, so that
2
vmin
= 2500 or
vmin = 50 mm/s 
2
vmax
= 22500 or
vmax = 150 mm /s 
cos t = 1 or
t = 2np s 
for vmax, cos t = -1, so that
(b)
When v = vmin:
where n = 0, 1, 2, . . .
Then
x = 25 {4(2np ) - 2 sin (2np )} or x = 200 np mm 
y = 25( 4 - 2(1)) = 50 or y = 50.0 mm 
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135
PROBLEM 11.92 (Continued)
Also,
v x = 25 {4 - 2(1)} = 50 mm/s
v y = 50 sin (2np ) = 0
When v = vmax :
q v min = 0 Æ 
cos t = -1
or t = (2n + 1)p s 
where n = 0, 1, 2,…
Then
x = 25 {4(2n + 1)p - 2 sin (2n + 1)p } y = 25 {4 - 2( -1)} Also,
or
x = 100(2n + 1)p mm 
or
y = 150 mm 
v x = 25{4 - 2( -1)} = 150 mm/s
v y = 50 sin (2n + 1)p = 0
q vmax = 0 Æ 
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136
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 cos2 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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137
PROBLEM 11.94
The damped motion of a vibrating particle is defined by the
position vector r = x1[1-1/(t + 1)]i +(y1e-pt/2 cos 2pt)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 ˆ
Ê
r = 30 Á1 i + 20(e -p t / 2 cos 2p t ) j
Ë t + 1˜¯
Then
v=
= 30
1
ˆ
Ê p
i + 20 Á - e -p t / 2 cos 2p t - 2p e -p t / 2 sin 2p t ˜ j
2
¯
Ë
2
(t + 1)
= 30
1
È
ˆ˘
Ê1
i - 20p Íe -p t / 2 Á cos 2p t + 2 sin 2p t ˜ ˙ j
2
¯˚
Ë2
(t + 1)
Î
a=
and
dr
dt
dv
dt
2
È p
˘
ˆ
Ê1
i - 20p Í- e -p t / 2 Á cos 2p t + 2 sin 2p t ˜ + e -p t / 2 ( -p sin 2p t + 4 cos 2p t ) ˙ j
3
¯
Ë
2
(t + 1)
Î 2
˚
60
=i + 10p 2 e -p t / 2 ( 4 sin 2p t - 7.5 cos 2pt ) j
3
(t + 1)
= -30
(a)
At t = 0:
Ê 1ˆ
r = 30 Á1 - ˜ i + 20(1) j
Ë 1¯
r = 20 mm ≠ 
or
È Ê1
ˆ˘
Ê 1ˆ
v = 30 Á ˜ i - 20p Í(1) Á + 0˜ ˙ j
¯˚
Ë 1¯
Î Ë2
or
v = 43.4 mm/s
46.3° 
a = 743 mm/s2
85.4° 
60
a = - i + 10p 2 (1)(0 - 7.5) j
(1)
or
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138
PROBLEM 11.94 (Continued)
(b)
At t = 1.5 s:
1 ˆ
Ê
r = 30 Á1 i + 20e -0.75p (cos 3p ) j
Ë 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/s2
86.9° 
30
ˆ
Ê1
i - 20p e -0.75p Á cos 3p + 0˜ j
2
¯
Ë2
(2.5)
= ( 4.80 mm/s)i + (2.9778 mm/s)j
v=
or
a=-
60
i + 10p 2 e -0.75p (0 - 7.5 cos 3p ) j
3
(2.5)
= ( -3.84 mm/s2 )i + (70.1582 mm/s2 )j
or
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139
PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r = (Rt cos wnt)i + ctj + (Rt sin wnt)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 w nt )i + ctj + ( Rt sin w nt )k
Then
v=
dr
= R(cos w n t - w nt sin w n t )i + cj + R(sin w nt + w nt cos w nt )k
dt
and
a=
dv
dt
(
)
+ R (w cos w t + w cos w t - w t sin w t ) k
= R ( -2w sin w t - w t cos w t ) i + R (2w cos w t - w t sin w t ) k
= R -w n sin w nt - w n sin w nt - w n2t cos w nt i
n
n
Now
n
n
n
2
n
2
n
n
n
n
n
2
n
n
n
v 2 = v x2 + v 2y + v z2
= [ R(cos w nt - w nt sin w nt )]2 + (c)2 + [ R(sin w nt + w nt cos w nt )]2
(
)
= R2 È cos2 w nt - 2w nt sin w nt cos w nt + w n2t 2 sin 2 w nt
Î
+ sin 2 w nt + 2w nt sin w nt cos w nt + w n2t 2 cos2 w nt ˘ + c 2
˚
(
(
)
)
= R2 1 + w n2t 2 + c 2
(
Also,
)
v = R2 1 + w n2t 2 + c 2 
or
a2 = ax2 + a2y + az2
(
)
cos w t - w t sin w t )˘
˚
2
= È R -2w n sin w nt - w n2t cos w nt ˘ + (0)2
Î
˚
(
(
2
2
+ È R 2w n
n
n
n
Î
= R2 È 4w n2 sin 2 w nt + 4w n3t sin w nt cos w nt + w n4 t 2 cos2 w nt
Î
+ 4w n2 cos2 w nt - 4w n3t sin w nt cos w nt + w n4 t 2 sin 2 w nt ˘
˚
(
(
= R2 4w n2 + w n4 t 2
)
)
)
a = Rw n 4 + w n2t 2 
or
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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 feet 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
y = A t 2 + 1 z = Bt sin t
sin t =
2
Ê yˆ
t2 = Á ˜ -1
Ë A¯
z
Bt
2
2
Ê zˆ
Ê xˆ
cos2 t + sin 2 t = 1 fi Á ˜ + Á ˜ = 1
Ë Bt ¯
Ë At ¯
Now
2
2
2
2
Ê zˆ
Ê xˆ
t2 = Á ˜ + Á ˜
Ë B¯
Ë A¯
or
2
Then
Ê zˆ
Ê xˆ
Ê yˆ
ÁË ˜¯ - 1 = ÁË ˜¯ + ÁË ˜¯
A
A
B
or
Ê zˆ
Ê xˆ
Ê yˆ
ÁË ˜¯ - ÁË ˜¯ - ÁË ˜¯ = 1 Q.E.D.
A
A
B
2
(a)
2
2

With A = 3 and B = 1, we have
dr
t
= 3(cos t - t sin t )i + 3
j + (sin t + t cos t )k
2
dt
t +1
Ê t ˆ
t 2 + 1 - t ÁË t 2 + 1 ˜¯
dv
a=
= 3( - sin t - sin t - t cos t )i + 3
j
dt
(t 2 + 1)
v=
and
+ (cos t + cos t - t sin t )k
= -3(2 sin t + t cos t )i + 3
1
j + (2 cos t - t sin t )k
(t + 1)3/2
2
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141
PROBLEM 11.96* (Continued)
At t = 0:
v = 3(1 - 0)i + (0) j + (0)k
v = v x2 + v 2y + v z2
v = 3 m/s 
or
a = -3(0)i + 3(1) j + (2 - 0)k
and
a2 = (0)2 + (3)2 + (2)2 = 13
Then
a = 3.61 m/s2 
or
(b)
If r and v are perpendicular, r ◊ v = 0
(
)
t ˆ
Ê
j + (sin t + t cos t )k ] = 0
[(3t cos t )i + 3 t 2 + 1 j + (t sin t )k ] ◊ [3(cos t - t sin t )i + 3
ÁË
˜
2
t + 1¯
t ˆ
Ê
+ (t sin t )(sin t + t cos t ) = 0
or
(3t cos t )[3(cos t - t sin t )] + (3 t 2 + 1) 3
ÁË
˜
t 2 + 1¯
Expanding
(9t cos2 t - 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0
10 + 8 cos2 t - 8t sin t cos t = 0
or ( with t π 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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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/s2 )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
PROBLEM 11.98
Three children are throwing snow‑
balls at each other. Child A throws
a snowball with a horizontal velocity v0. 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/s2 )t 2
2
At B:
or t B = 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/s2 ) 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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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 + ( v x )0 t = v0 t
At B:
y:
t B = 0.45152 s
or
Then
x:
y:
or
1
- 0.6 m = - (9.81 m/s2 )t 2
2
tC = 0.34975 s
or
Then
(2.1) m = ( v0 ) B (0.45152 s)
( v0 ) B = 4.651 m/s
or
At C:
1
-1 m = - (9.81 m/s2 )t 2
2
x:
3.7 m = ( v0 )C (0.34975 s)
( v0 )C = 10.579 m/s
4.65 m/s £ v0 £ 10.58 m/s 
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145
PROBLEM 11.100
A baseball pitching machine “throws” baseballs with a horizontal velocity v0. Knowing that height h varies
between 775 mm and 1050 mm, determine (a) the range of values of v0, (b) the values of a corresponding to
h = 775 mm and h = 1050 mm.
SOLUTION
(a)
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t -
1 2
gt
2
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = v0 t
When h = 775 mm, y = -0.725 m:
1
- 0.725 = - (9.81 m/s2 )t 2
2
t31 = 0.38446 s
or
Then
or
12 m = ( v0 )775 (0.38446 s)
( v0 )775 = 31.2126 m/s = 112.365 km/h
When h = 1050 mm, y = - 0.45 m:
1
- 0.45 m = - (9.81 m/s2 ) t 2
2
t 42 = 0.30289 s
or
Then
or
12 m = ( v0 )1050 (0.30289 s)
( v0 )1050 = 39.6183 m/s = 142.626 km/h
112.4 km/h £ v0 £ 142.6 km/h 
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146
PROBLEM 11.100 (Continued)
(b)
For the vertical motion
v y = (0) - gt
Now
tan a =
When h = 775 mm:
tan a =
| (v y ) B |
(vx ) B
=
gt
v0
(9.81 m/s2 )(0.38446 s)
31.226 m/s
= 0.12078
a 31 = 6.89∞ 
or
When h = 1050 mm:
or
2
(9.81 m/s )(0.30289 s)
39.6183 m/s
= 0.074999
tan a =
a 42 = 4.29∞ 
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147
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
( v x )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:
yc = 2.1 m + ( 4.5831 m/s)(0.71475 s)
1
- (9.81 m/s2 )(0.71475 s)2
2
= 2.87 m
yC > 2.43 m (height of net) fi ball clears net 
(b)
1 2
gt
2
At B, y = 0:
Solving
Then
1
0 = 2.1 m + ( 4.5831 m/s)t - (9.81 m/s2 )t 2
2
t B = 1.271175 s ( the other root is negative)
d = ( v x )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 
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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
( v x )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
or
x: 0.08 m = (0.91925 m/s) t or t B = 0.087028 s
y: 0.140 m = hB + ( - 0.77135 m/s)(0.087028 s)
1
- (9.81 m/s2 )(0.087028 s)2
2
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)
or
1
- (9.81 m/s2 )(0.158825 s)2
2
hC = 0.386 m
0.244 m £ h £ 0.386 m 
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149
PROBLEM 11.103
A golfer hits a golf ball with an
initial velocity of 50 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
( v x )0 = (50 m/s) cos 25∞
First note
( v y )0 = (50 m/s)sin 25∞
x B = d cos 5∞
and at B
Now
y B = - d sin 5∞
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
d cos 5∞ = (50 cos 25∞)t
At B
or t B =
cos 5∞
d
50 cos 25∞
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
( g = 9.81 m/s2 )
1 2
gt B
2
At B:
- d sin 5∞ = (50 sin 25∞)t B -
Substituting for t B
Ê cos 5∞ ˆ
1 Ê cos 5∞ ˆ 2
- d sin 5∞ = (50 sin 25∞) Á
d - gÁ
d
˜
2 Ë 50 cos 25∞ ˜¯
Ë 50 cos 25∞ ¯
2
or
or
2
(50 cos 25∞)2 (tan 5∞ + tan 25∞)
9.81 cos 5∞
= 232.733 m
d=
d = 233 m 
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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
( v x )0 = (0.75 m/s) cos 15∞ = 0.72444 m/s
First note
( v y )0 = -(0.75 m/s) sin 15∞ = - 0.19411 m/s
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
At the top of the trough, y = -3 + 0.36 = -2.64 m
or
1
- 2.64 m = ( - 0.19411 m/s) t - (9.81 m/s2 ) t 2
2
t BC = 0.71412 s (the other root is negative)
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
In time t BC
x BC = (0.72444 m/s)(0.71412 s) = 0.5173 m
Thus, the trough must be placed so that
x B £ 0.5173 m
xC ≥ 0.5173 m
0 £ d £ 0.5173 m 
Since the trough is 0.6 m wide, it then follows that
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151
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 a = 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
10 m = (0.9397 v0 ) t
y = ( v0 ) y t -
t = 10.6417 /v0 1 2
gt
2
1
- 6 m = v0 (0.3420) t - 9.81t 2
2
Ê 10.6417 ˆ
- 6 = (0.3420)(10.6417) - ( 4.905) Á
Ë v0 ˜¯
Ê 10.6417 ˆ
- 9.6395 = -( 4.905) Á
Ë v0 ˜¯
v02 =
(1)
2
2
( 4.905)(10.6417)2
= 57.6244 m 2 /s2
9.6395
v0 = 7.5911 m/s v0 = 7.59 m/s 
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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) 225 mm, (b) 425 mm.
SOLUTION
First note
( v x )0 = v0 cos 30∞
( v y )0 = v0 sin 30∞
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
At B:
(5 - d ) = ( v0 cos 30∞) t
or t B =
5-d
v0 cos 30∞
Vertical motion. (Uniformly accelerated motion)
1 2
gt ( g = 9.81 m/s2 )
2
1
0.9 = ( v0 sin 30∞) t B - gt B2
2
y = 0 + ( v y )0 t -
At B:
Substituting for tB
or
Ê 5-d ˆ 1 Ê 5-d ˆ
0.9 = ( v0 sin 30∞) Á
- g
Ë v0 cos 30∞ ˜¯ 2 ÁË v0 cos 30∞ ˜¯
v02 =
2 g (5 - d ) 2
È 1
˘
3 Í (5 - d ) - 0.9˙
Î 3
˚
2
d~m
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153
PROBLEM 11.106 (Continued)
2(9.81) (5 - 0.225)
2
(a)
d = 225 mm = 0.225 m
v02 =
or
(b)
d = 425 mm = 0.425 m
3È
Î
1
3
(5 - 0.225) - 0.9˘˚
2(9.81) (5 - 0.425)
2
v02 =
3È
Î
1
3
(5 - 0.425) - 0.9˘˚
or
v0 = 8.96 m/s 
v0 = 8.87 m/s 
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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
( v x )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 /s2 )
When the ball reaches the tire, t = t6
Ê
ˆ
6
y B ,C = ( v0 sin 3∞) Á
Ë v0 cos 3∞ ˜¯
ˆ
1 Ê
6
- gÁ
2 Ë v0 cos 3∞ ˜¯
2
or
v02 =
18(9.81)
cos 3∞(6 tan 3∞ - y B , C )
or
v02 =
177.065
0.314447 - y B ,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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155
PROBLEM 11.107 (Continued)
At C, y = -1.25 m:
or
v02 =
177.06 s
0.314447 - ( -1.25)
( v0 )C = 10.64 m/s
10.64 m/s £ v0 £ 14.48 m/s 
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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
( v x )0 = v0 cos 6∞
( v y )0 = - v0 sin 6∞
Horizontal motion. (Uniform)
x = x0 + ( v x )0 t
Vertical motion. (Uniformly accelerated motion)
y = y0 + ( v y )0 t At Point B:
1 2
gt
2
( g = 9.81 m/s2 )
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 Substituting for t B
1 2
gt B
2
Ê 0.050388 ˆ
Ê 0.050388 ˆ 1
- 0.032659 = ( - v0 sin 6∞) Á
- (9.81) Á
˜
Ë v0 cos 6∞ ˜¯
Ë v0 cos 6∞ ¯ 2
2
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157
PROBLEM 11.108 (Continued)
v02 =
or
2
1
2 (9.81)( 0.050388)
cos2 6∞(0.032659 - 0.050388 tan 6∞)
( v0 ) B = 0.678 m/s
or
x = (0.175 m) cos 30∞
y = (0.175 m) sin 30∞
At Point C:
x: 0.175 cos 30∞ = - 0.020 + ( v0 cos 6∞)t
tC =
or
0.171554
v0 cos 6∞
y: 0.175 sin 30∞ = 0.205 + ( - v0 sin 6∞)tC -
1 2
gtC
2
Substituting for tC
Ê 0.171554 ˆ
Ê 0.171554 ˆ 1
- 0.117500 = ( - v0 sin 6∞) Á
- (9.81) Á
˜
Ë v0 cos 6∞ ˜¯
Ë v0 cos 6∞ ¯ 2
or
or
v02 =
2
2
1
2 (9.81)( 0.171554)
cos2 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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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
( v x )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 At the tree limbs, t = t B ,C
1 2
gt
2
( g = 9.81 m/s2 )
Ê
ˆ 1 Ê
ˆ
5
5
y B ,C = ( v0 sin 65∞) Á
- gÁ
˜
Ë v0 cos 65∞ ¯ 2 Ë v0 cos 65∞ ˜¯
2
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159
PROBLEM 11.109 (Continued)
or
v02 =
=
1
2 (9.81)( 25)
cos2 65∞(5 tan 65∞ - y B , C )
686.566
5 tan 65∞ - y B , 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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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
( v x )0 = v0 sin 15∞
( v y )0 = v0 cos 15∞
Horizontal motion. (Uniform)
v x = ( v x )0 = v0 sin 15∞
Vertical motion. (Uniformly accelerated motion)
v y = ( v y )0 - gt
At Point B, v y < 0
Then
or
= v0 cos 15∞ - gt
tan 12∞ =
tB =
1 2
gt
2
1
= ( v0 cos 15∞) t - gt 2
2
y = 0 + ( v y )0 t -
(vx ) B
v0 sin 15∞
=
| ( v y ) B | gt B - v0 cos 15∞
ˆ
v0 Ê sin 15∞
+ cos 15∞˜
Á
g Ë tan 12∞
¯
g = 9.81 m /s2
= 0.22259 v0
Noting that
We have
or
y B = -0.2 m,
1
- 0.2 = ( v0 cos 15∞)(0.22259 v0 ) - (9.81)(0.22259 v0 )2
2
v0 = 2.67 m /s 
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161
PROBLEM 11.111
A model rocket is launched from Point A with an initial velocity v0 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 a 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 a
Vertical motion:
y = v0 t cos a cos a =
sin 2 a + cos2 a =
sin a =
1 2
gt
2
x
v0 t
1 Ê
1 2ˆ
ÁË y + gt ˜¯ 2
v0 t
(2)
2
1 È 2 Ê
1 2ˆ ˘
x
y
gt
+
+
Í
˜¯ ˙ = 1
ÁË
2
( v0 t )2 ÍÎ
˙˚
1
x 2 + y 2 + gyt 2 + g 2t 4 = v02t 2
4
(
)
1 2 4
g t - v02 - gy t 2 + ( x 2 + y 2 ) = 0 4
At Point B,
(1)
(3)
x 2 + y 2 = 120 m, x = 120 cos 30∞ m
y = -120 sin 30∞ = - 60 m
1
(9.81)2 t 4 - [752 - (9.81)( - 60)] t 2 + 1202 = 0
4
24.059 t 4 - 6213.6 t 2 + 14400 = 0
t 2 = 255.926 s2 and 2.33867 s2
t = 15.998 s and 1.5293 s
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162
PROBLEM 11.111 (Continued)
Restrictions on a :
0 £ a £ 120∞
120 cos 30∞
x
tan a =
=
= 0.08694
2
1
- 60 + ( 4.905)(15.998)2
y + 2 gt
a = 4.96877∞
and
120 cos 30∞
= - 2.14149
- 60 + ( 4.905)(1.5293)2
a = 115.031∞
Use a = 4.9687∞ corresponding to the steeper possible trajectory.
(a)
Angle a.
(b)
Maximum height.
a = 4.97∞ 
v y = 0 at
y = ymax
v y = v0 cos a - gt = 0
t=
v0 cos a
g
ymax = v0 t cos a =
v 2 cos2 a
1
gt = 0
2
2g
(75)2 cos2 4.9687∞
(2)(9.81)
ymax = 285 m 
(c)
Duration of the flight.
(time to reach B)
t = 16.00 s 
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163
PROBLEM 11.112
The initial velocity v0 of a hockey
puck is 160 km / h. Determine (a) the
largest value (less than 45°) of the
angle a for which the puck will enter
the net, (b) the corresponding time
required for the puck to reach the net.
SOLUTION
First note
and
400
m/s
9
ˆ
Ê 400
m/s˜ cos a
( v x )0 = v0 cos a = Á
¯
Ë 9
v0 = 160 km/h =
ˆ
Ê 400
m/s˜ sin a
( v y )0 = v0 sin a = Á
¯
Ë 9
(a)
Horizontal motion. (Uniform)
ˆ
Ê 400
x = 0 + ( v x )0 t = Á
cos a ˜ t
¯
Ë 9
At the front of the net,
Then
or
x=5m
ˆ
Ê 400
cos a ˜ t
5=Á
¯
Ë 9
9
tenter =
80 cos a
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
1
ˆ
Ê 400
=Á
sin a ˜ t - gt 2
¯
Ë 9
2
( g = 9.81 m/s2 )
At the front of the net,
1 2
ˆ
Ê 400
yfront = Á
sin a ˜ tenter - gtenter
¯
Ë 9
2
ˆ 1 Ê
ˆ
9
9
ˆÊ
Ê 400
=Á
- gÁ
sin a ˜ Á
˜
¯ Ë 80 cos a ¯ 2 Ë 80 cos a ˜¯
Ë 9
= 5 tan a Now
2
81g
12800 cos2 a
1
= sec2 a = 1 + tan 2 a
2
cos a
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164
PROBLEM 11.112 (Continued)
Then
or
yfront = 5 tan a tan 2 a -
81g
(1 + tan 2 a )
12800
ˆ
Ê 12800
64000
yfront ˜ = 0
tan a + Á1 +
¯
Ë
81g
81g
64000
81g
(
È
± Í - 64000
Î 81g
)
2
(
)
1/ 2
˘
- 4 1 + 12800
81g yfront ˙
˚
2
Then
tan a =
or
2
ÈÊ
Ê
ˆ˘
64000
64000 ˆ
12800
tan a =
± ÍÁ 1
+
y
fro
o
nt
ÁË
˜¯ ˙˙
162 ¥ 9.81 ÍË 162 ¥ 9.81˜¯
81 ¥ 9.81
˚
Î
or
tan a = 40.2713 ± [( 40.2713)2 - (1 + 16.1085 yfront )]1/ 2
1/ 2
Now 0 £ yfront £ 1.2 m so that the positive root will yield values of a > 45∞ for all values of yfront.
When the negative root is selected, a increases as yfront is increased. Therefore, for a max, set
yfront = yC = 1.2 m
(b)
Then
tan a = 40.2713 - [( 40.2713)2 - (1 + 16.1085 + 1.2)]1/ 2
or
a max = 14.2093∞ 9
tenter =
80 cos a
9
=
80 cos 14.2093∞
We had found
a max = 14.66∞ 
tenter = 0.1161 s 
or
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165
PROBLEM 11.113
The pitcher in a softball game
throws a ball with an initial
velocity v0 of 72 km / h at an angle
a with the horizontal. If the height
of the ball at Point B is 0.68 m,
determine (a) the angle a , (b) the
angle q that the velocity of
the ball at Point B forms with
the horizontal.
SOLUTION
First note
v0 = 72 km / h = 20 m /s
( v x )0 = v0 cos a = (20 m /s) cos a
and
( v y )0 = v0 sin a = (20 m /s) sin a
(a)
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = (20 cos a ) t
At Point B:
14 = (20 cos a )t
or t B =
7
10 cos a
Vertical motion. (Uniformly accelerated motion)
1 2
1
gt = (20 sin a )t - gt 2
2
2
1 2
0.08 = (20 sin a )t B - gt B
2
( g = 9.81 m /s2 )
y = 0 + ( v y )0 t -
At Point B:
Substituting for t B
or
Now Ê
ˆ
7 ˆ 1 Ê
7
0.08 = (20 sin a ) Á
- gÁ
˜
Ë 10 cos a ¯ 2 Ë 10 cos a ˜¯
2
1
49
8 = 1400 tan a - g
2 cos2 a
1
= sec2 a = 1 + tan 2 a
2
cos a
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166
PROBLEM 11.113 (Continued)
Then
or
Solving
8 = 1400 tan a - 24.5 g (1 + tan 2 a )
240.345 tan 2 a - 1400 tan a + 248.345 = 0
a = 10.3786∞ and a = 79.949∞
Rejecting the second root because it is not physically reasonable, we have
a = 10.38∞ 
(b)
We have
v x = ( v x )0 = 20 cos a
and
v y = ( v y )0 - gt = 20 sin a - gt
At Point B:
( v y ) B = 20 sin a - gt B
= 20 sin a -
7g
10 cos a
Noting that at Point B, v y < 0, we have
tan q =
=
=
| (v y ) B |
vx
7g
10 cos a
- 20 sin a
20 cos a
7
9.81
200 cos 10.3786∞
- sin 10.3786∞
cos 10.3786∞
q = 9.74∞ 
or
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167
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 a so that he lands at B.
SOLUTION
First note
( v x )0 = v0 cos a
( v y )0 = v0 sin a
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = ( v0 cos a ) t
At Point B:
or
1.8 = ( v0 cos a )t
tB =
1.8
v0 cos a
Vertical motion. (Uniformly accelerated motion)
At Point B:
Substituting for tB
or
1
y = 0 + ( v y )0 t - gt 2
2
1
= ( v0 sin a ) t - gt 2 ( g = 9.81 m/s2 )
2
1
-1.4 = ( v0 sin a ) t B - gt B2
2
Ê 1.8 ˆ 1 Ê 1.8 ˆ
-1.4 = ( v0 sin a ) Á
- g
Ë v0 cos a ˜¯ 2 ÁË v0 cos a ˜¯
v02 =
=
2
1.62 g
cos2 a (1.8 tan a + 1.4)
1.62 g
0.9 sin 2a + 1.4 cos2 a
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168
PROBLEM 11.114* (Continued)
Now minimize v02 with respect to a.
We have
dv02
-(1.8 cos 2a - 2.8 cos a sin a )
= 1.62 g
=0
da
(0.9 sin 2a + 1.4 cos2 a )2
1.8 cos 2a - 1.4 sin 2a = 0
or
tan 2a =
or
18
14
a = 26.0625∞ and a = 206.06∞
or
Rejecting the second value because it is not physically possible, we have
a = 26.1∞ 
Finally,
v02 =
1.62 ¥ 9.81
cos2 26.0625∞(1.8 tan 26.0625∞ + 1.4)
( v0 ) min = 2.94 m /s 
or
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169
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 a when (a) the
vegetables are just beginning to grow,
(b) the height h of the corn is 1.8 m.
SOLUTION
First note
( v x )0 = v0 cos a = (8 m /s) cos a
( v y )0 = v0 sin a = (8 m /s) sin a
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = (8 cos a ) t
At Point B:
x = d : d = (8 cos a ) t
tB =
or
d
8 cos a
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (8 sin a ) t - gt 2 ( g = 9.81 m /s2 )
2
1
0 = (8 sin a ) t B - gt B2
2
y = 0 + ( v y )0 t -
At Point B:
Simplifying and substituting for tB
0 = 8 sin a -
1 Ê d ˆ
g
2 ÁË 8 cos a ˜¯
64
sin 2a (1)
g
(a)When h = 0, the water can follow any physically possible trajectory. It then follows from Eq. (1) that d
is maximum when 2a = 90∞
or d =
a = 45∞ 
or
Then
d=
64
sin (2 ¥ 45∞)
9.81
dmax = 6.52 m 
or
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170
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 a increases in
value from 0 to 45° and then d decreases as a is further increased. Thus, dmax occurs for the value of a
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 a
Also, with ycorn = h, we have
h = (8 sin a ) tcorn -
1 2
gtcorn
2
Substituting for tcorn and noting h = 1.8 m,
Ê 1.5 ˆ 1 Ê 1.5 ˆ
1.8 = (8 sin a ) Á
- g
Ë 8 cos a ˜¯ 2 ÁË 8 cos a ˜¯
1.8 = 1.5 tan a -
or
Now
Then
or
Solving
2
2.25 g
128 cos2 a
1
= sec2 a = 1 + tan 2 a
cos2 a
2.25(9.81)
1.8 = 1.5 tan a (1 + tan 2 a )
128
0.172441 tan 2 a - 1.5 tan a + 1.972441 = 0
a = 58.229∞ and a = 81.965∞
From the above discussion, it follows that d = dmax when
a = 58.2∞ 
Finally, using Eq. (1)
d=
64
sin (2 ¥ 58.229∞)
9.81
dmax = 5.84 m 
or
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171
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 a.
SOLUTION
First note
( v x )0 = v0 cos a = (11.5 m /s) cos a
( v y )0 = v0 sin a = (11.5 m /s) sin a
By observation, dmax occurs when
ymax = 1.1 m
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (11.5 sin a ) - gt = (11.5 sin a ) t - gt 2
2
y = ymax at B, ( v y ) B = 0
v y = ( v y )0 - gt
When
Then
y = 0 + ( v y )0 t -
( v y ) B = 0 = (11.5 sin a ) - gt
11.5 sin a
g
( g = 9.81 m /s2 )
or
tB =
and
y B = (11.5 sin a ) t B -
1 2
gt B
2
Substituting for tB and noting yB = 1.1 m
Ê 11.5 sin a ˆ 1 Ê 11.5 sin a ˆ
1.1 = (11.5 sin a ) Á
˜¯
˜¯ - 2 g ÁË
g
g
Ë
=
or
sin 2 a =
2
1
(11.5)2 sin 2 a
2g
2.2 ¥ 9.81
11.52
a = 23.8265∞
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172
PROBLEM 11.116 (Continued)
(a)
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = (11.5 cos a ) t
At Point B:
where
Then
x = dmax
tB =
and t = t B
11.5
sin 23.8265∞ = 0.47356 s
9.81
dmax = (11.5)(cos 23.8265∞)(0.47356)
dmax = 4.98 m 
or
(b)
a = 23.8∞ 
From above
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173
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
q. Determine the velocity of portion CD of the belt
when (a) q = 45∞, (b) q = 60∞.
SOLUTION
We have
vCD = v A + vCD / A
where v A = (0.5 m /s)( - cos 65∞i - sin 65∞ j)
= (0.21131 m /s)i - (0.45315 m /s) j
vCD /A = (2 m /s)(cos q i + sin q j)
and
vCD = [( - 0.21131 + 2 cos q ) m/s]i + [( - 0.45315 + 2 sin q ) 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
38.3° 
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174
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 a
sin 15∞
a = 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 + 4.114j
= 5.05 m/s
55.8°
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175
PROBLEM 11.119
Shore-based radar indicates that a ferry leaves its slip with a velocity
v = 9.8 knots
70°, while instruments aboard the ferry indicate a speed of
10 knots 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 = 9.82 + 102 - 2(9.8)(10) cos 10∞
or
vR = 1.737147 knots
and
or
9.8 1.737 147
=
sin a
sin 10∞
a = 78.41∞
Noting that
v R = 1.737 knots
18.41° 
= 3.22 km / h
Alternatively one could use vector algebra.
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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 is vC/A
= 350 km / h
75°, and the relative velocity of C with respect to B
is vC/B = 390 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 36 km/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
vC = v B + vC / B
Then
v A + v C /A = v B + v C /B
or
v B - v A = v C /A - v C /B
Now
v B - v A = v B /A
so that
v B /A = v C /A - v C / B
or
v C /A = v C /B + v B /A
The graphical representation of the last equation is then as shown.
We have
v B2 /A = 3502 + 3902 - 2(350)(390) cos 65∞
or
v B /A = 399.0303 km / h
and
or
390 399.0303
=
sin a
sin 65∞
a = 62.3498∞
v B /A = 399 km / h
(b)
We have
v C = v A + v C /A
or
v A = (36 km/h) j - (350 km/h )( - cos 75∞i - sin 75∞ j )
12.65° 
v A = (90.587 km/h)i + (374.074 km/h ) j
v A = 385 km / h
or
76.4° 
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177
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 + vC t
rC /B = rC - rB = [(rC )0 - (rB )0 ] + ( vC - v B )t
= [(rC )0 - (rB )0 ] + vC /B t
Then
DrC /B = (rC /B )t2 - (rC /B )t1 = vC /B (t2 - t1 ) = vC /B Dt
For Dt = 15 min:
Ê1 ˆ
DrC /B = (390 km/h) Á h˜ = 97.5 km
Ë4 ¯
DrC /B = 97.5 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)
v B = v A + v B /A
We have
The graphical representation of this equation is then as shown.
Then
v 2B /A = 662 + 482 - 2(66)( 48) cos 155∞
or
v B /A = 111.366 km/h
and
48
111.366
=
sin a sin 155∞
a = 10.50∞
or
v B /A = 111.4 km/h
(b)
10.50° 
First note that
( 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.
at t = 3 min, A is (66 km/h )
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
PROBLEM 11.122
Knowing that the velocity of block B with respect to block A is
vB/A = 5.6 m /s
70°, determine the velocities of A and B.
SOLUTION
From the diagram
2 x A + 3x B = constant
Then
or
Now
2v A + 3v B = 0
2
| vB | = v A
3
v B = v A + v B /A
and noting that vA and vB 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
a = 180∞ - ( 40∞ + 30∞ + q B )
= 110∞ - q B
2
vA
vA
5.6
= 3
=
sin (110∞ - q B ) sin 40∞ sin (30∞ + q B )
v A sin 40∞ =
2
v A sin (110∞ - q B )
3
sin (110∞ - q B ) = 0.96418
or
q B = 35.3817∞
and
q B = 4.6183∞
For q B = 35.3817∞ :
vB =
or
v A = 5.94 m/s v B = 3.96 m/s
5.6 sin 40∞
2
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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180
PROBLEM 11.122 (Continued)
For q B = 4.6183∞ : v B =
or
5.6 sin 40∞
2
vA =
3
sin (30∞ + 4.6183∞)
v A = 9.50 m/s
v B = 6.34 m/s
v A = 9.50 m/s
v B = 6.34 m/s
30° 
4.62° 
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181
PROBLEM 11.123
Knowing that at the instant shown block A has a velocity of 200 mm /s
and an acceleration of 150 mm /s2 both directed down the incline,
determine (a) the velocity of block B, (b) the acceleration of block B.
SOLUTION
From the diagram
2x A + x B /A = constant
Then
2 v A + v B /A = 0
| v B /A | = 400 mm/s
or
2 a A + a B /A = 0
and
| aB /A | = 300 mm/s2
or
Note that vB/A and aB/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
v B2 = 2002 + 4002 - 2(200)( 400) cos 15∞
or
v B = 213.194 mm/s
and
or
200 213.194
=
sin a
sin 15∞
a = 14.05∞
v B = 213.194 mm/s
54.1° 
(b)The same technique that was used to determine vB can be used to determine aB. An alternative method is
as follows.
We have
a B = a A + a B /A
= (150i ) + 300( - cos 15∞i + sin 15∞ j)*
= - (139.778 mm/s2 )i + (77.646 mm/s2 ) j
a B = 159.9 mm/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 225 mm /s
and an acceleration of 375 mm /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 x B /A = constant
v A + 2 v B /A = 0 (1)
a A + 2 a B /A = 0 (2)
a A = 375 mm/s2 Ø
Data:
v A = 225 mm/s Ø
Eqs. (1) and (2)
a A = - 2 a B /A
v A = - 2 v B /A
375 = - 2aB /A
225 = - 2v B /A
aB /A = - 187.5 mm/s2
v B /A = - 112.5 mm/s
a B /A = 187.5 mm/s2
(a)
Velocity of B.
Law of cosines:
40°
v B /A = -112.5 mm/s
40°
v B = v A + v B /A
v B2 = (225)2 + (112.5)2 - 2(225)(112.5) cos 50∞
v B = 175.329 mm/s
Law of sines:
sin b sin 50∞
=
112.5 175.329
b = 29.44∞
a = 90∞ - b = 90∞ - 29.44∞ = 60.56∞
v B = 175.3 mm/s
60.6° 
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183
PROBLEM 11.124 (Continued)
(b)Acceleration of B. aB 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 = 375 mm/s2 Ø + 187.5 mm/s2
40°
= - 375 j - (187.5 cos 40∞)i + (187.5 sin 40∞) j
= - 375 j - 143.633i + 120.523j
a B = -143.633i - 254.477 j
a B = 292 mm/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)
a C = a B + a C /B
We have
The graphical representation of this equation is then as shown.
a = 180∞ - (20∞ + 105∞)
= 55∞
First note
Then
aC
2
=
sin 20∞ sin 55∞
aC = 0.83506 mm/s2
(b)
aC = 0.835 mm/s2
75° 
vC = 8.35 mm/s
75° 
For uniformly accelerated motion
vC = 0 + aC t
At t = 10 s:
vC = (0.83506 mm/s2 )(10 s)
= 8.3506 mm/s
or
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185
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)
a B = a A + a B /A
We have
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 /s2
and
0.5
0.95846
=
sin a sin 50∞
or
a = 23.6∞
(b)
a B = 0.958 m /s2
23.6° b
v B = 1.917 m /s
23.6° b
For uniformly accelerated motion
v B = 0 + aB t
At t = 2 s:
v B = (0.95846 m /s2 )(2 s)
= 1.91692 m /s
or
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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 0.75 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
[( v B ) x ]0 = ( v B )0 cos 30∞
= (0.75 m/s) cos 30∞
[( v B ) y ]0 = ( v B )0 sin 30∞
= (0.75 m/s)sin 30∞
Horizontal motion. (Uniform)
x = 0 + [( v B ) x ]0 t
( v B ) x = [( v B ) x ]0
= (0.75 cos 30∞) t
= 0.75 cos 30∞
Vertical motion. (Uniformly accelerated motion)
y = y0 + [( v B ) y ]0 t -
1 2
gt
2
= 0.45 + (0.75 sin 30∞)t -
( v B ) y = [( v B ) y ]0 - gt
1 2
gt
2
= 0.75 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
0.45 + (0.75 sin 30∞) t or
or
Solving
1 2
gt = [(0.75 cos 30∞) t ]tan 20∞
2
1
(9.81) t 2 + 0.75(cos 30∞ tan 20∞ - sin 30∞) t - 0.45 = 0
2
4.905t 2 - 0.13859t - 0.45 = 0
t = 0.31735 s and t = - 0.28909 s (Reject)
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187
PROBLEM 11.127 (Continued)
The velocity v B of the bag as it lands on the belt is then
v B = (0.75 cos 30∞)i + [0.75 sin 30∞ - 9.81(0.31735)]j
= (0.64952 m/s)i - (2.7382 m/s) j
Finally
or
v B = v A + v B /A
v B /A = (0.64952i - 2.7382 j) - 1.2(cos 20∞i + sin 20∞ j)
= - (0.47811 m/s)i - (3.1486 m/s) j
v B /A = 3.18 m/s
or
81.4° b
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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 = -2.5 m /s
0
vsy = 2 gh = (2)(9.81 m /s2 )(1.5 m) = 5.4249 m /s Ø
y-direction.
v S /B = v S - v B Relative velocity.
(a)
(1)
If vS /B is vertical,
- vS /B j = -2.5i - 5.4249 j - ( - v B cos 15i + v B sin 15 j)
= -2.5i - 5.4249 j + v B cos 15i - v B sin 15 j
Equate components.
i: 0 = -2.5 + v B cos 15∞
vB =
v B = 2.59 m/s
(b)
2.5
= 2.5882 m/s
cos 15∞
15° b
vS /C is as small as possible, so make vS /B ^ to v B into (1).
- vS /B sin 15∞i - vS /B cos 15∞ j = - 2.5i - 5.4249 j + v B cos 15∞i - v B sin 15∞ j
Equate components and transpose terms.
(sin 15∞) vS /B + (cos 15∞) v B = 2.5
(cos 15∞) vS /B - (sin 15∞) v B = 5.4249
Solving,
vS /B = 5.8871 m /s
v B = 1.0107 m /s
v B = 1.011 m/s
15° b
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189
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 = vship + v wind /ship
Case j
v w = v s + v w /s
v s = 9 km / h Æ; v w /s ≠
Case k
v s = 6 km / h ≠ ; v w /s 15
= 1.6667
9
a = 59.0∞
tan a =
vw = 92 + 152 = 17.49 km/h
v w = 17.49 km/h
59.0° b
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190
PROBLEM 11.130
When a small boat travels north at 5 km/h, a flag mounted on its
stern forms an angle q = 50∞ with the centerline of the boat as
shown. A short time later, when the boat travels east at 20 km / h,
angle q 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
q = 180∞ - (50∞ + 90∞ + a )
= 40∞ - a
We have
f = 180∞ - (50∞ + a )
= 130∞ - a
vW
vW
5
20
=
=
sin 50∞ sin ( 40∞ - a ) sin 50∞ sin (130∞ - a )
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191
PROBLEM 11.130 (Continued)
Therefore
or
or
or
Then
5
20
=
sin ( 40∞ - a ) sin (130∞ - a )
sin 130∞ cos a - cos 130∞ sin a = 4(sin 40∞ cos a - cos 40∞ sin a )
tan a =
sin 130∞ - 4 sin 40∞
cos 130∞ - 4 cos 40∞
a = 25.964∞
vW =
5 sin 50∞
= 15.79 km/h
sin ( 40∞ - 25.964∞)
vW = 15.79 km/h
26.0° b
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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 = v C + v D /C
The graphical representations of these equations are then as shown.
Then
vD
1
=
sin 8∞ sin (22∞ + a )
Equating the expressions for
vD
1
=
sin 7∞ sin (23∞ - a )
vD
1
sin 8∞
sin 7∞
=
sin (22∞ + a ) sin (23∞ - a )
or
or
or
Then
sin 8∞ (sin 23∞ cos a - cos 23∞ sin a ) = sin 7∞ (sin 22∞ cos a
+ cos 22∞ sin a )
tan a =
sin 8∞ sin 23∞ - sin 7∞ sin 22∞
sin 8∞ cos 23∞ + sin 7∞ cos 22∞
a = 2.0728∞
vD =
(1) sin 8∞
= 0.3412 m/s
sin (22∞ + 2.0728∞)
v D = 0.341 m/s
2.07° b
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193
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
a = -( vD cos a )i + ( vD sin a ) j
v D = v B + v D /B = v C + v D /C
0.866i + 0.5 j - (u1 cos 22∞)i - (u1 sin 22∞) j = 0.866i - 0.5 j - (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 = 1.3018 m/s
u2 = 1.3112 m/s
v D = 0.866i + 0.5 j - (1.3018 cos 22∞)i - (1.3112 sin 22∞) j
= -(0.341 m/s))i + (0.008158 m/s) j
v D = 0.341 m/s
2.07° b
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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 60 km / h. When
observed through a right-side window of an automobile traveling south at a speed of 45 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 = [60 + ( vR ) x ]tan 15∞
and
( vR ) y = [45 - ( vR ) x ]tan 30∞
Equating the expressions for ( vR ) y
[60 + ( vR ) x ]tan 15∞ = [45 - ( vR ) x ]tan 30∞
or
( vR ) x = 11.7163 km/h
Then
( vR ) y = (60 + 11.7163) tan 15∞ = 19.2163 km/h
v R = 22.5 km/h
58.6° b
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195
PROBLEM 11.133
Determine the peripheral speed of the centrifuge test cab A for
which the normal component of the acceleration is 10 g.
SOLUTION
an = 10 g = 10(9.81 m/s2 ) = 98.1 m/s2
an =
v2
r
v 2 = r an = (8 m)(98.1 m/s2 ) = 784.8 m2/s2
v = 28.0 m/s b
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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/s2 , (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
v = 72 km/h = 20 m/s
v2
r
Now
an =
or
d (20 m/s)2
=
2 3.2 m/s2
d = 250 m b
or
(b)
We have
v2
an =
r
Then
ˆ
Ê1
v 2 = (0.6 ¥ 9.81 m/s2 ) Á ¥ 180 m˜
¯
Ë2
or
v = 23.016 m/s
v = 82.9 km/h b
or
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197
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 0.8 m/s2 .
SOLUTION
an =
v2
r
an = 0.8 m/s2
v = 72 km/h = 20 m/s
0.8 m/s2 =
(20 m/s)2
r
r = 500 m b
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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
r
Then
( vmax )2AB = (3 ¥ 9.81 m/s2 )(24 m)
or
( vmax ) AB = 26.577 m/s
( vmax ) AB = 95.7 km/h b
or
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199
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/s2, 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 ( aA )n = 0
aA = ( aA )t
aA = 20 mm/s2 b
or
(b)
We have uniformly accelerated motion
v A = 0 + ( a A )t t
At t = 2 s:
Now
Finally,
v A = (20 mm/s2 )(2 s) = 40 mm/s
( aA )n =
v 2A ( 40 mm/s)2
=
= 17.7778 mm/s2
rA
90 mm
a2A = ( aA )t2 + ( aA )2n
= (20)2 + (17.7778)2
aA = 26.8 mm/s2 b
or
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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/s2 , 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 r = 400 m,
an =
v 2 (20)2
=
= 1.000 m/s2
r
400
a = an2 + at2
But
at = a2 - an2 = (1.5)2 - (1.000)2 = ± 1.11803 m/s2
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 b
Corresponding tangential acceleration.
at = 1.118 m/s2 b
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201
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 30 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 D s12
Substituting
or
Also
At t = 2 s:
Now
(7 m/s)2 = ( 4 m/s)2 + 2at (30 m)
at = 0.55 m/s2
v = v1 + at t
v = 4 m/s + (0.55 m/s2 )(2 s) = 5.1 m/s
v2
an =
r
(5.1 m/s)2
= 0.41616 m/s2
62.5 m
At t = 2 s:
an =
Finally
a2 = at2 + an2
At t = 2 s:
a2 = 0.552 + 0.416162
a = 0.690 m/s2 b
or
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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/s2 . 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/s2 , 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 v B = 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
v B2 /A = 4502 + 5402 - 2( 450)(540) cos 60∞
or
v B /A = 501.10 km/h
and
or
540 501.10
=
sin a sin 60∞
a = 68.9∞
v B /A = 501 km/h
(b)
First note
Now
a A = 8 m/s2 Æ (a B )t = 3 m/s2
( aB ) n =
v 2B
rB
=
60∞
(150 m/s)2
300 m
(a B ) n = 75 m/s2
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/s2 )i - (34.902 m/s2 ) j
Finally
or
a B = a A + a B /A
a B /A = ( -66.452i - 34.902 j) - (8i )
= -(74.452 m/s2 )i - (34.902 m/s2 ) j
a B /A = 82.2 m/s2
or
25.1∞ 
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203
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 170 m. He continues to decelerate at the
same constant rate so that 10 s after entering the ramp, his
speed has decreased to 30 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
25
m/s
3
While the car is on the straight portion of the highway.
First note
v10 = 30 km/h =
a = astraight = at
and for the circular exit ramp
a = at2 + an2
where
an =
v2
r
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 fi a = an =
a=
Then
or
astraight
1
ast.
4
(
2
v10
r
)2
25
2
m/s
v10
1
= at fi at =
= 3
r
4
170 m
at = -1.63399 m/s2
(The car is decelerating; hence the minus sign.)
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204
PROBLEM 11.141 (Continued)
Then at t = 10 s:
or
Then at t = 0:
25
m/s = v0 + ( -1.63399 m/s2 )(10 s)
3
v0 = 24.6732 m/s
amax =
at2
Ê v2 ˆ
+Á 0 ˜
Ë r¯
2
1/ 2
2
Ï
È (24.6732 m/s)2 ˘ ¸Ô
Ô
= Ì( -1.63399 m/s2 )2 + Í
˙ ˝
170 m
Î
˚ Ô˛
ÔÓ
amax = 3.94 m/s2 
or
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205
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/s2 , and the speed of B is increasing at the rate of
2 m/s2 . 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
v B = 144 km/h = 40 m/s
(a)
v B = v A + v B /A
We have
The graphical representation of this equation is then as shown.
We have
v B2 /A = 1622 + 1442 - 2(162)(144) cos165∞
or
v B /A = 303.39 km/h
and
or
144
303.39
=
sin a sin 165∞
a = 7.056∞
v B /A = 303 km/h
(b)
First note
(a A )t = 7 m/s2
60∞
(a B )t = 2 m/s2
45∞
52.9∞ 
v2
r
Now
an =
Then
( aA )n =
or
(a A ) n = 6.75 m/s2
( 45 m/s)2
300 m
( aB ) n =
30∞
( 40 m/s)2
250 m
(a B ) n = 6.40 m/s2
45°
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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/s2 )i - (9.4372 m/s2 ) j
and
a B = 2(cos 45∞i - sin 45∞ j) + 6.40(cos 45∞i + sin 45∞ j)
= (5.9397 m/s2 )i + (3.1113 m/s2 ) j
Finally
or
a B = a A + a B /A
a B /A = (5.9397i + 3.1113j) - ( -2.3457i - 9.4372 j)
= (8.2854 m/s2 )i + (12.5485 m/s2 ) j
a B /A = 15.04 m/s2
or
56.6∞ 
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207
PROBLEM 11.143
A golfer hits a golf ball 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
( aA )n =
rA =
v 2A
rA
(50 m/s)2
(9.81 m/s2 ) cos 25∞
r A = 281 m 
or
(b)
We have
( aB ) n =
v B2
rB
where Point B is the highest point of the trajectory, so that
v B = ( v A ) x = v A cos 25∞
Then
rB =
[(50 m/s) cos 25∞]2
9.81 m/s2
r B = 209 m 
or
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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
( aA )n =
v 2A
rA
v 2A = (9.81 cos 40∞)(8.5 m)
= 63.8766 m2 /s2
v A = 7.99 m/s
or
(b)
We have
( aB ) n =
v B2
40∞ 
rB
where Point B is the highest point of the trajectory, so that
v B = ( v A ) x = v A cos 40∞
Then
rB =
(63.8766 m2 /s2 ) cos2 40∞
9.81 m/s2
r B = 3.82 m 
or
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209
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
( aA )n =
(a) We have
or r A =
v 2A
rA
(2 m/s)2
(9.81 m/s2 ) sin 15∞
= 1.5754 m
r A = 1.575 m 
or
( aB ) n =
(b) We have
v B2
rB
where Point B is the highest point of the trajectory, so that
v B = ( v A ) x = v A sin 15∞
rB =
Then
[(2 m/s) sin 15∞]2
9.81 m/s2
= 0.02713 m
r B = 0.0271 m 
or
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210
PROBLEM 11.146
Coal is discharged from the tailgate A of a dump truck with an
initial velocity v A = 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)
( aA )n =
We have
v 2A
rA
(2 m/s)2
or
rA =
or
r A = 0.6341 m
(9.81 m/s2 ) cos 50∞
Horizontal motion. (Uniform)
( v B ) x = ( v A ) x = (2 m/s) cos 50∞ = 1.28558 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∞ = 1.53209 m/s
At Point B, y = -1 m:
(v B )2y = (1.53209 m/s)2 - 2(9.81 m/s2 )( -1 m)
or
( v B ) y = 4.6869 m/s
Then
and
or
Now
or
v B = ( v B )2x + ( v B )2y = (1.28558)2 + ( 4.6869)2 = 4.8600 m/s
tan q =
vy
vx
=
4.6869
1.28558
q = 74.661∞
( aB ) n =
rB =
v B2
rB
( 4.86 m/s)2
(9.81 m/s2 ) cos(74.661∞)
or
r B = 9.10 m 
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211
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
v B2
rB
We have
( aB ) n =
where
( aB ) n = aB cosq = g cosq
Noting that the horizontal motion is uniform, we have
(vB ) x = v A
where
( v B ) x = v B cos q
cosq =
Then
rB =
vA
vB
v B2
g
( )
vA
vB
rB =
or
v B2

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 y B = yC and v B = vC .
( aA )n =
Now
v 2A
rA
or
rA =
v 2A
g cos 25∞
Then
rB =
v 2A
3
3
rA =
4
4 g cos 25∞
( aB ) n =
We have
( aB ) n = g cos q
where
so that
or
v B2
rB
v 2A
v B2
3
=
4 g cos 25∞ g cos q
v B2 =
3 cos q 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∞ ( v B ) x = v B cos q
v A cos 25∞ = v B cos q
cos q =
vA
cos 25∞
vB
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213
PROBLEM 11.148 (Continued)
Substituting for cos q in Eq. (1), we have
or
v B2 =
ˆ v 2A
3 Ê vA
cos
25
∞
˜¯ cos 25∞
4 ÁË v B
v B3 =
3 3
vA
4
3
v B = 3 v A = 18.17 m/s
4
cos q = 3
4
cos 25∞
3
q = ± 4.04∞
v B = 18.17 m/s
4.04∞ 
and
v B = 18.17 m/s
4.04∞ 
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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 q 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 r = rmin /cos3 q .
SOLUTION
For the arbitrary Point C, we have
( aC ) n =
or
rC =
vC2
rC
vC2
g cos q
Noting that the horizontal motion is uniform, we have
( v A ) x = ( vC ) x
where
Then
( v A ) x = v0 cos a
( vC ) x = vC cos q
v0 cos a = vC cos q
cos a
v0
cos q
or
vC =
so that
v02 cos2 a
1 Ê cos a ˆ
v
=
rC =
0
g cos q ÁË cos q ˜¯
g cos3 q
2
(a)In the expression for rC , v0 , a , and g are constants, so that rC is minimum where cosq is maximum.
By observation, this occurs at Point B where q = 0.
rmin = r B =
(b)
rC =
rC =
v02 cos2 a
g
Q.E.D. 
Q.E.D. 
Ê v02 cos2 a ˆ
Á
˜
g
cos3 q Ë
¯
1
rmin
cos3 q
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215
PROBLEM 11.150
A projectile is fired from Point A with an initial
velocity v0 which forms an angle a with the
horizontal. Express the radius of curvature of
the trajectory of the projectile at Point C in
terms of x, v0 , a , and g.
SOLUTION
We have
or
( aC ) n =
rC =
vC2
rC
vC2
g cos q
Noting that the horizontal motion is uniform, we have
( v A ) x = ( vC ) x
where ( v A ) x = v0 cos a
Then
or
so that
x = 0 + (v0 ) x t = ( v0 cos a )t
( vC ) x = vC cos q
v0 cos a = vC cos q
cos q =
rC =
and
( vC ) x = v0 cos a (1)
v0
cos a
vC
vC3
g v0 cos a
For the uniformly accelerated vertical motion have
( vC ) y = ( v0 ) y - gt = v0 sin a - gt
From above
Then
Now
x = ( v0 cos a )t
( vC ) y = v0 sin a - g
or t =
x
v0 cos a
x
v0 cos a
(2)
vC2 = ( vC )x2 + ( vC )2y
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216
PROBLEM 11.150 (Continued)
Substituting for ( vC ) x [Eq. (1)] and ( vC ) y [Eq. (2)]
vC2
Ê
x ˆ
= ( v0 cos a ) + Á v0 sin a - g
v0 cos a ˜¯
Ë
2
2
Ê 2 gx tan a
g 2 x2 ˆ
= v02 Á1 +
˜
v02
v04 cos2a ¯
Ë
or
vC3
=
v03
Ê 2 gx tan a
g 2 x2 ˆ
1
+
Á
˜
v02
v04 cos2a ¯
Ë
3/ 2
Finally, substituting into the expression for rC , obtain
v02 Ê 2 gx tan a
g 2 x2 ˆ
r=
1
+
˜
g cos a ÁË
v02
v04 cos2 a ¯
3/ 2

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217
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
and
dr
= R(cos w n t - w n t sin w n t )i + cj + R(sin w n t + w n t cos w n t )k
dt
dv
a=
= R - w n sin w nt - w n sin w nt - w n2t cos w nt i
dt
v=
(
)
(
)
+ R w n cos w nt + w n cos w nt - w n2t sin w nt k
or
a = w n R [-(2 sin w nt + w nt cos w nt )i
+ (2 cos w nt - w nt sin w nt ) k ]
Now
v 2 = R2 (cos w nt - w nt sin w nt )2 + c 2
+ R2 (sin w nt + w nt cos w nt )2
(
)
= R2 1 + w n2t 2 + c 2
Then
(
)
1/ 2
v = È R2 1 + w n2t 2 + c 2 ˘
Î
˚
and
R2w n2t
dv
=
dt È R2 1 + w 2t 2 + c 2 ˘1/ 2
n
Î
˚
Now
a2 = at2 + an2
(
2
)
Ê v2 ˆ
Ê dv ˆ
=Á ˜ +Á ˜
Ë dt ¯
Ë r¯
At t = 0:
2
dv
=0
dt
a = w n R(2 k ) or a = 2w n R
v 2 = R2 + c 2
Then, with
we have
or
dv
= 0,
dt
v2
a=
r
2w n R =
R2 + c 2
r
r=
or
R2 + c 2

2w n R
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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=
(
)
Ê 3t ˆ
dr
= 3(cos t - t sin t )i + Á 2
˜ j + (sin t + t cos t )k
dt
Ë t + 1¯
È
Ê
2
dv
Á
a=
= 3( - sin t - sin t - t cos t )i + 3 Í t + 1 - t Ë
Í
dt
t2 + 1
ÍÎ
+ (cos t + cos t - t sin t )k
and
= - 3(2 sin t + t cos t )i + 3
t
t
2
ˆ˘
˜
+ 1¯ ˙ j
˙
˙˚
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(cos2 t + t 4 sin 2 t ) - 8(t 3 + t )sin 2t
v = [t 4 + 19t 2 + 1 + 8(cos2 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)siin 2t + 2(t 3 + t ) cos 2t ]
=
dt
2[t 4 + 19t 2 + 1 + 8(cos2 t + t 4 sin 2 t ) - 8(t 3 + t )sin 2t]1/ 2
2
a =
at2
+
an2
2
Ê v2 ˆ
Ê dv ˆ
=Á ˜ +Á ˜
Ë dt ¯
Ë r¯
2
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219
PROBLEM 11.152* (Continued)
At t = 0:
a = 3 j + 2k
or
a = 13 m/s2
dv
=0
dt
v 2 = 9 (m/s)2
Then, with
dv
= 0,
dt
we have
a=
or
r=
v2
r
9 m2 /s2
13 m/s2
r = 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/s2 , R = 6161 km.
SOLUTION
an = g
We have
R2
=
and an =
v2
r
v2
r
Then
g
or
vcirc. = R
r2
R2
r2
g
r
where r = R + h
For the given data
vcirc. = 6161 km
8.53 m/s2
3
(6161 + 160) ¥ 10 m
¥
3600 s
1h
vcirc. = 25.8 ¥ 103 km/h 
or
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221
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/s2 , R = 3332 km.
SOLUTION
an = g
We have
R2
r2
and an =
v2
r
R2 v 2
=
r
r2
Then
g
or
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 3332 km
3.83 m/s2
3
(3332 + 160) ¥ 10 m
¥
3600 s
1h
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/s2 , R = 69, 893 km.
SOLUTION
an = g
We have
R2
r2
and an =
v2
r
R2 v 2
=
r
r2
Then
g
or
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 69, 893
26.0 m/s2
3
(69, 893 + 160) ¥ 10 m
¥
3600 s
1h
vcirc. = 153.3 ¥ 103 km/h 
or
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223
PROBLEM 11.156
Knowing that the diameter of the sun is 1.391 ¥ 106 km and that the acceleration of gravity at its surface is
274 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 = 1.072 ¥ 105 km / h
SOLUTION
an = g
We have
Then
or
g
R2
r2
=
R2
v2
and
a
=
n
r
r2
v2
r
Ê Rˆ
r = gÁ ˜
Ë v¯
2
where R =
1
d
2
For the given data:
g = 274 m/s2 = 274 ¥
(3600)2
km/h 2 = 3551040 km/h 2
1000
ˆ
Ê1
¥ 1.391 ¥ 106 km
˜
2Á 2
rearth = 3551040 km/h Á
˜
5
ÁË 1.072 ¥ 10 km/h ˜¯
2
rearth = 149.5 ¥ 106 km 
or
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224
PROBLEM 11.157
Knowing that the diameter of the sun is 1.391 ¥ 106 km and that the acceleration of gravity at its surface is
274 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 = 3.47 ¥ 104 km / h
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v 2
=
r
r2
Ê Rˆ
r = gÁ ˜
Ë v¯
2
where R =
For the given data:
2
g = 274 m / s = 274 ¥
1
d
2
(3600)2
1000
km / h 2 = 3551040 km / h 2
ˆ
Ê1
¥ 1.391 ¥ 106 km
˜
2Á 2
rsaturn = 3551040 km / h Á
˜
4
.
km/h
3
47
¥
10
˜¯
ÁË
2
rsaturn = 1.427 ¥ 109 km 
or
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225
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 v 2
=
r
r2
v=R
g
r
where
r = R+h
The circumference s of the circular orbit is equal to
s = 2p r
Assuming that the speed of the telescope is constant, we have
s = vtorbit
Substituting for s and v
2p r = R
or
torbit =
=
g
torbit
r
2p r 3/2
R g
[(6370 + 590) km]3/2
1h
2p
¥
6370 km [9.81 ¥ 10 -3 km/s2 ]1/2 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 270 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 3370 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
=
R2
r2
and an =
v2
r
v2
r
v=R
g
r
where
r = R+h
The circumference s of a circular orbit is equal to
s = 2p r
Assuming that the speed of the satellite in each orbit is constant, we have
s = vtorbit
Substituting for s and v
2p r = R
or
torbit =
=
Now
or
or
g
torbit
r
2p r 3/2
R g
2p ( R + h)3/2
R
g
(torbit )2 = 1.1(torbit )1
2p ( R + h2 )3/2
2p ( R + h1 )3/2
= 1.1
R
R
g
g
h2 = (1.1)2/3 ( R + h1 ) - R
= (1.1)2/3 (3370 + 270) km - (3370 km)
= 508.792 km
h2 = 509 km 
or
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227
PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around the
earth at altitudes of 180 and 300 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
an = g
We have
Then
g
R2
r
2
=
R2
r2
and an =
v2
r
or
v2
r
v=R
g
r
r = R+h
where
The circumference s of a circular orbit is
equal to
s = 2p r
Assuming that the speeds of the satellites are constant, we have
s = vT
Substituting for s and v
2p r = R
or
Now
T=
g
T
r
2p r 3/ 2 2p ( R + h)3/ 2
=
R g
R
g
hB > hA fi (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
È 2p ( R + hA )3/ 2 ˘
È 2p ( R + hB )3/ 2 ˘
=
+
NÍ
N
(
1
)
˙
Í
˙
g
g
˙˚
ÍÎ R
˙˚
ÍÎ R
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228
PROBLEM 11.160 (Continued)
or
N=
=
Then
( R + hA )3/ 2
( R + hB )
(
3/ 2
1
)
- ( R + hA )
6370 + 300 3 / 2
6370 + 180
TC = N (T ) B = N
-1
3/ 2
=
1
( )
R + hB 3 / 2
R + hA
-1
= 36.223 orbits
2p ( R + hB )3/ 2
R
g
2p [(6370 + 300) km ]
1h
¥
/
1
2
6370 m 9.81 m/s2 ¥ 1 km
3600 s
1000 m
3/ 2
= 36.223
(
)
TC = 54.6 h 
or
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229
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 pt) and q = pt, where t and q 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 q for which the
magnitude of the velocity is maximum.
SOLUTION
We have
r = b(2 + cos p t )
q = pt
Then
r = -p b sin p t
q = p
and
r = -p 2 b cos p t
q = 0
Now
v = re r + rq eq = -(p b sin p t ) e r + p b(2 + cos p t ) eq
and
a = ( r - rq 2 ) e r + ( rq + 2rq ) eq
= [-p 2 b cos p t - p 2 b(2 + cos p t )]e r
+ (0 - 2p 2 b sin p t )eq
= -2p 2 b[(1 + cos p t ) e r + (sin p t ) eq ]
(a)
At t = 2 s:
v = -(0)e r + p b(2 + 1)eq
v = 3p beq 
or
a = -2p 2 b[(1 + 1) e r + (0) eq ]
a = -4p 2 be r 
or
(b)
We have
v = p b ( - sin p t )2 + (2 + cos p t )2
= p b 5 + 4 cos p t
q = pt
= p b 5 + 4 cos q
By observation,
v = vmax
when cos q = 1
q = 2np , n = 0, 1, 2, . . . 
or
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230
PROBLEM 11.162
The two-dimensional motion of a particle is defined by the relation r = 2b cos wt and q = wt where b and w
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 w t
r = -2bw sin w t
(a)
Velocity.
q = wt
q = w
r = -2bw 2 cos w t q = 0
vr = r = -2bw sin w t vq = rq = 2bw cos w t
v 2 = vr2 + vq2 = (2bw )2 [( - sin w t)2 + (cos w t)2 ] = (2bw )2 Acceleration.
v = 2bw 
ar = r - rq 2 = -2bw 2 cos w t - (2b cos w t)(w )2
= - 4bw 2 cos w t
aq = rq + 2rq = (2b cos w t)(0) + 2( - 2bw sin w t)(w )
= - 4bw 2 sin w t
a2 = ar2 + aq2 = ( - 4bw 2 )2 (cos2 w t + sin 2 w t) = ( 4bw 2 )2 (b)
Since
Thus:
a = 4bw 2 
v = 2bw = constant, at = 0
an = a = 4bw 2
an =
v2
v 2 (2bw )2
=
=b r=b
; r=
an
r
4bw 2
For the path, r = constant.
Thus, path is a circle.

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231
PROBLEM 11.163
The rotation of rod OA about O is defined by the relation q = p(4t2 - 8t), where q
and t are expressed in radians and seconds, respectively. Collar B slides along the
rod so that its distance from O is r = (250 + 150 sin p t), where r and t are expressed
in mm 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 = 25(10 + 6 sin p t )
q = p ( 4t 2 - 8t )
Then
r = 150p cos p t
q = qp (t - 1)
and
r = -150p 2 sin p t
q = qp
At t = 1 s:
r = 250 mm
q = -4p rad
r = -150p mm/s
q = 0
r=0
q = 8p rad/s2
(a)
We have
v B = re r + rq eq
v B = -(150p mm /s)e r 
so that
(b)
We have
a B = ( r - rq 2 )e r + ( rq + 2rq )eq
= (250)(8p )eq
a B = (2000p mm/s2 )eq 
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 q = (2 /p) (sin pt), where
q and t are expressed in radians and seconds, respectively. Collar B slides along the
rod so that its distance from O is r = 625/(t + 4) where r and t are expressed in mm 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
625
t+4
r=
Then
r = -
and
r=
At t = 1 s
r = 125 mm
q =0
r = -25 mm/s
q = -2 rad/s
r = 10 mm/s2
q = 0
(a)
We have
q=
2
sin p t
p
We have
625
(t + 4) 2
1250
( t + 4 )3
We have
q = -2p sin p t
v B = re r + rq eq = ( -1)e r + (5)( -2)eq
v A = -(25 mm/s)e r - (250 mm/s)eq 
or
(b)
q = 2 cos p t
a B = ( r - rq 2 )e r + ( rq + 2rq )eq
= [10 - (125)(2)2 ]e r + [0 + 2( -25)( -2)]eq
a B = -( 490 mm/s2 )e r + (100 mm/s2 )eq 
or
(c)
We have
a B /OA = r
a B /OA = (10 mm/s2 )e r 
so that
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233
PROBLEM 11.165
The path of particle P is the ellipse defined by the relations r = 2 / (2 - cos pt)
and q = pt, where r is expressed in meters, t is in seconds, and q is in
radians. Determine the velocity and the acceleration of the particle when
(a) t = 0, (b) t = 0.5 s.
SOLUTION
2
2 - cos p t
We have
r=
Then
r =
and
r = -2p
-2p sin p t
At t = 0:
Now
q = p
(2 - cos p t )2
p cos p t (2 - cos p t ) - sin p t (2p sin p t )
= -2p 2
(a)
q = pt
(2 - cos p t )3
2 cos p t - 1 - sin 2 p t
(2 - cos p t )3
r=2m
q =0
r = 0
q = p rad/s
r = -2p 2 m/s2
q = 0
v = re r + rq eq = (2)(p )eq
v = (2p m/s)eq 
or
and
q = 0
a = ( r - rq 2 )e r + ( rq + 2rq )eq
= [-2p 2 - (2)(p )2 ]e r
a = -( 4p 2 m/s2 )e r 
or
(b)
At t = 0.5 s:
r =1 m
r =
q=
p
-2p
= - m/s
2
2
(2)
r = -2p 2
-1 - 1 p 2
m/s2
=
2
(2)3
p
rad
2
q = p rad/s
q = 0
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234
PROBLEM 11.165 (Continued)
Now
Ê pˆ
v = re r + rq eq = Á - ˜ e r + (1)(p )eq
Ë 2¯
or
and
ˆ
Êp
v = - Á m/s˜ e r + (p m/s)eq 
¯
Ë2
r - rq 2 )e r + ( rq + 2rq )eq
a = ( Èp 2
˘
È Ê pˆ
˘
= Í - (1)(p )2 ˙ e r + Í2 Á - ˜ (p )˙ eq
Î Ë 2¯
˚
Î2
˚
Êp2
ˆ
m/s2 ˜ e r - (p 2 m/s2 )eq 
a = -Á
Ë 2
¯
or
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235
PROBLEM 11.166
The two-dimensional motion of a particle is defined by the relations r = 2a cos q and q = bt2 / 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)
We have
r = 2a cosq
1
q = bt 2
2
Then
r = -2aq sin q
q = bt
and
r = -2a(q sin q + q 2 + cos q )
q = b
Substituting for q and q
r = -2abt sin q
r = -2ab(sin q + bt 2 cos q )
Now
Then
vq = rq = 2abt cos q
vr = r = -2abt sin q
v = vr2 + vq2 = 2abt [( - sin q )2 + (cos q )2 ]1/ 2
v = 2abt 
or
Also
ar = r - rq 2 = -2ab(sin q + bt 2 cos q ) - 2ab2t 2 cos q
= -2ab(sin q + 2bt 2 coss q )
and
aq = rq + 2rq = 2ab cos q - 4ab2t 2 sin q
= -2ab(cos q - 2bt 2 sin q )
Then
a = ar2 + aB2 = 2ab[(sin q + 2bt 2 cos q )2
+ (cos q - 2bt 2 sin q )2 ]1/ 2
a = 2ab 1 + 4b2t 4 
or
(b)
Now
2
Ê v2 ˆ
Ê dv ˆ
a2 = at2 + an2 = Á ˜ + Á ˜
Ë dt ¯
Ë r¯
Then
dv d
= (2abt ) = 2ab
dt dt
2
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236
PROBLEM 11.166 (Continued)
so that
or
(
2ab 1 + 4b2t 4
) = (2ab) + a
2
2
2
n
4a2 b2 (1 + 4b2t 4 ) = 4 a2 b2 + an2
or
an = 4ab2t 2
Finally
an =
v2
(2abt )2
fi r =
r
4 ab2t 2
r=a 
or
Since the radius of curvature is a constant, the path is a circle of radius a.

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237
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, q, and q.
SOLUTION
We have
r=
b
cos q
Then
r =
bq sin q
cos2 q
We have
v 2 = vr2 + vq2 = ( r)2 + ( rq )2
2
2
Ê bq sin q ˆ
Ê bq ˆ
=Á
˜ + ÁË cos q ˜¯
Ë cos2q ¯
=
ˆ b2q 2
b2q 2 Ê sin 2q
+
1
Á
˜=
cos2q Ë cos2q ¯ cos 4q
v=±
or
bq
cos2q
For the position of the car shown, q is decreasing; thus, the negative root is chosen.
v=
bq

cos2q
v=-
bq

cos2q
Alternative solution.
r = - v sinq
From the diagram
or
bq sin q
cos2q
= - v sin q
or
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238
PROBLEM 11.168
Determine the magnitude of the acceleration of the race car of
and q.
Problem 11.167 in terms of b, q, q,
SOLUTION
We have
r=
b
cos q
Then
r =
bq sin q
cos2 q
For rectilinear motion
a=
dv
dt
From the solution to Problem 11.167
v=Then
a=
bq
cos2q
dÊ
bq ˆ
Á
dt Ë cos2q ˜¯
= -b
q cos2q - q ( -2q cos q sin q )
cos 4q
a=-
or
b
(q + 2q 2 tan q ) 
cos2q
Alternative solution
b
cos q
From above
r=
Then
r=b
r =
bq sin q
cos2 q
(q sin q + q 2 cos q )(cos2q ) - (q sin q )( -2q cos q sin q )
cos 4q
Èq sin q q 2 (1 + sin 2 q ) ˘
= bÍ
+
˙
2
cos3q
Î cos q
˚
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239
PROBLEM 11.168 (Continued)
Now
a2 = ar2 + aq2
where
Èq sin q q 2 (1 + sin 2 q ) ˘ bq 2
ar = r - rq 2 = b Í
+
˙2
cos2q
Î cos q
˚ cos q
=
ar =
and
Then
2q 2 sin 2 q ˆ
b Ê sin
+
q
q
Á
cos q ˜¯
cos2q Ë
b sin q (q + 2q 2 tan q )
2
cos q
bq 2 sin q
bq
aq = rq + 2rq =
+2
cos q
cos2q
b cos q (q + 2q tann q )
=
cos2q
a=±
b
(q + 2q 2 tan q )[(sin q )2 + (cos q )2 ]1/ 2
2
cos q
For the position of the car shown, q is negative; for a to be positive, the negative root is chosen.
a=-
b
(q + 2q 2 tan q ) 
2
cos q
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240
PROBLEM 11.169
After taking off, a helicopter climbs in
a straight line at a constant angle b. Its
flight is tracked by radar from Point A.
Determine the speed of the helicopter in
terms of d, b, q, and q.
SOLUTION
From the diagram
r
d
=
sin (180∞ - b ) sin ( b - q )
or
d sin b = r (sin b cos q - cos b sin q )
tan b
tan b cos q - sin q
or
r=d
Then
r = d tan b
-( - tan b sin q - cos q ) q
(tan b cos q - sin q )2
= dq tan b
tan b sin q + cos q
(tan b cos q - sin q )2
From the diagram
vr = v cos ( b - q )
where
vr = r
Then
dq tan b
tan b sin q + cos q
(tan b cos q - sin q )2
= v(cos b cos q + sin b sin q )
= v cos b (tan b sin q + cos q )
v=
or
dq tan b sec b
(tan b cosq - sin q )2

Alternative solution.
We have
v 2 = vr2 + vq2 = ( r)2 + ( rq )2
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241
PROBLEM 11.169 (Continued)
Using the expressions for r and r from above
È
tan b sin q + cos q ˘
v = Ídq tan b
˙
(tan b cos q - sin q )2 ˚
Î
2
1/ 2
or
v=±
È (tan b sin q + cos q )2 ˘
dq tan b
+ 1˙
Í
(tan b cos q - sin q ) Î (tan b cos q - sin q )2 ˚
1/ 2
=±
È
˘
tan 2 b + 1
dq tan b
Í
˙
(tan b cos q - sin q ) Î (tan b cos q - sin q )2 ˚
Note that as q increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v=
dq tan b sec b
(tan b cos q - sin q )2

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242
PROBLEM 11.170*
Pin P is attached to BC and slides freely in the slot of OA.
Determine the rate of change q of the angle q, knowing that
BC moves at a constant speed v0. Express your answer in terms
of v0, h, b, and q.
SOLUTION
From the diagram
r
h
=
sin (90∞ - b ) sin ( b + q )
or
r (sin b cos q + cos b sin q ) = h cos b
or
Also,
Then
r=
h
tan b cos q + sin q
vq = v0 sin ( b + q ) where vq = rq
hq
= v0 (sin b cos q + cos b sin q )
tan b cos q + sin q
v cos b
q = 0
(tan b cos q + sin q )2 
h
or
Alternative solution.
h
tan b cos q + sin q
From above
r=
Then
r = h
tan b sin q - cos q q
(tan b cos q + sin q )2
Now
v02 = vr2 + vq2 = ( r)2 + ( rq )2
or
v02
È
tan b sin q - cos q ˘
= Íhq
2˙
Î (tan b cos q + sin q ) ˚
2
2
Ê
ˆ
hq
+Á
Ë tan b cos b + sinn q ˜¯
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243
PROBLEM 11.170* (Continued)
or
v0 = ±
hq
tan b cos q + sin q
1/ 2
È (tan b sin q - cos q )2 ˘
+Í
+ 1˙
2
Î (tan b cos q + sin q )
˚
1/ 2
È
˘
tan 2 b + 1
hq
=±
Í
˙
tan b cos q + sin q Î (tan b cos q + sin q )2 ˚
Note that as q increases, member BC moves in the indicated direction. Thus, the positive root is chosen.
v cos b
q = 0
(tan b cos q + sin q )2 
h
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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 q = 60∞ to the position
q = 35∞. Knowing that b = 25 m, determine the average speed of
the car during the 0.5-s interval.
SOLUTION
From the diagram:
Dr12 = 25 tan 60∞ - 25 tan 35∞
= 25.796 m
Now
Dr12
Dt12
25.796 m
=
0.5 s
= 51.592 m/s
vave =
vave = 185.7 km/h 
or
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245
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 = 1000 m and q = 20∞, respectively.
Four seconds later, the radar station
sighted the helicopter at r = 1100 m and
q = 23.1∞. Determine the average speed
and the angle of climb b of the helicopter
during the 4-s interval.
SOLUTION
We have
r0 = 1000 m q 0 = 20∞
r4 = 1100 m q 4 = 23.1∞
From the diagram:
Dr 2 = 10002 + 11002
- 2(1000)(1100) cos (23.1∞ - 20∞)
or
Now
Dr = 114.975 m
Dr
Dt
114.975 m
=
4s
= 28.74375 m/s
vave =
vave = 103.5 km/h 
or
Also,
or
Dr cos b = r4 cos q 4 - r0 cos q 0
cos b =
1100 cos 23.1∞ - 1000 cos 20∞
114.975
b = 51.2∞ 
or
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246
PROBLEM 11.173
A particle moves along the spiral shown; determine the magnitude of the velocity
of the particle in terms of b, q, and q.
SOLUTION
Hyperbolic spiral.
r=
b
q
dr
b dq
b
=- 2
= - 2 q
dt
q
q dt
b b
vr = r = - 2 q vq = rq = q
q
q
r =
2
Ê 1ˆ
Ê 1ˆ
v = vr2 + vq2 = bq Á - 2 ˜ + Á ˜
Ëq¯
Ë q ¯
bq
= 2 1+q2
q
2
v=
b
1 + q 2 q 
q2
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247
PROBLEM 11.174
A particle moves along the spiral shown; determine the magnitude of the velocity
of the particle in terms of b, q, and q.
SOLUTION
Logarithmic spiral.
r = e bq
r =
dr
dq
= be bq
= be bq q
dt
dt
vr = r = be bq q vq = rq = e bq q
v = vr2 + vq2 = e bq q b2 + 1
v = e bq 1 + b2 q 
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248
PROBLEM 11.175
A particle moves along the spiral shown. Knowing that q is constant and denoting
this constant by w, determine the magnitude of the acceleration of the particle in
terms of b, q, and w.
SOLUTION
b
q
Hyperbolic spiral.
r=
From Problem 11.173
r = -
b q
q2
r=-
b 2b 2
q + 3q
q2
q
ar = r - rq 2 = -
b 2b 2 b 2
q + 3q - q
q
q2
q
b
b
b
Ê b ˆ
aq = rq + 2rq = q + 2 Á - 2 q ˜ q = q - 2 2 q 2
¯
Ë
q
q
q
q
Since q = w = constant q = 0, and we write:
b 2 bw 2
2
w
w = 3 (2 - q 2 )
q
q3
q
2
b
bw
aq = -2 2 w 2 = - 3 (2q )
q
q
ar = +
2b
a = ar2 + aq2 =
bw 2
bw 2
2 2
2
2
+
2
=
4 - 4q 2 + q 4 + 4q 2
(
q
)
(
q
)
q3
q3
a=
bw 2
4 +q4 
q3
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249
PROBLEM 11.176
A particle moves along the spiral shown. Knowing that q is constant and
denoting this constant by w, determine the magnitude of the acceleration of the
particle in terms of b, q, and w.
SOLUTION
Logarithmic spiral.
r = e bq
dr
= be bq q
dt
r = be bq q + b2 e bq q 2 = be bq (q + bq 2 )
r =
ar = r - rq 2 = be bq (q + bq 2 ) - e bq q 2
a = rq + 2rq = e bq q + 2(be bq q )q
q
Since q = w = constant, q = q , and we write
ar = be bq (bw 2 ) - e bq w 2 = e bq (b2 - 1)w 2
aq = 2be bq w 2
a = ar2 + aq2 = e bq w 2 (b2 - 1)2 + (2b)2
= e bq w 2 b 4 - 2b2 + 1 + 4b2 = e bq w 2 b 4 + 2b2 + 1
= e bq w 2 (b2 + 1)2 = e bq w 2 (b2 + 1)
a = (1 + b2 )w 2 e bq 
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250
PROBLEM 11.177
Show that r = hf sin q knowing that at the instant shown, step AB of
the step exerciser is rotating counterclockwise at a constant rate f .
SOLUTION
From the diagram
r 2 = d 2 + h2 - 2dh cosf
Then
Now
or
2rr = 2dhf sin f
r
d
=
sin f sin q
r=
d sin f
sin q
Substituting for r in the expression for r
Ê d sin f ˆ
˜ r = dhf sin f
ÁË
sin q ¯
or
r = hf sin q
Q.E.D. 
Alternative solution.
First note
a = 180∞ - (f + q )
Now
v = v r + vq = re r + rq eq
With B as the origin
vP = df
( d = constant fi d = 0)
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251
PROBLEM 11.177 (Continued)
With O as the origin
( vP )r = r
where
( vP )r = vP sin a
Then
Now
or
r = df sin a
h
d
=
sin a sin q
d sin a = h sin q
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252
PROBLEM 11.178
The motion of a particle on the surface of a right circular cylinder is
defined by the relations R = A, q = 2pt, and z = At2 /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
q = 2p t
z=
1 2
At
4
Then
R = 0
q = 2p
z =
1
At
2
and
= 0
R
q = 0
z =
1
A
2
Now
v 2 = vR2 + vq2 + v z2
0
R ) 2 + ( Rq )2 + ( z )2
= (
Ê1 ˆ
= 0 + ( A + 2p )2 + Á At ˜
Ë2 ¯
2
1 ˆ
Ê
= A2 Á 4p 2 + t 2 ˜
Ë
4 ¯
v=
or
and
1
A 16p 2 + t 2 
2
a2 = aR2 + aq2 + az2
0
0
0- Rq 2 )2 + ( R
R
R q )2 + ( z )2
= (
q +2
Ê1 ˆ
= [- A(2p )2 ]2 + 0 + Á A˜
Ë2 ¯
2
1ˆ
Ê
= A2 Á16p 4 + ˜
Ë
4¯
a=
or
1
A 64p 4 + 1 
2
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253
PROBLEM 11.179
The three-dimensional motion of a particle is defined by the cylindrical coordinates (see Figure 11.26)
R = A / (t + 1), q = 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 = -
q = Bt ,
A
,
(t + 1)2
z=
q = B
Ct
t +1
z = C
=
=
R
and
2A
(t + 1)3
q = 0
z = -
(t + 1) - t
(t + 1)2
C
(t + 1)2
zC
(t + 1)3
Now
v 2 = ( vR )2 + ( vq )2 + ( v z )2
= ( R )2 + ( Rq )2 + ( z )2
and
a2 = ( aR )2 + ( aq )2 + ( az )2
0
- Rq 2 )2 + ( R
= (R
q + 2 Rq )2 + ( z )2
(a)
At t = 0:
R= A
R = -A
q = B
z = C
= 2 A
R
Then
z = -2C
v 2 = ( - A)2 + ( AB)2 + (C )2
v = (1 + B 2 ) A2 + C 2 
or
and
a2 = (2 A - AB 2 )2 + [ z ( - A)( B)]2 + ( -2C )2
ÈÊ 1 2 ˆ 2
C2 ˘
= 4 A ÍÁ1 - B ˜ + B 2 + 2 ˙
Ë 2 ¯
A ˙˚
ÎÍ
ÈÊ 1 ˆ
˘
= 4 ÍÁ1 + B 4 ˜ A2 + C 2 ˙
¯
Ë
4
Î
˚
2
Ê 1 ˆ
a = 2 Á1 + B 4 ˜ A2 + C 2 
Ë 4 ¯
or
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254
PROBLEM 11.179 (Continued)
(b)
As t Æ •:
R=0
R = 0
= 0
R
q = B
z = 0
z = 0
v=0 
a=0 
and
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255
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
Then
and
r = ( Rt cos w nt )i + ctj + ( Rt sin w nt )k
dr
= R(cos w nt - w nt sin w nt )i + cj + R(sin w nt + w nt cos w nt )k
dt
dv
a=
dt
v=
(
)
cos w t - w t sin w t ) k
= R -w n sin w nt - w n sin w nt - w n2t cos w nt i
(
+ R w n cos w nt + w n
n
2
n
n
= w n R[-(2 sin w nt + w nt cos w nt )i + (2 cos w nt - w nt sin w nt)k ]
It then follows that the vector (v ¥ a) is perpendicular to the osculating plane.
i
j
k
( v ¥ a) = w n R R(cos w nt - w nt sin w nt ) c R(sin w nt + w nt cos w nt )
-(2 sin w nt + w nt cos w nt ) 0 (2 cos w nt - w nt sin w nt )
= w n R{c(2 cos w nt - w nt sin w nt )i + R[-(sin w nt + w nt cos w nt )(2 sin w nt + w nt cos w nt )
- (cos w nt - w nt sin w nt )(2 cos w nt - w nt sin w nt )]j + c(2 sin w nt + w nt cos w nt )k
(
)
= w n R Èc(2 cos w nt - w nt sin w nt )i - R 2 + w n2t 2 j + c(2 sin w nt + w nt cos w nt )k ˘
Î
˚
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256
PROBLEM 11.180* (Continued)
The angle a formed by the vector (v ¥ a) and the y axis is found from
cos a =
where
( v ¥ a) ◊ j
| ( v ¥ a ) || j |
|j| =1
(
( v ¥ a) ◊ j = -w n R2 2 + w n2t 2
)
(
|( v ¥ a) | = w n R ÈÍc 2 (2 cos w nt - w nt sin w nt )2 + R2 2 + w n2t 2
Î
)
2
1/ 2
+ c 2 (2 sin w nt + w nt cos w nt )2 ˘˚
(
)
(
)
1/ 2
2
= w n R ÈÍc 2 4 + w n2t 2 + R2 2 + w n2t 2 ˘˙
Î
˚
Then
(
)
w R ÈÍc ( 4 + w t ) + R (2 + w t ) ˘˙
Î
˚
- R (2 + w t )
=
Èc 4 + w t + R 2 + w t ˘
) (
) ˙˚
ÍÎ (
-w n R2 2 + w n2t 2
cosa =
2
n
2 2
n
2
1/ 2
2 2 2
n
2 2
n
2
2 2
n
2
1/ 2
2 2 2
n
The angle b that the osculating plane forms with y axis (see the above diagram) is equal to
b = a - 90∞
Then
cos a = cos ( b + 90∞) = - sin b
- sin b =
Then
tan b =
(
)
) + R (2 + w t ) ˘˚˙
- R 2 + w n2t 2
(
Èc 2 4 + w 2 t 2
n
ÎÍ
(
R 2 + w n2t 2
c
2
1/ 2
2 2 2
n
)
4 + w n2t 2
(
)
È R 2 + w n2t 2 ˘
˙ 
b = tan
Í c 4 + w 2t 2 ˙
n
˚
Î
-1 Í
or
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257
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 = p / 2 s.
SOLUTION
(
)
r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k
Given:
r - m, t - s;
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
= 3(cos t - t sin t )i +
3t
t2 +1
j + (sin t + t cos t )k
( )
t
2
dv
t 2 +1
a=
= 3( - sin t - sin t - t cos t )i + 3 t + 12- t
j
dt
t +1
+ (cos t + cos t - t sin t )k
3
= -3(2 sin t + t cos t )i + 2
j + (2 cos t - t sin t )k
(t + 1)3/ 2
and
(a)
At t = 0:
v = (3 m/s)i
a = (3 m/s2 ) j + (2 m/s2 )k
Then
and
Then
or
v ¥ a = 3i ¥ (3j + 2k )
= 3( -2 j + 3k )
| v ¥ a | = 3 ( -2)2 + (3)2 = 3 13
3( -2 j + 3k )
1
=
( -2 j + 3k )
3 13
13
2
3
cos q x = 0
cos q y = cos q z =
13
13
eb =
q x = 90∞
q y = 123.7∞
q z = 33.7∞ 
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258
PROBLEM 11.181* (Continued)
(b)
At t =
p
s:
2
ˆ
ˆ Ê 3p
Ê 3p
v = -Á
m/s˜ i + Á
m/s˜ j + (1 m/s)k
¯ Ë p2 + 4
Ë 2
¯
È
˘ Êp
24
ˆ
m/s2 ˙ j - Á m/s2 ˜ k
a = -(6 m/s2 )i + Í 2
3/ 2
¯
Ë
2
Î (p + 4)
˚
Then
i
3p
v¥a = 2
-6
j
3p
(p 2 + 4)1/ 2
24
2
(p + 4)3/ 2
k
1
-
p
2
È
˘ Ê
3p 2
24
3p 2 ˆ
+
= -Í
+
i
j
6
˙ Á
2
1/ 2
4 ˜¯
(p 2 + 4)3/ 2 ˚ Ë
Î 2(p + 4)
È
˘
18p
36p
+ Í- 2
k
+ 2
3/ 2
1/ 2 ˙
(p + 4) ˚
Î (p + 4)
= -4.43984i - 13.40220 + 12.99459k
and
Then
| v ¥ a | = [( -4.43984)2 + ( -13.40220)2 + (12.99459)2 ]1/ 2
= 19.18829
eb =
1
( -4.43984i - 13.40220 j + 12.99459k )
19.1829
cos q x = or
13.40220
4.43984
cos q y = 19.18829
19.18829
q x = 103.4∞
q y = 134.3∞
cos q z =
12.99459
19.18829
q z = 47.4∞ 
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259
PROBLEM 11.182
The motion of a particle is defined by the relation x = 2t3 - 15t2 + 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
a=
= 12t - 30
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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260
PROBLEM 11.183
The acceleration of a particle is defined by the relation a = - 60x -1.5, where a and x are expressed in m /s2 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
We have
When x = 4 m, v = 0:
v
dv
= a = - 60 x -1.5
dx
x
Ú0 vdv = Ú4 (-60 x
-1.5
)dx
1 2
v = 120[ x -0.5 ]4x
2
or
Ê 1 1ˆ
v 2 = 240 Á
Ë x 2 ˜¯
or
(a)
v
When x = 2 m:
Ê 1 1ˆ
v 2 = 240 Á
Ë 2 2 ˜¯
v = -7.05 m/s 
or
(b)
When x = 1 m:
Ê 1ˆ
v 2 = 240 Á1 - ˜
Ë 2¯
v = -10.95 m/s 
or
(c)
When x = 0.1 m:
1ˆ
Ê 1
v 2 = 240 Á
Ë 0.1 2 ˜¯
v = -25.3 m/s 
or
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261
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 m, 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:
Ê 1ˆ
a = -2700 Á ˜ (270 m/s - 0)
Ë 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
dx
t
Ú0 v0 - kx = Ú 0 dt
1
- [ln( v0 - kx )]0x = t
k
or
t=
1 Ê v0 ˆ 1 Ê 1
ln
= ln Á
k ÁË v0 - kx ˜¯ k Ë 1 - vk
0
When x = 97.5 mm = 0.0975 m
t=
ˆ
˜
x¯
1
È
˘
1
ln Í 2700 1/s
˙
1
2700 s Î1 - 270 m/s (0.0975 m ) ˚
t = 1.366 ¥ 10 -3 s 
or
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262
PROBLEM 11.185
A freight elevator moving upward with a constant velocity of 2 m /s passes a passenger elevator which is
stopped. Four seconds later, the passenger elevator starts upward with a constant acceleration of 0.8 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 ≥ 0:
yF = 0 + vF t
t ≥ 4 s:
1
yP = 0 + 0(t - 4) + aP (t - 4)2
2
yF = yP
1
(2 m/s)t = (0.8 m/s2 )(t - 4)2
2
When
Expanding and simplifying
t 2 - 13t + 16 = 0
Solving
t = 1.3765 s and t = 11.6235 s
Must require t > 4 s
At t = 11.6235 s:
t = 11.62 s 
yF = (2 m/s)(11.6235 s) = 23.247 m
yF = yP = 23.2 m 
or
(b)
For t ≥ 4 s:
vP = 0 + aP (t - 4)
At t = 11.6235 s:
vP = (0.8 m/s2 )(11.6235 - 4) s
vP = 6.10 m/s 
or
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263
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 = 2yC + 2yD + constant
L2 = yA + yB + (yB - yD) + constant
Eliminate yD .
L1 + 2 L2 = 2 yC + 2 yD + 2 y A + 2 y B + 2( y B - yD ) + constant
2(yC + y A + 2y B ) = constaant
Differentiate to obtain relationships for velocities and accelerations, positive downward.
vC + v A + 2v B = 0 (1)
aC + aA + 2aB = 0 (2)
Initial velocities.
( vC )0 = 0, ( v A )0 = 75 mm/s
From (1),
1
( v B )0 = - [( vC )0 + ( vA )0 ] = -37.5 mm/s
2
Accelerations.
aC = -25 mm/s aA = 0
From (2),
aB = -
1
(ac + aA ) = 12.5 mm/s
2
Motion of block B.
v B = ( v B )0 + aB t = -37.5 + 12.5t
1
Dy B = ( v B )0 t + aB t 2 = -37.5t + 6.25t 2
2
-37.5 + 12.5t = 0 (a)
Time at v B = 0.
(b)
Corresponding position.
Dy B = ( -37.5)(3.00) + (6.25)(3.00)2 t = 3.00 s 
Dy B = -56.25 mm 
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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)
( y B - yD ) + ( yC - yD ) = constant
v B + vC - 2vD = 0 Then
(2)
Combining Eqs. (1) and (2) to eliminate vD
2v A + v B + vC = 0 (3)
Now
v A/C = v A - vC = -300 mm/s (4)
and
v B /A = v B - v A = 200 mm/s (5)
(3) + ( 4) - (5) fi
Then
(2v A + v B + vC ) + ( v A - vC ) - ( v B - v A ) = ( -300) - (200)
v A = 125 mm/s ≠ 
or
v B - ( -125) = 200
and using Eq. (5)
v B = 75 mm/s Ø 
or
-125 - vC = -300
Eq. (4)
vC = 175 mm/s Ø 
or
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265
PROBLEM 11.188
An oscillating water sprinkler at Point A rests on an incline
which forms an angle a with the horizontal. The sprinkler
discharges water with an initial velocity v0 at an angle f with
the vertical which varies from - f0 to + f0. Knowing that
v0 = 10 m /s f0 = 40∞, and a = 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 f = (10 m/s)sin f
( v0 ) y = v0 cos f = (10 m/s) cos f
Also, along incline CAB
y = x tan10∞
Horizontal motion. (Uniform)
x = 0 + ( v0 ) x t = (10 sin f ) t
or t =
x
10 sin f
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v0 ) y t Substituting for t
Ê x ˆ 1 Ê x ˆ
y = (10 cos f ) Á
- g
Ë 10 sin f ˜¯ 2 ÁË 10 sin f ˜¯
=
At B:
f = 40∞, x = d B :
1 2
1
gt = (10 cos f )t - gt 2
2
2
d B tan 10∞ =
x
g x2
tan f 200 sin 2 f
( g = 9.81 m/s2 )
dB
9.81 d B2
tan 40∞ 200 sin 2 40∞
d B = 8.55 m 
or
At C:
2
f = -40∞, x = - dC :
- dC tan 10∞ =
- dC
9.81 ( - dC )2
tan ( -40∞) 200 sin 2 ( -40∞)
dC = 11.52 m 
or
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266
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
v T = v A + v T /A
Using this equation, the two cases are then graphically represented as shown.
From the diagram
( vT ) x = 25 - 15 sin 45∞
= 14.3934 km/h
(vT ) y = 15 sin 45∞
= 10.6066 km/h
vT = 17.88 km/h
36.4∞ 
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267
PROBLEM 11.190
The driver of an automobile decreases her speed at a constant rate from 70 to 45 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
175
m/s
9
25
v2 = 45 km/h =
m/s
2
v1 = 70 km/h =
We have uniformly decelerated motion
v 2 = v12 + 2at ( s - s1 )
2
When v = v2 :
2
ˆ
Ê 175
ˆ
Ê 25
m/s˜ = Á
m/s˜ + 2at (225 m)
ÁË
¯
Ë 9
¯
2
or
at = -0.49297 m/s2
Then when Ds = 150 m
ˆ
Ê 175
m/s˜ + 2( -0.49297 m/s2 )(150 m)
v2 = Á
¯
Ë 9
2
= 230.195 m2 /s2
Now
an =
v2
r
230.195 m2 /s2
450 m
= 0.51154 m/s2
=
Finally
a2 = at2 + an2
= ( -0.49297 m/s2 )2 + (0.51154 m/s2 )2
a = 0.710 m/s2 
or
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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
( v x )0 = v0 cos 40∞
( v y )0 = v0 sin 40∞
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
At B:
4.2 = ( v0 cos 40∞) t
or t B =
4.2
v0 cos 40∞
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t At B:
1 2
gt
2
0.45 = ( v0 sin 40∞) t B -
( g = 9.81 m/s2 )
1 2
gt B
2
Substituting for t B
Ê
4.2 ˆ 1 Ê
4.2 ˆ
0.45 = ( v0 sin 40∞) Á
- gÁ
˜
Ë v0 cos 40∞ ¯ 2 Ë v0 cos 40∞ ˜¯
or
v02
=
2
1
2 (9.81)(17.64) / cos
2
40∞
-0.45 + 4.2 tan 40∞
v0 = 6.93 m/s 
or
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269
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
( aA )n =
v 2A
rA
or
È4
˘
v 2A = Í (9.81 m/s2 )˙ (25 m)
Î5
˚
or
v A = 14.0071 m/s
vA = 14.01 m/s
(b)
We have
where
Then
( aB ) n =
v B2
rB
vB = (v A ) x =
rB
36.9° 
4
vA
5
2
4
(
5 ¥ 14.0071 m/s )
=
9.81 m/s2
r B = 12.80 m 
or
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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 values
of r, r , q and q for this instant?
SOLUTION
Geometry. The polar coordinates are
Ê 600 ˆ
= 36.87∞
v = (800)2 + (600)2 = 1000 m q = tan -1 Á
Ë 800 ˜¯
Velocity Analysis.
v = 150 m /s Æ
vr = 150 cos q = 120 m/s
vq = -150 sin q = -90 m/s
vr = r r = 120 m/s 
v
90
vq = rq q = q = r
1000
q = -0.0900 rad/s 
Acceleration analysis.
at = 25 m/s2
an =
v 2 (150)2
=
= 11.25 m/s2
r
2000
a = 25 m /s2 Æ + 11.25 m /s2 ≠ = 27.41 m /s2
24.23°
b = 24.23∞
q - b = 12.64∞
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271
PROBLEM 11.193 (Continued)
ar = a cos (q - b ) = 27.41 cos 12.64∞ = 26.74 m/s2
aq = - a sin (q - b ) = -27.41 sin 12.64∞ = -6.00 m/s2
a = r - rq 2 r = a + rq 2
r
r
r = 26.74 + (1000)(0.0900)2
a = rq + 2rq
r = 34.8 m/s2 
q
a
2rq
q = q r
r
=
-6.00 (2)(120)( -0.0900)
1000
1000
q = -0.0156 rad/s2 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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272