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11.1 The motion of a particle is defined by the relation x 5 1.5t4 230t2 1 5t 1 10, where x and t are expressed in meters and sec-onds, 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 218t2 1 2t 1 5, where x and t are expressed in meters and sec-onds, 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 5

2 t2 2

30t 1 8x, where x and t are expressed in feet and seconds, respectively. Determine the time, the position, and the accelera-tion 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 accelera-tion 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. Deter-mine (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, respec-tively. 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 posi-tion 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.

PROBLEMS†

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614 Kinematics of Particles 11.11 The acceleration of a particle is directly proportional to the time t. 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. De termine (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, respec-tively, 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 posi-tion of the particle when v 5 4.5 m/s, (c) the maximum velocity of the particle.

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 mate-rial. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.

v

Fig. P11.19

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615Problems 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 0.82v2 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 2k1v, 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 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, deter-mine the velocity of the ball when it strikes the bottom of the lake.

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.

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 parti-cle 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.

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, deter-mine (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, respec-tively, 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.

30 ft

Fig. P11.24

v

Fig. P11.27

v

x

Fig. P11.28

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9.183 ( a ) K 1 5 2.26g ta 4 y g ; K 2 5 17.27g ta 4 y g ; K 3 5 19.08g ta 4 y g .( b ) ( u x ) 1 5 85.0°, ( u y ) 1 5 36.8°, ( u z ) 1 5 53.7°;( u x ) 2 5 81.7°, ( u y ) 2 5 54.7°, ( u z ) 2 5 143.4°;( u x ) 3 5 9.70°, ( u y ) 3 5 99.0°, ( u z ) 3 5 86.3°.

9.185 I x 5 ab 3 y28; I y 5 a 3 b y20. 9.187 4 a 3 b y15; ay15. 9.188 I x 5 4 a 4 ; I y 5 16 a 4 y3. 9.189 ( a ) 3.13 3 10 6 mm 4 . ( b ) 2.41 3 10 6 mm 4 . 9.190 I x 5 634 3 10 6 mm 4 ; I y 5 38.0 3 10 6 mm 4 . 9.191 I xy 5 22.81 in 4 . 9.193 ( a ) 7 ma 2 y18. (b) 0.819 ma 2 . 9.195 I x 5 0.877 kg ? m 2 ; I y 5 1.982 kg ? m 2 ; I z 5 1.652 kg ? m 2 . 9.C1 u 5 20°: I x 9 5 14.20 in 4 , I y 9 5 3.15 in 4 , I x 9 y 9 5 23.93 in 4 . 9.C3 ( a ) Ix¿ 5 371 3 103 mm4, Iy¿ 5 64.3 3 103 mm4;

kx¿ 5 21.3 mm, ky¿ 5 8.87 mm. ( b ) Ix¿ 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.

9.C5 ( a ) 5.99 3 10 23 kg ? m 2 . ( b ) 77.4 3 10 23 kg ? m 2 . 9.C6 ( a ) 74.0 3 10 26 lb ? ft ? s 2 . ( b ) 645 3 10 26 lb ? ft ? s 2 .

( c ) 208 3 10 −6 lb ? ft ? s 2 .

CHAPTER 10 10.1 82.5 Nw. 10.2 120 lb y. 10.3 49.5 N ? m i. 10.4 1200 lb ? in. l. 10.7 ( a ) 60.0 N C , 8.00 mmw. ( b ) 300 N C , 40.0 mmw. 10.8 ( a ) 120.0 N C , 16.00 mmw. ( b ) 300 N C , 40.0 mmw. 10.9 Q 5 2 P sin u ycos ( u y2). 10.10 Q 5 2 P cos u ycos (u y2). 10.11 Q 5 (3 P y2) tan u 10.12 Q 5 P [( l y a )cos 3 u 2 1]. 10.15 M 5 1

2 Wl tan a sin u.

10.16 M 5 Pl y2 tan u. 10.17 M 5 7 Pa cos u 10.18 ( a ) M 5 Pl sin 2 u . ( b ) M 5 3 Pl cos u. ( c ) M 5 Pl sin u. 10.21 85.2 lb ? ft i. 10.22 22.8 lb d 70.0°. 10.24 36.4°. 10.25 38.7°. 10.26 68.0°. 10.28 19.81° and 51.9°. 10.30 25.0°. 10.31 39.7° and 69.0°. 10.32 52.2° . 10.33 40.2° . 10.35 22.6°. 10.36 51.1°. 10.37 52.4°. 10.38 19.40°. 10.39 59.0°. 10.40 78.7°, 324°, 379°. 10.43 12.03 kN q . 10.44 20.4°. 10.45 2370 lb r . 10.46 2550 lb r . 10.47 h 5 1y(1 1 m cot a) 10.49 37.6 N, 31.6 N.

10.51 300 N ? m, 81.8 N ? m. 10.52 h 5 tan uytan (u 1 fs) . 10.53 7.75 kNx. 10.54 H 5 1.361 kNx; M H 5 550 N ? m l. 10.57 0.833 in.w. 10.58 0.625 in. y. 10.66 19.40°. 10.67 Equilibrium is neutral. 10.69 u 5 0 and u 5 180.0°, unstable;

u 5 75.5° and u 5 284°, stable. 10.70 u 5 90.0° and u 5 270° unstable;

u 5 22.0° and u 5 158.0°, stable. 10.71 u 5 245.0°, unstable; u 5 135.0°, stable. 10.72 u 5 263.4°, unstable; u 5 116.6°, stable. 10.73 59.0°, stable. 10.74 78.7°, stable; 324°, unstable; 379°, stable. 10.78 9.39° and 90.0°, stable; 34.2°, unstable. 10.79 357 mm. 10.80 252 mm. 10.81 17.11°, stable; 72.9°, unstable. 10.83 49.1°. 10.85 54.8°. 10.86 37.4°. 10.88 16.88 m. 10.90 k . 6.94 lbyin. 10.91 15.00 in. 10.92 P , 2 kL y9. 10.93 P , kL y18. 10.94 P , k ( l 2 a )2y2l. 10.96 P , 160.0 N. 10.98 P , 764 N. 10.100 ( a ) P , 10.00 lb. ( b ) P , 20.0 lb. 10.101 60.0 lbw. 10.102 600 lb ? in. i. 10.104 ( a ) 20.0 N. ( b ) 105.0 N. 10.106 39.2°. 10.107 60.4°. 10.108 7.13 in. 10.110 ( a ) 0, unstable. ( b ) 137.8°, stable. 10.112 ( a ) 22.0°. ( b ) 30.6°. 10.C1 u 5 60°: 2.42 in.; u 5 120°: 1.732 in.;

(MyP) max 5 2.52 in. at u 5 73.7°. 10.C2 u 5 60°: 171.1 N C . For 32.5° # u # 134.3°, | F | # 400 N. 10.C3 u 5 60°: 296 N T . For u # 125.7°, | F | # 400 N. 10.C4 ( b ) u 5 60°, datum at C : V 5 2294 in ? lb.

( c ) 34.2°, stable; 90°, unstable; 145.8°, stable 10.C5 ( b ) u 5 50°, datum at E: V 5 100.5 J. d Vyd u 5 22.9 J.

( c ) u 5 0, unstable; 30.4°. 10.C6 ( b ) u 5 60°, datum at B: 30.0 J.

( c ) u 5 0, unstable; 41.4°, stable. 10.C7 ( b ) u 5 60°, datum at u 5 0: 237.0 J. ( c ) 52.2°, stable.

CHAPTER 11 11.1 266.0 m, 149.0 mys, 228 mys 2 . 11.2 3.00 m, 27.00 mys. 11.3 3.00 s, 259.5 ft, 25.0 ftys 2 . 11.4 248 in., 72.0 in.ys, 2383 in.ys 2 . 11.5 0.667 s, 0.259 m, 28.56 mys. 11.6 ( a ) 1.000 s and 4.00 s. ( b ) 1.500 m, 24.5 m. 11.9 ( a ) 4.00 s. ( b ) 256.0 mys, 260 m.

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11.10 x 5 t 4 y108 1 10t 1 24, v 5 t 3 y27 1 10. 11.11 233.0 in.ys, 2.00 s, 87.7 in. 11.12 ( a ) 3.00 ftys 4 . ( b ) v 5 (t 3 2 32) ftys,

x 5 (t 4 y4 2 32t 1 64) ft. 11.15 ( a ) 5.89 ftys. ( b ) 1.772 ft. 11.16 236.8 ft 2 , 1.832 s 22 . 11.17 ( a ) 0.0900 s 22 . ( b ) 616.97 mmys. 11.18 ( a ) 48.0 m 3 ys 2 . ( b ) 21.6 m. ( c ) 4.90 mys. 11.21 ( a ) 22.5 m. ( b ) 38.4 mys. 11.22 ( a ) 29.3 mys. ( b ) 0.947 s. 11.23 ( a ) 50.0 in. ( b ) . ( c ) 0.866 s. 11.24 3.33 ftys. 11.25 ( a ) 0.1457 sym. ( b ) 145.2 m. ( c ) 6.86 mys. 11.26 ( a ) 3.33 m. ( b ) 2.22 s. ( c ) 1.667 s. 11.27 ( a ) 7.15 mi. ( b ) 22.75 3 10 26 ftys 2 . ( c ) 49.9 min. 11.28 ( a ) 20.0525 mys 2 . ( b ) 6.17 s. 11.31 ( a ) 2.36 v 0 T, pv 0 yT. ( b ) 0.363 v 0 . 11.33 ( a ) 1.500 mys 2 . ( b ) 10.00 s. 11.34 ( a ) 25.0 mys. ( b ) 19.00 mys. ( c ) 36.8 m. 11.35 ( a ) 2.71 s. ( b ) 50.4 miyh. 11.36 ( a ) 252 ftys. ( b ) 1076 ft. 11.39 ( a ) 0.500 km. ( b ) 42.9 kmyh. 11.40 ( a ) 22.10 mys 2 , 2.06 mys 2 . ( b ) 2.59 s before A reaches the

exchange zone. 11.41 ( a ) 15.05 s, 734 ft. ( b ) 42.5 miyh, 23.7 miyh. 11.42 ( a ) 5.50 ftys 2 . ( b ) 9.25 ftys 2 . 11.43 ( a ) 3.00 s. ( b ) 4.00 ftys 2 . 11.44 ( a ) 20.250 mys 2 , 0.300 mys 2 . ( b ) 20.8 s. ( c ) 85.5 kmyh. 11.46 ( a ) 17.36 ftys 2 b, 3.47 ftys 2 b. ( b ) 20.1 ft. ( c ) 9.64 ftys. 11.47 ( a ) 2.00 mysx. ( b ) 2.00 mysw. ( c ) 8.00 mysx. 11.48 ( a ) 20.0 mys 2 y, 6.67 mys 2 w. (b) 13.33 mysw, 13.33 mw. 11.49 ( a ) 30.0 ftysx. ( b ) 15.00 ftysx. ( c ) 45.0 ftysx. ( d ) 30.0 ftysx. 11.50 ( a ) 2.40 ftys 2 x, 4.80 ftys 2 w. ( b ) 12.00 ftysx. 11.53 ( a ) 200 mmys y. ( b ) 600 mmys y. ( c ) 200 mmys z.

( d ) 400 mmys y. 11.54 ( a ) 13.33 mmys 2 z, 20.0 mmys 2 z. ( b ) 13.33 mmys 2 y.

( c ) 70.0 mmys y, 440 mm y. 11.55 ( a ) 10.00 mmys y, ( b ) 6.00 mmys 2 y, 2.00 mmys 2 x.

( c ) 175 mmx. 11.56 ( a ) 240 mmys 2 w, 345 mmys 2 x. ( b ) 130 mmys y, 43.3 mmysx.

( c ) 728 mm y. 11.57 ( a ) 2.00 in.ys 2 x, 3.00 inys 2 w. ( b ) 0.667 s. ( c ) 0.667 in.x. 11.58 ( a ) (1 2 6t 2 )y4 in.ys 2 . ( b ) 9.06 in. 11.61 ( 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

11.62 See Prob. 11.61 for plots. ( a ) 30.0 ftys. ( b ) 30 ftys, 114 ft. 11.63 ( a ) 0 , t , 10 s, a 5 0; 10 s , t , 26 s, a 5 25 ftys 2 ;

26 s , t , 41 s, a 5 0; 41 s , t , 46 s, a 5 3 ftys 2 ; 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.

11.64 ( a ) Same as Prob. 11.63. ( b ) 420 ft. ( c ) 10.69 s, 40.0 s. 11.65 ( a ) 44.8 s. ( b ) 103.3 mys 2 x. 11.66 207 mmys 11.67 ( a ) 10.5 s. ( b ) v-t and x-t curves. 11.69 3.96 mys 2 . 11.70 ( a ) 0.600 s. ( b ) 0.200 mys, 2.84 m. 11.71 9.39 s. 11.72 8.54 s, 58.3 miyh. 11.73 1.525 s. 11.74 ( a ) 50.0 mys, 1194 m. ( b ) 59.25 mys.

11.77 ( a ) 18.00 s. ( b ) 178.8 m, ( c ) 34.7 kmyh. 11.78 (b) 3.75 m. 11.79 ( a ) 2.00 s. ( b ) 1.200 ftys, 0.600 ftys. 11.80 ( a ) 5.01 min. ( b ) 19.18 miyh. 11.83 ( a ) 2.96 s. ( b ) 224 ft. 11.84 (a) 163.0 in.ys 2 . (b) 114.3 in.ys 2 . 11.86 104 ft. 11.89 ( a ) 8.60 mmys c 35.5°, 17.20 mmys 2 a 35.5°.

( b ) 33.4 mmys a 8.6°, 39.3 mmys 2 a 14.7°. 11.90 ( a ) 0, 159.1 mys 2 b 82.9°. ( b ) 6.28 mys y, 157.9 mys 2 w. 11.91 ( a ) 5.37 mys. ( b ) t 5 2.80 s, x 5 27.56 m, y 5 5.52 m,

v 5 5.37 mys 2 b 63.4°. 11.92 ( a ) 2.00 in.ys, 6.00 in.ys. ( b ) For v min , 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.

11.95 2R2(1 1 vn2 t2) 1 c2, Rvn24 1 v n

2t2 . 11.96 ( a ) 3.00 ftys, 3.61 ftys 2 . ( b ) 3.82 s. 11.97 353 m. 11.98 ( a ) 15.50 mys. ( b ) 5.12 m. 11.99 15.38 ftys # v 0 # 35.0 ftys. 11.100 ( a ) 70.4 miyh # v 0 # 89.4 miyh. ( b ) 6.89°, 4.29°. 11.101 ( a ) 2.87 m . 2.43 m. ( b ) 7.01 m from the net. 11.102 0.244 m # h # 0.386 m. 11.103 726 ft or 242 yd. 11.104 0 # d # 1.737 ft. 11.105 23.8 ftys. 11.106 ( a ) 29.8 ftys. ( b ) 29.6 ftys. 11.107 10.64 mys # v 0 # 14.48 mys. 11.108 0.678 mys # v 0 # 1.211 mys. 11.111 ( a ) 4.90°. ( b ) 963 ft. ( c ) 16.24 s. 11.112 (a) 14.66°. ( b ) 0.1074 s. 11.113 ( a ) 10.38°. ( b ) 9.74°. 11.115 ( a ) 45.0°, 6.52 m. ( b ) 58.2°, 5.84 m. 11.117 ( a ) 1.540 mys a 38.6°. ( b ) 1.503 mys a 58.3°. 11.118 5.05 mys b 55.8°. 11.119 1.737 knots c 18.41°. 11.120 ( a ) 2.67 miyh d 12.97°. ( b ) 258 miyh a 76.4°.

( c ) 65 m c 40°. 11.123 ( a ) 8.53 in.ys b 54.1°. ( b ) 6.40 in.ys 2 b 54.1°. 11.124 ( a ) 7.01 in.ys d 60°. ( b ) 11.69 in.ys 2 d 60.6°. 11.125 ( a ) 0.835 mmys 2 b 75°. ( b ) 8.35 mmys b 75°. 11.126 ( a ) 0.958 mys 2 c 23.6°. ( b ) 1.917 mys c 23.6°. 11.127 10.54 ftys d 81.3°. 11.128 (a) 5.18 ftys b 15°. (b) 1.232 ftys b 15°. 11.129 17.49 kmyh a 59.0°. 11.130 15.79 kmyh c 26.0°. 11.133 28.0 mys. 11.134 ( a ) 250 m. ( b ) 82.9 kmyh. 11.135 1815 ft. 11.136 59.9 miyh. 11.137 ( a ) 20.0 mmys 2 . ( b ) 26.8 mmys 2 . 11.138 ( a ) 178.9 m. ( b ) 1.118 mys 2 . 11.139 2.53 ftys 2 . 11.141 15.95 ftys 2 . 11.143 ( a ) 281 m. ( b ) 209 m. 11.144 ( a ) 7.99 mys a 40°. ( b ) 3.82 m. 11.145 ( a ) 6.75 ft. ( b ) 0.1170 ft. 11.146 ( a ) 1.739 ft. ( b ) 27.9 ft. 11.147 r B 5 v B

2 y9v A . 11.148 18.17 mys a 4.04° and 18.17 mys c 4.04°. 11.151 (R 2 1 c 2 )y2v n R.

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11.152 2.50 ft. 11.153 25.8 3 10 3 kmyh. 11.154 12.56 3 10 3 kmyh. 11.155 153.3 3 10 3 kmyh. 11.156 92.9 3 10 6 mi. 11.157 885 3 10 6 mi. 11.158 1.606 h. 11.161 ( a ) 3pb e u , 24p 2 b e r . ( b ) u 5 2Np, N 5 0, 1, 2, . . . . 11.162 ( a ) 2bv, 4bv 2 . ( b ) r 5 b, a circle. 11.163 ( a ) 2(6p in.ys 2 ) e r , (80 p in.ys 2 ) e u . ( b ) 0. 11.165 ( a ) (2p mys) e u , 2(4p 2 mys 2 ) e r

( b ) 2(py2 mys) e r 1 (p mys) e u , 2(p 2 y2 mys 2 ) e r 2 (p 2 mys 2 ) e u . 11.166 ( a ) 2abt, 2ab21 1 4b2t4 . ( b ) r 5 a(circle). 11.169 du

. tan b sec by(tan b cos u 2 sin u) 2 .

11.170 v 0 cos b (tan b cos u 1 sin u) 2 yh. 11.171 185.7 kmyh. 11.172 61.8 miyh, 49.7°. 11.175 (bv2yu 3 ) 24 1 u4 . 11.176 (1 1 b 2 ) v 2 e bu . 11.180 tan 21 [R(2 1 v N

2 t 2 )y c24 1 vN2 t2 ]

11.181 ( a ) u x 5 90°, u y 5 123.7°, u z 5 33.7°. ( b ) u x 5 103.4°, u y 5 134.3°, u z 5 47.4° .

11.182 ( a ) 1.00 s, 4.00 s. ( b ) 1.50 m, 24.5 m. 11.184 ( a ) 22.43 3 10 6 ftys 2 . ( b ) 1.366 3 10 23 s. 11.185 ( a ) 11.62 s, 69.7 ft. ( b ) 18.30 ftys. 11.186 ( a ) 3.00 s. ( b ) 56.25 mm above its initial position. 11.187 v A 5 12.5 mmysx, v B 5 75 mmysw,

v C 5 175 mmysw. 11.189 17.88 kmyh a 36.4°. 11.190 2.44 ftys 2 . 11.193 r. 5 120 mys, r 5 34.8 mys2, u

.5 20.0900 radys,

u 5 20.0156 radys2 .

CHAPTER 12 12.1 ( a ) 4.987 lb at 0°, 5.000 lb at 45°, 5.013 lb at 90°. ( b ) 5.000 lb

at all latitudes. ( c ) 0.1554 lb ? s 2 yft at all latitudes. 12.2 ( a ) 3.24 N. ( b ) 2.00 kg. 12.3 1.300 3 10 6 kg ? mys. 12.5 ( a ) 6.67 mys. ( b ) 0.0755. 12.6 ( a ) 225 kmyh. ( b ) 187.1 kmyh. 12.7 0.242 mi. 12.8 ( a ) 135.3 ft. ( b ) 155.8 ft. 12.9 419 N to start and 301 N during sliding. 12.10 0.414 mys 2 c 15°. 12.11 ( a ) A: 2.49 mys 2 y, B: 0.831 mys 2 w. ( b ) 74.8 N. 12.12 ( a ) A: 0.698 mys 2 y, B: 0.233 mys 2 w. ( b ) 79.8 N. 12.15 ( a ) 0.986 mys 2 b 25°. (b) 51.7 N. 12.16 ( a ) 1.794 mys 2 b 25°. ( b ) 58.2 N. 12.17 ( a ) 0.997 ftys 2 a 15°, 1.619 ftys 2 a 15°. 12.19 System 1: ( a ) 10.73 ftys 2 . (b) 14.65 ftys. (c) 1.864 s.

System 2: ( a ) 16.10 ftys 2 . ( b ) 17.94 ftys. ( c ) 1.242 s. System 3: ( a ) 0.749 ftys 2 . ( b ) 3.87 ftys. ( c ) 26.7 s.

12.20 ( a ) 1.962 mys 2 x. ( b ) 39.1 N. 12.21 ( a ) 6.63 mys 2 z. ( b ) 0.321 m y. 12.22 ( a ) 19.53 mys 2 a 65°. ( b ) 4.24 mys 2 d 65°. 12.24 0.347 m 0 v 0

2 yF 0 . 12.26 2kym (2l2 1 x0

2 2 l) . 12.27 119.5 miyh. 12.28 ( a ) 33.6 N. ( b ) a A 5 4.76 mys 2 y, a B 5 3.08 mys 2 w,

a C 5 1.401 mys 2 z.

12.29 ( a ) 36.0 N. ( b ) a A 5 5.23 mys 2 y, a B 5 2.62 mys 2 w. a C 5 0. 12.30 ( a ) a A 5 a B 5 a D 5 2.76 ftys 2 w, a C 5 11.04 ftys 2 x.

( b ) 18.80 lb. 12.31 ( a ) 24.2 ftysw. ( b ) 17.25 ftysx. 12.36 ( a ) 80.4 N. ( b ) 2.30 mys. 12.37 ( a ) 49.9°. ( b ) 6.85 N. 12.38 8.25 ftys. 12.40 2.77 mys , v , 4.36 mys. 12.42 9.00 ftys , v C , 12.31 ftys. 12.43 2.42 ftys , v , 13.85 ftys. 12.44 ( a ) 131.7 N. ( b ) 88.4 N. 12.45 ( a ) 553 N. ( b ) 659 N. 12.46 ( a ) 668 ft. ( b ) 120.0 lbx . 12.47 ( a ) 6.95 ftys 2 c 20°. ( b ) 8.87 ftys 2 c 20°. 12.48 ( a ) 2.905 N. ( b ) 13.09°. 12.49 1126 N b 25.6°. 12.50 24.1° # u # 155.9°. 12.51 ( a ) 43.9°. ( b ) 0.390. ( c ) 78.8 kmyh. 12.53 ( a ) 0.1858 W. ( b ) 10.28°. 12.55 468 mm. 12.56 2.36 mys # v # 4.99 mys. 12.57 ( a ) 0.1904, motion impending downward.

( b ) 0.349, motion impending upward. 12.58 ( a ) Does not slide. 1.926 lb b 80°.

( b ) Slides downward. 1.123 lb b 40°. 12.61 ( a ) 0.1834. ( b ) 10.39° for impending motion to the left.,

169.6° for impending motion to the right. 12.62 ( a ) 2.98 ftys. ( b ) 19.29° for impending motion to the left.

160.7° for impending motion to the right. 12.64 1.054 2eV/mv0

2 . 12.65 1.333 l. 12.66 (a) F r 5 210.73 N, F u 5 0.754 N.

( b ) F r 5 24.44 N, F u 5 1.118 N. 12.67 F r 5 0.0523 N, F u 5 0.432 N. 12.68 ( a ) F r 5 21.217 lb, F u 5 0.248 lb.

( b ) F r 5 20.618 lb, F u 5 20.0621 lb. 12.69 ( a ) mc 2 (r 0 2 kt) t 2 . ( b ) mc(r 0 2 3kt). 12.70 2.00 s. 12.71 P 5 (5.76 N) tan u sec 3 u b u

Q 5 (5.76 N) tan 3 u sec 3 u y 12.76 v r 5 v 0 sin 2uy 1cos 2u. vu 5 v01cos 2u. 12.79 (a) r 5 (gt 2 R 2 y4p 2 ) 1y3 . (b) g 5 24.8 mys 2 . 12.80 ( a ) 35800 km and 22240 mi. (b) 3070 mys and 10090 ftys. 12.81 4.13 3 10 21 lb ? s 2 yft. 12.82 ( a ) 1 hr 57 min. ( b ) 3380 km. 12.84 ( a ) 86.9 3 10 24 kg. ( b ) 436000 km. 12.86 ( a ) 5280 ftys. ( b ) 8000 ftys. 12.87 ( a ) 1551 mys. ( b ) 15.8 mys . 12.88 5000 mys. 12.89 53 ftys. 12.90 (a) At A (a A ) r 5 0, (a A ) u 5 0. ( b ) 1536 in.ys 2 . ( c ) 32.0 in.ys. 12.91 ( a ) 24.0 in.ys. ( b ) a r 5 2258 in.ys 2 , a u 5 0. ( c ) 2226 in.ys 2 . 12.98 10.42 kmys. 12.99 ( a ) 10.13 kmys. ( b ) 2.97 kmys. 12.103 ( a ) 26.3 3 10 3 ftys. ( b ) 448 ftys. 12.104 22y(2 1 a). 12.105 ( a ) 52.4 3 10 3 ftys. ( b ) 1318 ftys at A , 3900 ftys at B . 12.108 98.0 h. 12.109 4.95 h. 12.110 54.0°.

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• 1 6 C H A P T E R 1 2 K I N E M AT I C S O F A PART 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.

12-3. 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 mis, 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.

-12-9. The acceleration of a particle traveling along a straight line is a = kjv, where k is a constant. If S = 0, v = Va when ( = 0, determine the velocity of the particle as a function of time t.

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.

60 ft/s --

B 1l17'it 1------ 6000 ft------- I

Prob. 12-10

12-11. A particle travels along a straight line with a velocity v = (12 - 3(2) mis, 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 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 1 7

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, please don't drive !

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.

-12-21. Two particles A and B start from rest at the origin • s = 0 and move along a straight line such that aA = (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/(3s1/3 + 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).

• 1 8 C H A P T E R 1 2 K I N E M AT I C S O F A PART 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.

12-30. The velocity of a particle traveling along a straight line is v = Vo - 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.

-12-37. Ball A is thrown vertically upwards with a velocity of Vo. Ball 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 Yo from the earth's surface. With what velocity does the particle strike the earth if it is released from rest at an altitude Yo = 500 km? Use the numerical data in Prob. 12-38.

*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.

Answe rs to Se lected P ro b l ems Chapter 1 2 12-1. v2 = VB + 2ac( S - so)

ac = 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 = 127 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. dt = dv a

v = Vr-2k-t-+-v--"6 12-10. SA = 3200 ft 12-11. a = -24 m/s2

�s = -880 m ST = 912 m

12-13. �s = 2 m ST = 6 m vavg = 0.333 m/s (vsp)avg = 1 m/s

12-14. vavg = 0.222 m/s (vsp)avg = 2.22 m/s

12-15. d = 517 ft d = 616 ft

12-17. 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

12-18. S = 1708 m vavg = 22.3 m/s

12-19. alH = 1 .06 m/s2 12-21. VA = (3t2 - 3t) ft/s

VB = (4t3 - 8t) ft/s t = 0 s and = 1 s B stops

12-22.

12-23.

12-25.

12-26.

12-27.

12-29.

t = 0 s t = Yz s SABl r=4 s = 152 ft (ST)A = 41 ft (ST)B = 200 ft Choose the root greater than 10 m S = 1 1 .9 m v = 0.250 m/s v = (20e-2r) m/s a = ( -40e-2r) m/s2 S = 1O(1 - e-2r ) m

1 (g + kV02) S = 2k In g + kv2 1 ( k 2) hmax = 2k In 1 + :gYo

v = 4 .11 m/s a = 4.13 m/s2 v = 1 .29 m/s S Ir=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 = � ( 1 - e-kr)

12-31.

12-33.

12-34.

12-35.

12-37.

a = -kvoe-kr

t = vf

In (Vf + V)

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

699