bee29400_ch11_600-689.indd Page 613 11/26/08 8:35:56 PM user-s173 /Volumes/204/MHDQ077/work%0/indd%0 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. 613 bee29400_ch11_600-689.indd Page 614 614 11/26/08 8:35:57 PM user-s173 /Volumes/204/MHDQ077/work%0/indd%0 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. bee29400_ch11_600-689.indd Page 615 11/26/08 8:35:58 PM user-s173 /Volumes/204/MHDQ077/work%0/indd%0 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. P11.28 615 bee29400_ans_1305-1328.indd Page 1315 1/5/09 10:14:55 PM user-s172 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. /Volumes/204/MHDQ078/work%0/indd%0 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. 1315 bee29400_ans_1305-1328.indd Page 1316 1/6/09 2:56:37 AM user-s172 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. /Volumes/204/MHDQ078/work%0/indd%0 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 11.154 11.155 11.156 11.157 11.158 11.161 11.162 11.163 11.165 11.166 11.169 11.170 11.171 11.172 11.175 11.176 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, 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 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, 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. 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

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