# x 2 t onds, respectively. Determine the position, the velocity, and the

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PROBLEMS†
11.1 The motion of a particle is defined by the relation x 5 1.5t4 2
30t2 1 5t 1 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 5 4 s.
11.2 The motion of a particle is defined by the relation x 5 12t3 2
18t2 1 2t 1 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.
11.3 The motion of a particle is defined by the relation x 5 53 t3 2 52 t2 2
30t 1 8x, where x and t are expressed in feet and seconds,
respectively. Determine the time, the position, and the acceleration when v 5 0.
11.4 The motion of a particle is defined by the relation x 5 6t2 2
8 1 40 cos pt, where x and t are expressed in inches and seconds,
respectively. Determine the position, the velocity, and the acceleration when t 5 6 s.
11.5 The motion of a particle is defined by the relation x 5 6t4 2 2t3 2
12t2 1 3t 1 3, where x and t are expressed in meters and seconds,
respectively. Determine the time, the position, and the velocity
when a 5 0.
11.6 The motion of a particle is defined by the relation x 5 2t3 2 15t2 1
24t 1 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.
11.7 The motion of a particle is defined by the relation x 5 t3 2 6t2 2
36t 2 40, where x and t are expressed in feet and seconds, respectively. Determine (a) when the velocity is zero, (b) the velocity, the
acceleration, and the total distance traveled when x 5 0.
11.8 The motion of a particle is defined by the relation x 5 t3 2 9t2 1
24t 2 8, where x and t are expressed in inches and seconds,
respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.
11.9 The acceleration of a particle is defined by the relation a 5 28 m/s2.
Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when
v 5 16 m/s, determine (a) the time when the velocity is zero,
(b) the velocity and the total distance traveled when t 5 11 s.
11.10 The acceleration of a particle is directly proportional to the square
of the time t. When t 5 0, the particle is at x 5 24 m. Knowing
that at t 5 6 s, x 5 96 m and v 5 18 m/s, express x and v in
terms of t.
†Answers to all problems set in straight type (such as 11.1) are given at the end of the
book. Answers to problems with a number set in italic type (such as 11.7 ) are not given.
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11.11 The acceleration of a particle is directly proportional to the time t.
Kinematics of Particles
At t 5 0, the velocity of the particle is v 5 16 in./s. Knowing that
v 5 15 in./s and that x 5 20 in. when t 5 1 s, determine the
velocity, the position, and the total distance traveled when t 5 7 s.
11.12 The acceleration of a particle is defined by the relation a 5 kt2.
(a) Knowing that v 5 232 ft/s when t 5 0 and that v 5 132 ft/s
when t 5 4 s, determine the constant k. (b) Write the equations of
motion, knowing also that x 5 0 when t 5 4 s.
11.13 The acceleration of a particle is defined by the relation a 5 A 2 6t2,
where A is a constant. At t 5 0, the particle starts at x 5 8 m with
v 5 0. Knowing that at t 5 1 s, v 5 30 m/s, determine (a) the
times at which the velocity is zero, (b) the total distance traveled
by the particle when t 5 5 s.
11.14 It is known that from t 5 2 s to t 5 10 s the acceleration of a
particle is inversely proportional to the cube of the time t. When
t 5 2 s, v 5 215 m/s, and when t 5 10 s, v 5 0.36 m/s. Knowing
that the particle is twice as far from the origin when t 5 2 s as it
is when t 5 10 s, determine (a) the position of the particle when
t 5 2 s and when t 5 10 s, (b) the total distance traveled by the
particle from t 5 2 s to t 5 10 s.
11.15 The acceleration of a particle is defined by the relation a 5 2k/x.
It has been experimentally determined that v 5 15 ft/s when
x 5 0.6 ft and that v 5 9 ft/s when x 5 1.2 ft. Determine
(a) the velocity of the particle when x 5 1.5 ft, (b) the position of
the particle at which its velocity is zero.
11.16 A particle starting from rest at x 5 1 ft is accelerated so that its
velocity doubles in magnitude between x 5 2 ft and x 5 8 ft.
Knowing that the acceleration of the particle is defined by the
relation a 5 k[x 2 (A/x)], determine the values of the constants A
and k if the particle has a velocity of 29 ft/s when x 5 16 ft.
11.17 A particle oscillates between the points x 5 40 mm and
x 5 160 mm with an acceleration a 5 k(100 2 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 5 100 mm and is
zero at both x 5 40 mm and x 5 160 mm. Determine (a) the value
of k, (b) the velocity when x 5 120 mm.
11.18 A particle starts from rest at the origin and is given an acceleration
a 5 k/(x 1 4)2, where a and x are expressed in m/s2 and m, respectively, and k is a constant. Knowing that the velocity of the particle
is 4 m/s when x 5 8 m, determine (a) the value of k, (b) the position of the particle when v 5 4.5 m/s, (c) the maximum velocity
of the particle.
v
Fig. P11.19
11.19 A piece of electronic equipment that is surrounded by packing mate-
rial is dropped so that it hits the ground with a speed of 4 m/s. After
impact the equipment experiences an acceleration of a 5 2kx,
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.
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Problems
11.20 Based on experimental observations, the acceleration of a particle
is defined by the relation a 5 2(0.1 1 sin x/b), where a and x are
expressed in m/s2 and meters, respectively. Knowing that
b 5 0.8 m and that v 5 1 m/s when x 5 0, determine (a) the
velocity of the particle when x 5 21 m, (b) the position where the
velocity is maximum, (c) the maximum velocity.
11.21 Starting from x 5 0 with no initial velocity, a particle is given an
acceleration a 5 0.8 2v2 1 49, where a and v are expressed in
m/s2 and m/s, respectively. Determine (a) the position of the particle
when v 5 24 m/s, (b) the speed of the particle when x 5 40 m.
11.22 The acceleration of a particle is defined by the relation a 5 2k 1v,
where k is a constant. Knowing that x 5 0 and v 5 81 m/s at
t 5 0 and that v 5 36 m/s when x 5 18 m, determine (a) the
velocity of the particle when x 5 20 m, (b) the time required for
the particle to come to rest.
11.23 The acceleration of a particle is defined by the relation a 5 20.8v
30 ft
where a is expressed in in./s2 and v in in./s. Knowing that at t 5 0
the velocity is 40 in./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.
11.24 A bowling ball is dropped from a boat so that it strikes the surface
of a lake with a speed of 25 ft/s. Assuming the ball experiences a
downward acceleration of a 5 10 2 0.9v2 when in the water, determine the velocity of the ball when it strikes the bottom of the
lake.
Fig. P11.24
11.25 The acceleration of a particle is defined by the relation a 5 0.4(1 2
kv), where k is a constant. Knowing that at t 5 0 the particle starts
from rest at x 5 4 m and that when t 5 15 s, v 5 4 m/s, determine
(a) the constant k, (b) the position of the particle when v 5 6 m/s,
(c) the maximum velocity of the particle.
v
11.26 A particle is projected to the right from the position x 5 0 with
an initial velocity of 9 m/s. If the acceleration of the particle is
defined by the relation a 5 20.6v3/2, where a and v are expressed
in m/s2 and m/s, respectively, determine (a) the distance the particle will have traveled when its velocity is 4 m/s, (b) the time when
v 5 1 m/s, (c) the time required for the particle to travel 6 m.
Fig. P11.27
11.27 Based on observations, the speed of a jogger can be approximated
by the relation v 5 7.5(1 2 0.04x)0.3, where v and x are expressed
in mi/h and miles, respectively. Knowing that x 5 0 at t 5 0, determine (a) the distance the jogger has run when t 5 1 h, (b) the jogger’s
acceleration in ft/s2 at t 5 0, (c) the time required for the jogger to
run 6 mi.
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 5 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 5
3.6 m/s, determine (a) the acceleration of the air at x 5 2 m, (b) the
time required for the air to flow from x 5 l to x 5 3 m.
v
x
Fig. P11.28
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9.183 (a) K1 5 2.26g ta4yg; K 2 5 17.27g ta4yg; K3 5 19.08g ta4yg.
9.185
9.187
9.188
9.189
9.190
9.191
9.193
9.195
9.C1
9.C3
9.C5
9.C6
(b) (ux)1 5 85.0°, (uy)1 5 36.8°, (uz)1 5 53.7°;
(ux)2 5 81.7°, (uy)2 5 54.7°, (uz)2 5 143.4°;
(ux)3 5 9.70°, (uy)3 5 99.0°, (uz)3 5 86.3°.
Ix 5 ab3y28; Iy 5 a3 by20.
4 a3 by15; ay 15.
Ix 5 4 a4; Iy 5 16 a4y3.
(a) 3.13 3 10 6 mm4. (b) 2.41 3 10 6 mm4.
Ix 5 634 3 10 6 mm4; Iy 5 38.0 3 10 6 mm4.
Ixy 5 22.81 in4.
(a) 7ma 2y18. (b) 0.819 ma 2.
Ix 5 0.877 kg ? m2; Iy 5 1.982 kg ? m2; Iz 5 1.652 kg ? m2.
u 5 20°: Ix9 5 14.20 in4, Iy9 5 3.15 in4, Ix9 y9 5 23.93 in4.
(a) I x¿ 5 371 3 103 mm4, Iy¿ 5 64.3 3 103 mm4;
kx¿ 5 21.3 mm, ky¿ 5 8.87 mm. (b) I x¿ 5 40.4 in4,
Iy¿ 5 46.5 in4; kx¿ 5 1.499 in., ky¿ 5 1.607 in.
(c) kx 5 2.53 in., ky 5 1.583 in. (d) kx 5 1.904 in.,
ky 5 0.950 in.
(a) 5.99 3 1023 kg ? m2. (b) 77.4 3 1023 kg ? m2.
(a) 74.0 3 1026 lb ? ft ? s2. (b) 645 3 1026 lb ? ft ? s2.
(c) 208 3 10−6 lb ? ft ? s2.
CHAPTER 10
10.1
10.2
10.3
10.4
10.7
10.8
10.9
10.10
10.11
10.12
10.15
10.16
10.17
10.18
10.21
10.22
10.24
10.25
10.26
10.28
10.30
10.31
10.32
10.33
10.35
10.36
10.37
10.38
10.39
10.40
10.43
10.44
10.45
10.46
10.47
10.49
82.5 Nw.
120 lb y.
49.5 N ? m i.
1200 lb ? in. l.
(a) 60.0 N C, 8.00 mmw. (b) 300 N C, 40.0 mmw.
(a) 120.0 N C, 16.00 mmw. (b) 300 N C, 40.0 mmw.
Q 5 2 P sin uycos (uy2).
Q 5 2 P cos uycos (uy2).
Q 5 (3Py2) tan u
Q 5 P[(lya)cos3 u 2 1].
M 5 12 Wl tan a sin u.
M 5 Ply2 tan u.
M 5 7Pa cos u
(a) M 5 Pl sin 2 u. (b) M 5 3Pl cos u. (c) M 5 Pl sin u.
85.2 lb ? ft i.
22.8 lb d 70.0°.
36.4°.
38.7°.
68.0°.
19.81° and 51.9°.
25.0°.
39.7° and 69.0°.
52.2°.
40.2°.
22.6°.
51.1°.
52.4°.
19.40°.
59.0°.
78.7°, 324°, 379°.
12.03 kN q .
20.4°.
2370 lb r.
2550 lb r.
h 5 1y(1 1 m cot a)
37.6 N, 31.6 N.
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10.51
10.52
10.53
10.54
10.57
10.58
10.66
10.67
10.69
10.70
10.71
10.72
10.73
10.74
10.78
10.79
10.80
10.81
10.83
10.85
10.86
10.88
10.90
10.91
10.92
10.93
10.94
10.96
10.98
10.100
10.101
10.102
10.104
10.106
10.107
10.108
10.110
10.112
10.C1
10.C2
10.C3
10.C4
10.C5
10.C6
10.C7
300 N ? m, 81.8 N ? m.
h 5 tan uytan (u 1 fs).
7.75 kNx.
H 5 1.361 kNx; MH 5 550 N ? m l.
0.833 in.w.
0.625 in. y.
19.40°.
Equilibrium is neutral.
u 5 0 and u 5 180.0°, unstable;
u 5 75.5° and u 5 284°, stable.
u 5 90.0° and u 5 270° unstable;
u 5 22.0° and u 5 158.0°, stable.
u 5 245.0°, unstable; u 5 135.0°, stable.
u 5 263.4°, unstable; u 5 116.6°, stable.
59.0°, stable.
78.7°, stable; 324°, unstable; 379°, stable.
9.39° and 90.0°, stable; 34.2°, unstable.
357 mm.
252 mm.
17.11°, stable; 72.9°, unstable.
49.1°.
54.8°.
37.4°.
16.88 m.
k . 6.94 lbyin.
15.00 in.
P , 2 kLy9.
P , kLy18.
P , k(l 2 a)2y2l.
P , 160.0 N.
P , 764 N.
(a) P , 10.00 lb. (b) P , 20.0 lb.
60.0 lbw.
600 lb ? in. i.
(a) 20.0 N. (b) 105.0 N.
39.2°.
60.4°.
7.13 in.
(a) 0, unstable. (b) 137.8°, stable.
(a) 22.0°. (b) 30.6°.
u 5 60°: 2.42 in.; u 5 120°: 1.732 in.;
(MyP)max 5 2.52 in. at u 5 73.7°.
u 5 60°: 171.1 N C. For 32.5° # u # 134.3°, |F| # 400 N.
u 5 60°: 296 N T. For u # 125.7°, |F| # 400 N.
(b) u 5 60°, datum at C: V 5 2294 in ? lb.
(c) 34.2°, stable; 90°, unstable; 145.8°, stable
(b) u 5 50°, datum at E: V 5 100.5 J. dVydu 5 22.9 J.
(c) u 5 0, unstable; 30.4°.
(b) u 5 60°, datum at B: 30.0 J.
(c) u 5 0, unstable; 41.4°, stable.
(b) u 5 60°, datum at u 5 0: 237.0 J. (c) 52.2°, stable.
CHAPTER 11
11.1
11.2
11.3
11.4
11.5
11.6
11.9
266.0 m, 149.0 mys, 228 mys2.
3.00 m, 27.00 mys.
3.00 s, 259.5 ft, 25.0 ftys2.
248 in., 72.0 in.ys, 2383 in.ys2.
0.667 s, 0.259 m, 28.56 mys.
(a) 1.000 s and 4.00 s. (b) 1.500 m, 24.5 m.
(a) 4.00 s. (b) 256.0 mys, 260 m.
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11.10 x 5 t4y108 1 10t 1 24, v 5 t3y27 1 10.
11.11 233.0 in.ys, 2.00 s, 87.7 in.
11.12 (a) 3.00 ftys4. (b) v 5 (t3 2 32) ftys,
11.15
11.16
11.17
11.18
11.21
11.22
11.23
11.24
11.25
11.26
11.27
11.28
11.31
11.33
11.34
11.35
11.36
11.39
11.40
11.41
11.42
11.43
11.44
11.46
11.47
11.48
11.49
11.50
11.53
11.54
11.55
11.56
11.57
11.58
11.61
11.62
11.63
11.64
11.65
11.66
11.67
11.69
11.70
11.71
11.72
11.73
11.74
1316
x 5 (t4y4 2 32t 1 64) ft.
(a) 5.89 ftys. (b) 1.772 ft.
236.8 ft 2 , 1.832 s22.
(a) 0.0900 s22. (b) 616.97 mmys.
(a) 48.0 m3ys2. (b) 21.6 m. (c) 4.90 mys.
(a) 22.5 m. (b) 38.4 mys.
(a) 29.3 mys. (b) 0.947 s.
(a) 50.0 in. (b) `. (c) 0.866 s.
3.33 ftys.
(a) 0.1457 sym. (b) 145.2 m. (c) 6.86 mys.
(a) 3.33 m. (b) 2.22 s. (c) 1.667 s.
(a) 7.15 mi. (b) 22.75 3 1026 ftys2. (c) 49.9 min.
(a) 20.0525 mys2. (b) 6.17 s.
(a) 2.36 v 0T, pv 0 yT. (b) 0.363 v 0.
(a) 1.500 mys2. (b) 10.00 s.
(a) 25.0 mys. (b) 19.00 mys. (c) 36.8 m.
(a) 2.71 s. (b) 50.4 miyh.
(a) 252 ftys. (b) 1076 ft.
(a) 0.500 km. (b) 42.9 kmyh.
(a) 22.10 mys2 , 2.06 mys2. (b) 2.59 s before A reaches the
exchange zone.
(a) 15.05 s, 734 ft. (b) 42.5 miyh, 23.7 miyh.
(a) 5.50 ftys2. (b) 9.25 ftys2.
(a) 3.00 s. (b) 4.00 ftys2.
(a) 20.250 mys2 , 0.300 mys2. (b) 20.8 s. (c) 85.5 kmyh.
(a) 17.36 ftys2 b, 3.47 ftys2 b. (b) 20.1 ft. (c) 9.64 ftys.
(a) 2.00 mysx. (b) 2.00 mysw. (c) 8.00 mysx.
(a) 20.0 mys2 y, 6.67 mys2 w. (b) 13.33 mysw, 13.33 mw.
(a) 30.0 ftysx. (b) 15.00 ftysx. (c) 45.0 ftysx. (d) 30.0 ftysx.
(a) 2.40 ftys2x, 4.80 ftys2 w. (b) 12.00 ftysx.
(a) 200 mmys y. (b) 600 mmys y. (c) 200 mmys z.
(d) 400 mmys y.
(a) 13.33 mmys2 z, 20.0 mmys2 z. (b) 13.33 mmys2 y.
(c) 70.0 mmys y, 440 mm y.
(a) 10.00 mmys y, (b) 6.00 mmys2 y, 2.00 mmys2x.
(c) 175 mmx.
(a) 240 mmys2 w, 345 mmys2x. (b) 130 mmys y, 43.3 mmysx.
(c) 728 mm y.
(a) 2.00 in.ys2x, 3.00 inys2 w. (b) 0.667 s. (c) 0.667 in.x.
(a) (1 2 6t 2)y4 in.ys2. (b) 9.06 in.
(a) Corresponding values of (t, v, x) are (0, 218 ftys, 0),
(4 s, 26 ftys, 245 ft), (10 s, 30 ftys, 24 ft), (20 s, 220 ftys,
74 ft). (b) 12 ftys, 74 ft, 176 ft., 20.0 ftys
See Prob. 11.61 for plots. (a) 30.0 ftys. (b) 30 ftys, 114 ft.
(a) 0 , t , 10 s, a 5 0; 10 s , t , 26 s, a 5 25 ftys2;
26 s , t , 41 s, a 5 0; 41 s , t , 46 s, a 5 3 ftys2;
t . 46 s, a 5 0; x 5 2540 ft at t 5 0, x 5 60 ft at t 5 10 s,
x 5 380 ft at t 5 26 s, x 5 80 ft at t 5 41 s, x 5 17.5 ft at
t 5 46 s, x 5 22.5 ft at t 5 50 s. (b) 1383 ft. (c) 9.00 s, 49.5 s.
(a) Same as Prob. 11.63. (b) 420 ft. (c) 10.69 s, 40.0 s.
(a) 44.8 s. (b) 103.3 mys2x.
207 mmys
(a) 10.5 s. (b) v-t and x-t curves.
3.96 mys2.
(a) 0.600 s. (b) 0.200 mys, 2.84 m.
9.39 s.
8.54 s, 58.3 miyh.
1.525 s.
(a) 50.0 mys, 1194 m. (b) 59.25 mys.
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11.77
11.78
11.79
11.80
11.83
11.84
11.86
11.89
11.90
11.91
11.92
11.95
11.96
11.97
11.98
11.99
11.100
11.101
11.102
11.103
11.104
11.105
11.106
11.107
11.108
11.111
11.112
11.113
11.115
11.117
11.118
11.119
11.120
11.123
11.124
11.125
11.126
11.127
11.128
11.129
11.130
11.133
11.134
11.135
11.136
11.137
11.138
11.139
11.141
11.143
11.144
11.145
11.146
11.147
11.148
11.151
(a) 18.00 s. (b) 178.8 m, (c) 34.7 kmyh.
(b) 3.75 m.
(a) 2.00 s. (b) 1.200 ftys, 0.600 ftys.
(a) 5.01 min. (b) 19.18 miyh.
(a) 2.96 s. (b) 224 ft.
(a) 163.0 in.ys2. (b) 114.3 in.ys2.
104 ft.
(a) 8.60 mmys c 35.5°, 17.20 mmys2 a 35.5°.
(b) 33.4 mmys a 8.6°, 39.3 mmys2 a 14.7°.
(a) 0, 159.1 mys2 b 82.9°. (b) 6.28 mys y, 157.9 mys2 w.
(a) 5.37 mys. (b) t 5 2.80 s, x 5 27.56 m, y 5 5.52 m,
v 5 5.37 mys2 b 63.4°.
(a) 2.00 in.ys, 6.00 in.ys. (b) For vmin, t 5 2Np s, x 5 8Np in.,
y 5 2 in., v 5 2.00 in.ys y or 2.00 in.ys z.
For vmax, t 5 (2N 1 1)p s, x 5 4(2N 1 1)p, y 5 6 in.,
v 5 6.00 in.ys y or 6.00 in.ys z.
2R2 (1 1 vn2 t2 ) 1 c2, Rvn 24 1 v 2n t2.
(a) 3.00 ftys, 3.61 ftys2. (b) 3.82 s.
353 m.
(a) 15.50 mys. (b) 5.12 m.
15.38 ftys # v 0 # 35.0 ftys.
(a) 70.4 miyh # v 0 # 89.4 miyh. (b) 6.89°, 4.29°.
(a) 2.87 m . 2.43 m. (b) 7.01 m from the net.
0.244 m # h # 0.386 m.
726 ft or 242 yd.
0 # d # 1.737 ft.
23.8 ftys.
(a) 29.8 ftys. (b) 29.6 ftys.
10.64 mys # v 0 # 14.48 mys.
0.678 mys # v 0 # 1.211 mys.
(a) 4.90°. (b) 963 ft. (c) 16.24 s.
(a) 14.66°. (b) 0.1074 s.
(a) 10.38°. (b) 9.74°.
(a) 45.0°, 6.52 m. (b) 58.2°, 5.84 m.
(a) 1.540 mys a 38.6°. (b) 1.503 mys a 58.3°.
5.05 mys b 55.8°.
1.737 knots c 18.41°.
(a) 2.67 miyh d 12.97°. (b) 258 miyh a 76.4°.
(c) 65 m c 40°.
(a) 8.53 in.ys b 54.1°. (b) 6.40 in.ys2 b 54.1°.
(a) 7.01 in.ys d 60°. (b) 11.69 in.ys2 d 60.6°.
(a) 0.835 mmys2 b 75°. (b) 8.35 mmys b 75°.
(a) 0.958 mys2 c 23.6°. (b) 1.917 mys c 23.6°.
10.54 ftys d 81.3°.
(a) 5.18 ftys b 15°. (b) 1.232 ftys b 15°.
17.49 kmyh a 59.0°.
15.79 kmyh c 26.0°.
28.0 mys.
(a) 250 m. (b) 82.9 kmyh.
1815 ft.
59.9 miyh.
(a) 20.0 mmys2. (b) 26.8 mmys2.
(a) 178.9 m. (b) 1.118 mys2.
2.53 ftys2.
15.95 ftys2.
(a) 281 m. (b) 209 m.
(a) 7.99 mys a 40°. (b) 3.82 m.
(a) 6.75 ft. (b) 0.1170 ft.
(a) 1.739 ft. (b) 27.9 ft.
rB 5 vB2 y9vA .
18.17 mys a 4.04° and 18.17 mys c 4.04°.
(R 2 1 c2)y2vnR.
bee29400_ans_1305-1328.indd Page 1317 1/6/09 2:56:39 AM user-s172
11.152
11.153
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11.180
11.181
11.182
11.184
11.185
11.186
11.187
11.189
11.190
11.193
2.50 ft.
25.8 3 103 kmyh.
12.56 3 103 kmyh.
153.3 3 103 kmyh.
92.9 3 10 6 mi.
885 3 10 6 mi.
1.606 h.
(a) 3pb e u, 24p2 b er. (b) u 5 2Np, N 5 0, 1, 2, . . . .
(a) 2bv, 4bv2. (b) r 5 b, a circle.
(a) 2(6p in.ys2) er, (80 p in.ys2) e u. (b) 0.
(a) (2p mys) e u, 2(4p2 mys2) er
(b) 2(py2 mys) er 1 (p mys) eu, 2(p2y2 mys2)er 2 (p2 mys2) eu.
(a.) 2abt, 2ab 21 1 4b2t4. (b) r 5 a(circle).
du tan b sec by(tan b cos u 2 sin u)2.
v 0 cos b (tan b cos u 1 sin u)2yh.
185.7 kmyh.
61.8 miyh, 49.7°.
(bv2yu3) 24 1 u4.
(1 1 b2) v2 ebu.
tan21[R(2 1 vN2 t 2)yc24 1 v2N t2]
(a) ux 5 90°, uy 5 123.7°, uz 5 33.7°. (b) ux 5 103.4°,
uy 5 134.3°, uz 5 47.4°.
(a) 1.00 s, 4.00 s. (b) 1.50 m, 24.5 m.
(a) 22.43 3 10 6 ftys2. (b) 1.366 3 1023 s.
(a) 11.62 s, 69.7 ft. (b) 18.30 ftys.
(a) 3.00 s. (b) 56.25 mm above its initial position.
vA 5 12.5 mmysx, vB 5 75 mmysw,
vC 5 175 mmysw.
17.88 kmyh a 36.4°.
2.44 ftys2.
.
.
r 5 120 mys, r¨ 5 34.8 mys2, u 5 20.0900 radys,
CHAPTER 12
12.1 (a) 4.987 lb at 0°, 5.000 lb at 45°, 5.013 lb at 90°. (b) 5.000 lb
12.2
12.3
12.5
12.6
12.7
12.8
12.9
12.10
12.11
12.12
12.15
12.16
12.17
12.19
12.20
12.21
12.22
12.24
12.26
12.27
12.28
at all latitudes. (c) 0.1554 lb ? s2yft at all latitudes.
(a) 3.24 N. (b) 2.00 kg.
1.300 3 10 6 kg ? mys.
(a) 6.67 mys. (b) 0.0755.
(a) 225 kmyh. (b) 187.1 kmyh.
0.242 mi.
(a) 135.3 ft. (b) 155.8 ft.
419 N to start and 301 N during sliding.
0.414 mys2 c 15°.
(a) A: 2.49 mys2 y, B: 0.831 mys2 w. (b) 74.8 N.
(a) A: 0.698 mys2 y, B: 0.233 mys2 w. (b) 79.8 N.
(a) 0.986 mys2 b 25°. (b) 51.7 N.
(a) 1.794 mys2 b 25°. (b) 58.2 N.
(a) 0.997 ftys2 a 15°, 1.619 ftys2 a 15°.
System 1: (a) 10.73 ftys2. (b) 14.65 ftys. (c) 1.864 s.
System 2: (a) 16.10 ftys2. (b) 17.94 ftys. (c) 1.242 s.
System 3: (a) 0.749 ftys2. (b) 3.87 ftys. (c) 26.7 s.
(a) 1.962 mys2x. (b) 39.1 N.
(a) 6.63 mys2 z. (b) 0.321 m y.
(a) 19.53 mys2 a 65°. (b) 4.24 mys2 d 65°.
0.347 m0 v 02 yF0.
2kym ( 2l 2 1 x20 2 l) .
119.5 miyh.
(a) 33.6 N. (b) a A 5 4.76 mys2 y, a B 5 3.08 mys2 w,
aC 5 1.401 mys2 z.
/Volumes/204/MHDQ078/work%0/indd%0
12.29 (a) 36.0 N. (b) a A 5 5.23 mys2 y, a B 5 2.62 mys2 w. aC 5 0.
12.30 (a) a A 5 a B 5 a D 5 2.76 ftys2 w, aC 5 11.04 ftys2x.
12.31
12.36
12.37
12.38
12.40
12.42
12.43
12.44
12.45
12.46
12.47
12.48
12.49
12.50
12.51
12.53
12.55
12.56
12.57
12.58
12.61
12.62
12.64
12.65
12.66
12.67
12.68
12.69
12.70
12.71
12.76
12.79
12.80
12.81
12.82
12.84
12.86
12.87
12.88
12.89
12.90
12.91
12.98
12.99
12.103
12.104
12.105
12.108
12.109
12.110
(b) 18.80 lb.
(a) 24.2 ftysw. (b) 17.25 ftysx.
(a) 80.4 N. (b) 2.30 mys.
(a) 49.9°. (b) 6.85 N.
8.25 ftys.
2.77 mys , v , 4.36 mys.
9.00 ftys , vC , 12.31 ftys.
2.42 ftys , v , 13.85 ftys.
(a) 131.7 N. (b) 88.4 N.
(a) 553 N. (b) 659 N.
(a) 668 ft. (b) 120.0 lbx.
(a) 6.95 ftys2 c 20°. (b) 8.87 ftys2 c 20°.
(a) 2.905 N. (b) 13.09°.
1126 N b 25.6°.
24.1° # u # 155.9°.
(a) 43.9°. (b) 0.390. (c) 78.8 kmyh.
(a) 0.1858 W. (b) 10.28°.
468 mm.
2.36 mys # v # 4.99 mys.
(a) 0.1904, motion impending downward.
(b) 0.349, motion impending upward.
(a) Does not slide. 1.926 lb b 80°.
(b) Slides downward. 1.123 lb b 40°.
(a) 0.1834. (b) 10.39° for impending motion to the left.,
169.6° for impending motion to the right.
(a) 2.98 ftys. (b) 19.29° for impending motion to the left.
160.7° for impending motion to the right.
1.054 2eV/mv20.
1.333 l.
(a) Fr 5 210.73 N, Fu 5 0.754 N.
(b) Fr 5 24.44 N, Fu 5 1.118 N.
Fr 5 0.0523 N, Fu 5 0.432 N.
(a) Fr 5 21.217 lb, Fu 5 0.248 lb.
(b) Fr 5 20.618 lb, Fu 5 20.0621 lb.
(a) mc2(r 0 2 kt) t 2. (b) mc(r0 2 3kt).
2.00 s.
P 5 (5.76 N) tan u sec3 u b u
Q 5 (5.76 N) tan3 u sec3 u y
vr 5 v 0 sin 2uy 1cos 2u. vu 5 v0 1cos 2u.
(a) r 5 (gt 2 R 2y4p 2)1y3. (b) g 5 24.8 mys2.
(a) 35800 km and 22240 mi. (b) 3070 mys and 10090 ftys.
4.13 3 1021 lb ? s2yft.
(a) 1 hr 57 min. (b) 3380 km.
(a) 86.9 3 1024 kg. (b) 436000 km.
(a) 5280 ftys. (b) 8000 ftys.
(a) 1551 mys. (b) 15.8 mys.
5000 mys.
53 ftys.
(a) At A (a A)r 5 0, (a A)u 5 0. (b) 1536 in.ys2. (c) 32.0 in.ys.
(a) 24.0 in.ys. (b) ar 5 2258 in.ys2 , a u 5 0. (c) 2226 in.ys2.
10.42 kmys.
(a) 10.13 kmys. (b) 2.97 kmys.
(a) 26.3 3 103 ftys. (b) 448 ftys.
22y(2 1 a).
(a) 52.4 3 103 ftys. (b) 1318 ftys at A, 3900 ftys at B.
98.0 h.
4.95 h.
54.0°.
1317
•
16
CHAPTER 1 2
K I N E M AT I C S O F A P A RT I C L E
PRO B L E M S
-12-1. A car starts from rest and with constant
acceleration achieves a velocity of 15 m/s when it travels a
distance of 200 m. Determine the acceleration of the car
and the time required.
12-2. A train starts from rest at a station and travels with a
constant acceleration of 1 m/s2 . Determine the velocity of the
train when ( = 30s and the distance traveled during this time.
12-10. Car A starts from rest at ( = 0 and travels along a
straight road with a constant acceleration of 6 ft/ S2 until it
reaches a speed of 80 ft/s. Afterwards it maintains this
speed. Also, when t = 0, car B located 6000 ft down the
road is traveling towards A at a constant speed of 60 ft/ s.
Determine the distance traveled by car A when they pass
each other.
12-3. An elevator descends from rest with an acceleration
of 5 ft/s2 until it achieves a velocity of 15 ft/ s. Determine the
time required and the distance traveled.
*12-4. A car is traveling at 15 mi s, when the traffic light
50 m ahead turns yellow. Determine the required constant
deceleration of the car and the time needed to stop the car
at the light.
-12-5. A particle is moving along a straight line with the
acceleration a = (12( - 3(1/2) ft/s2 , where ( is in seconds.
Determine the velocity and the position of the particle as a
function of time. When ( = 0, v = 0 and S = 15 ft.
12-6. A ball is released from the bottom of an elevator
which is traveling upward with a velocity of 6 ft/s. If the ball
strikes the bottom of the elevator shaft in 3 s, determine the
height of the elevator from the bottom of the shaft at the
instant the ball is released. Also, find the velocity of the ball
when it strikes the bottom of the shaft.
12-7. A car has an initial speed of 25 m/s and a constant
deceleration of 3 m/ s2 . Determine the velocity of the car
when ( = 4 S. What is the displacement of the car during the
4-s time interval? How much time is needed to stop the car?
*12-8. If a particle has an initial velocity of va = 12 ft/s to
the right, at Sa = 0, determine its position when ( = 10 s, if
a = 2 ft/ S2 to the left.
Va
-12-9. The acceleration of a particle traveling along a
straight line is a = kjv, where k is a constant. If S = 0, v =
when ( = 0, determine the velocity of the particle as a
function of time t.
60 ft/s
--
1l17'it
B
1------ 6000 ft -------
Prob. 12-10
I
12-11. A particle travels along a straight line with a velocity
v = (12 - 3(2 ) mi s, where t is in seconds. When ( = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from t = 0
to t = 10 s, and the distance the particle travels during this
time period.
*12-12. A sphere is fired downwards into a medium with
an initial speed of 27 m/s. If it experiences a deceleration of
a = (-6t) m/s2 , where ( is in seconds, determine the
distance traveled before it stops.
-12-13. A particle travels along a straight line such
that in 2 s it moves from an initial position SA = +0.5 m to
a position SB = - 1 .5 m. Then in another 4 s it moves from
SB to Sc = + 2.5 m. Determine the particle's average
velocity and average speed during the 6-s time interval.
12-14. A particle travels along a straight-line path such
that in 4 s it moves from an initial position SA = -8 m to a
position SB = +3 m. Then in another 5 s it moves from SB to
Sc = -6 m. Determine the particle's average velocity and
average speed during the 9-s time interval.
1 2.2
12-15. Tests reveal that a normal driver takes about 0.75 s
before he or she can react to a situation to avoid a collision. It
takes about 3 s for a driver having 0.1 % alcohol in his system
to do the same. If such drivers are traveling on a straight road
at 30 mph (44 ft/s) and their cars can decelerate at 2 ft/S2 ,
determine the shortest stopping distance d for each from the
moment they see the pedestrians. Moral: If you must drink,
Vj = 44 [tis
-
Prob. 12-15
*12-16. As a train accelerates uniformly it passes successive
kilometer marks while traveling at velocities of 2 m/s and
then 10 m/s. Determine the train's velocity when it passes
the next kilometer mark and the time it takes to travel the
2-km distance.
-12-17. A ball is thrown with an upward velocity of 5 m/s
from the top of a 10-m high building. One second later
another ball is thrown vertically from the ground with a
velocity of 10 m/s. Determine the height from the ground
where the two balls pass each other.
12-18. A car starts from rest and moves with a constant
acceleration of 1 .5 m/s2 until it achieves a velocity of 25 m/s.
It then travels with constant velocity for 60 seconds.
Determine the average speed and the total distance
traveled.
12-19. A car is to be hoisted by elevator to the fourth floor
of a parking garage, which is 48 ft above the ground. If the
elevator can accelerate at 0.6 ft/s2 , decelerate at 0.3 ft/S2 ,
and reach a maximum speed of 8 ft/s, determine the shortest
time to make the lift, starting from rest and ending at rest.
*12-20. A particle is moving along a straight line such that
its speed is defined as v = ( -4s2) mis, where s is in meters.
If s = 2 m when t = 0, determine the velocity and
acceleration as functions of time.
RECTILINEAR KINEMATICS: CONTINUOUS MOTION
17
-12-21. Two particles A and B start from rest at the origin
s = 0 and move along a straight line such that
a A = (6t - 3) ft/S2 and aB = (12t2 - 8) ft/S2 , where t is in
seconds. Determine the distance between them when
t = 4 s and the total distance each has traveled in t = 4 s.
12-22. A particle moving along a straight line is subjected
to a deceleration a = (-2v3 ) m/s2 , where v is in m/s. If it
has a velocity v = 8 m/s and a position s = 1 0 m when
t = 0, determine its velocity and position when t = 4 s.
12-23. A particle is moving along a straight line such that
its acceleration is defined as a = (-2v) m/s2 , where v is in
meters per second. If v = 20 m/s when s = 0 and t = 0,
determine the particle's position, velocity, and acceleration
as functions of time.
*12-24. A particle starts from rest and travels along a
straight line with an acceleration a = (30 - 0.2v) ft/S2 ,
where v is in ft/s. Determine the time when the velocity of
the particle is v = 30 ft/s.
-12-25. When a particle is projected vertically upwards
with an initial velocity of vo, it experiences an acceleration
a = - (g + kv2) , where g is the acceleration due to gravity,
k is a constant and v is the velocity of the particle.
Determine the maximum height reached by the particle.
12-26. The acceleration of a particle traveling along a
straight line is a = (0.02el) m/s2 , where t is in seconds. If
v = 0, s = 0 when t = 0, determine the velocity and
acceleration of the particle at s = 4 m .
12-27. A particle moves along a straight line with an
acceleration of a = 5/(3s 1/3 + s5/2) m/s2 , where s is in
meters. Determine the particle's velocity when s = 2 m, if it
starts from rest when s = 1 m. Use Simpson's rule to
evaluate the integral.
*12-28. If the effects of atmospheric resistance are
accounted for, a falling body has an acceleration defined by
the equation a = 9.81[1 - v2 (1O-4)] m/s2 , where v is in m/s
and the positive direction is downward. If the body is
released from rest at a very high altitude, determine (a) the
velocity when t = 5 s, and (b) the body's terminal or
maximum attainable velocity (as t � (0).
•
•
18
CHAPTER 1 2
K I N E M AT I C S O F A P A RT I C L E
-12-29. The position of a particle along a straight line is
given by s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in
seconds. Determine the position of the particle when
t = 6 s and the total distance it travels during the 6-s time
interval. Hint: Plot the path to determine the total distance
traveled.
Vo
12-30. The velocity of a particle traveling along a straight
line is v =
- ks, where k is constant. If s = 0 when t = 0,
determine the position and acceleration of the particle as a
function of time.
12-31. The acceleration of a particle as it moves along a
straight line is given by a = (2t - 1 ) m/s2 , where t is in
seconds. If s = 1 m and v = 2 m/s when t = 0, determine
the particle's velocity and position when t = 6 s. Also,
determine the total distance the particle travels during this
time period.
*12-32. Ball A is thrown vertically upward from the top
of a 30-m-high-building with an initial velocity of 5 m/s. At
the same instant another ball B is thrown upward from the
ground with an initial velocity of 20 m/s. Determine the
height from the ground and the time at which they pass.
-12-33. A motorcycle starts from rest at t = 0 and travels
along a straight road with a constant acceleration of 6 ft/S2
until it reaches a speed of 50 ft/s. Afterwards it maintains
this speed. Also, when t = 0, a car located 6000 ft down the
road is traveling toward the motorcycle at a constant speed
of 30 ft/s. Determine the time and the distance traveled by
the motorcycle when they pass each other.
12-34. A particle moves along a straight line with a
velocity v = (200s) mm/s, where s is in millimeters.
Determine the acceleration of the particle at s = 2000 mm.
How long does the particle take to reach this position if
s = 500 mm when t = O?
-12-35. A particle has an initial speed of 27 m/s. If it
experiences a deceleration of a = ( -6t) m /s2 , where t is in
seconds, determine its velocity, after it has traveled 10 m.
How much time does this take?
*12-36. The acceleration of a particle traveling along a
straight line is a = (8 - 2s) m/s2 , where s is in meters. If
v = 0 at s = 0, determine the velocity of the particle at
s = 2 m, and the position of the particle when the velocity
is maximum.
Vo.
-12-37. Ball A is thrown vertically upwards with a velocity
of
B all B is thrown upwards from the same point with
the same velocity t seconds later. Determine the elapsed
time t < 2vo/g from the instant ball A is thrown to when the
balls pass each other, and find the velocity of each ball at
this instant.
12-38. As a body is projected to a high altitude above the
earth's surface, the variation of the acceleration of gravity
with respect to altitude y must be taken into account.
Neglecting air resistance, this acceleration is determined
from the formula a = -go[R2/(R + yf] , where go is the
constant gravitational acceleration at sea level, R is the
radius of the earth, and the positive direction is measured
upward. If go = 9.81 m/s2 and R = 6356 km, determine the
minimum initial velocity (escape velocity) at which a
projectile should be shot vertically from the earth's surface
so that it does not fall back to the earth. Hint: This requires
that v = O as y ----> 00 .
12-39. Accounting for the vanatlOn of gravitational
acceleration a with respect to altitude y (see Prob. 12-38),
derive an equation that relates the velocity of a freely
falling particle to its altitude. Assume that the particle is
released from rest at an altitude from the earth's surface.
With what velocity does the particle strike the earth if it is
released from rest at an altitude
= 500 km? Use the
numerical data in Prob. 12-38.
Yo
Yo
*12-40. When a particle falls through the air, its initial
acceleration a = g diminishes until it is zero, and
thereafter it falls at a constant or terminal velocity vf . If
this variation of the acceleration can be expressed as
a = (g/v2f ) (v2f - v2), determine the time needed for the
velocity to become v = vf/2 . Initially the particle falls
from rest.
-12-41. A particle is moving along a straight line such that
its position from a fixed point is s = (12 - 15t2 + 5t3 ) m,
where t is in seconds. Determine the total distance traveled
by the particle from t = 1 s to t = 3 s. Also, find the average
speed of the particle during this time interval.
An swe rs to S e l e cted P ro b l e m s
Chapter 1 2
12-1. v2 = VB + 2ac( S - so)
a c = 0.5625 m/s2
v = Vo + act
t = 26.7 s
12-2. v = 0 + 1 (30) = 30 m/s
S = 450 m
12-3. t = 3 s
S = 22.5 ft
12-5. dv = a dt
v = ( 6t2 - 2t3/2) ft/ s
ds = v dt
S = (2t3 - � t5/2 + 1 5) ft
12-6. h = 1 27 ft
v = -90.6 ft/ s = 90. 6 ft / s �
12-7. v = 13 m/s
�s = 76 m
t = 8.33 s
12-9.
12-10.
12-11.
12-13.
12-14.
12-15.
12-17.
12-18.
12-19.
12-21.
dv
a
2-t
k+-v--"6
v = VrSA = 3200 ft
a = -24 m/ s2
�s = -880 m
ST = 9 1 2 m
�s = 2 m
ST = 6 m
vavg = 0.333 m/ s
(vsp)avg = 1 m/ s
vavg = 0.222 m/ s
(vsp)avg = 2.22 m/ s
d = 5 1 7 ft
d = 616 ft
h = 5t' - 4.905 ( t, )2 + 1 0
h = 19.81t' - 4.905 (t, )2 - 14.905
t' = 1 . 682 m
h = 4.54 m
S = 1708 m
vavg = 22.3 m/ s
al H = 1.06 m/ s2
VA = (3 t 2 - 3t) ft/ s
VB = (4t3 - 8t) ft/s
t = 0 s and = 1 s
B stops
dt =
t = 0s
t = Yz s
SAB l r =4 s = 152 ft
(ST)A = 41 ft
(ST)B = 200 ft
12-22. Choose the root greater than 10 m S
v = 0.250 m/s
12-23. v = (20e-2r) m/ s
a = ( -40e-2r ) m/ s2
12-25.
(
1O(1 - e -2r) m
g + kV02
1
S = In
2k
S
=
=
11.9 m
)
g + kv2
1
k
hmax = I n ( 1 + :gYo
2)
2k
12-26. v = 4.11 m/s
a = 4.13 m/ s2
12-27. v = 1.29 m/s
12-29. S I r = 6 s = -27.0 ft
v = 4.50t2 - 27.0t + 22.5
The times when the particle stops are
t = 1 s and t = 5 s.
Stot = 69.0 ft
12-30. S =
12-33.
12-34.
12-35.
12-37.
( )
-kvoe- kr
Vf + V
vf
t = In
2g vf - v
Distance between motorcycle and car 5541.67 ft
t = 77.6 s
Sm = 3.67(1W ft
a = 80 km/ s2
t = 6.93 ms
vavg = 10 m/s <aavg = 6 m/ s2 <­
ball A
h = vot' - !r t,2
2
VA = Vo - gt'
h = vo (t' - t) - !r et' - t)2
2
VB = Vo - get' - t)
2vo +--=gt---'
t' = --'2g
a
12-31.
� ( 1 - e - kr)
=
699
```