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Unformatted text preview: )/ t4 + 5t2 + 3 so when t = 1, aT = 7/3. 47. aN = κ(ds/dt)2 = (1/ρ)(ds/dt)2 = (1/1)(2.9 × 105 )2 = 8.41 × 1010 km/s2 48. a = (d2 s/dt2 )T + κ(ds/dt)2 N where κ = |d2 y/dx2 | . If d2 y/dx2 = 0, then κ = 0 and [1 + (dy/dx)2 ]3/2 a = (d2 s/dt2 )T so a is tangent to the curve. 49. aN = κ(ds/dt)2 = [2/(1 + 4x2 )3/2 ](3)2 = 18/(1 + 4x2 )3/2 50. y = ex , aN = κ(ds/dt)2 = [ex /(1 + e2x )3/2 ](2)2 = 4ex /(1 + e2x )3/2 517 Chapter 14 51. a = aT T + aN N; by the Pythagorean Theorem aN = 52. As in Exercise 51, a 2 = a2 + a2 , 81 = 9 + a2 , aN = T N N 2 2 − a2 = T √ 9−9=0 √ √ 72 = 6 2. √√ 12 c , so c2 = 1000aN , c ≤ 10 10 1.5 ≈ 38.73 m/s. 1000 53. Let c = ds/dt, aN = κ ds dt 54. 10 km/h is the same as 1 100 m/s, so F = 500 36 15 , aN = a 100 36 2 ≈ 257.20 N. √ 55. (a) v0 = 320, α = 60◦ , s0 = 0 so x = 160t, y = 160 3t − 16t2 . √ √ (b) dy/dt = 160 3 − 32t, dy/dt = 0 when t = 5 3 so √√ √2 ymax = 160 3(5 3) − 16(5 3) = 1200 ft. √ √ √ √ (c) y = 16t(10 3 − t), y = 0 when t = 0 or 10 3 so xmax = 160(10 3) = 1600 3 ft. √ √ √ √ (d) v(t) = 160i + (160 3 − 32t)j, v(10 3) = 160(i − 3j), v(10 3) = 320 ft/s. √ √ 56. (a) v0 = 980, α = 45◦ , s0 = 0 so x = 490 2 t, y = 490 2t − 4.9t2 √ √ (b) dy/dt = 490 2 − 9.8t, dy/dt = 0 when t = 50 2 so √ √ √2 ymax = 490 2(50 2) − 4.9(50 2) = 24, 500 m. √ √ (c) y = 4.9t(100 2 − t), y = 0 when t = 0 or 100 2 so √ √ xmax = 490 2(100 2) = 98, 000 m. √ √ √ √ √ (d) v(t) = 490 2 i + (490 2 − 9.8t)j, v(100 2) = 490 2(i − j), v(100 2) = 980 m/s. √ 57. v0 = 80, α = −60◦ , s0 = 168 so x = 40t, y = 168 − 40 3 t − 16t2 ; y = 0 when √ √ √ √ t = −7 3/2 (invalid) or t = 3 so x( 3) = 40 3 ft. √ 58. v0 = 80, α = 0◦ , s0 = 168 so x = 80t, y = 168 − 16t2 ; y = 0 when t = − 42/2 (invalid) or √ √ √ t = 42/2 so x( 42/2) = 40 42 ft. √ 59. α = 30◦ , s0 = 0 so x = 3v0 t/2, y = v0 t/2 − 16t2 ; dy/dt = v0 /2 − 32t, dy/dt = 0 when t = v0 /64 2 so ymax = v0 /256 = 2500, v0 = 800 ft/s. 60. α = 45◦ , s0 = 0 so x = √ √ √ 2 v0 t/2, y = 2v0 t/2 − 4.9t2 ; y = 0 when t = 0 or 2v0 /9.8 so 2 xmax = v0 /9.8 = 24, 500, v0 = 490 m/s. 61. v0 = 800, s0 = 0 so x = (800 cos α)t, y = (800 sin α)t − 16t2 = 16t(50 sin α − t); y = 0 when t = 0 or 50 sin α so xmax = 40, 000 sin α cos α = 20, 000 sin 2α = 10, 000, 2α = 30◦ or 150◦ , α = 15◦ or 75◦ . 62. (a) v0 = 5, α = 0◦ , s0 = 4 so x = 5t, y = 4 − 16t2 ; y = 0 when t = −1/2 (invalid) or 1/2 so it takes the ball 1/2 s to hit the ﬂoor. √ (b) v(t) = 5i − 32tj, v(1/2) = 5i − 16j, v(1/2) = 281 so the ball hits the ﬂoor with a speed √ of 281 ft/s. (c) v0 = 0, α = −90◦ , s0 = 4 so x = 0, y = 4 − 16t2 ; y = 0 when t = 1/2 so both balls would hit the ground at the same instant. Exercise Set 14.6 518 √ 3 63. (a) v0 = 40, α = 60, s0 = 4, so x = 20t, y = 4 + 20 3t − 16t2 ; when x = 15, t = , 4 2 √3 3 ≈ 20.98 ft, so the water clears the corner point A with 0.98 ft to y = 4 + 20 3 − 16 4 4 spare. √ (b) y = 20 when −16t2 + 25 3t − 16 = 0, t = 0.668 (reject) or 1.500, x(1.500) ≈ 30 ft, so the water hits the roof. (c) about 15 ft √ 64. x = (v0 /2)t, y = 4 + (v0 3/2)t − 16t2 , solve x = 15, y = 20 simultaneously for v0 and t, 15 √ 3 − 1, t ≈ 0.7898, v0 ≈ 30/0.7898 ≈ 37.98 ft/s. v0 /2 = 15/t, t2 = 16 √ √ 65. (a) x = (35 2/2)t, y = (35 2/2)t − 4.9t2 , from Exercise 17a in Section 14.5 κ= √ |x y − x y | 9.8 √ = 0.004 2 ≈ 0.00566 , κ(0) = 22 [(x )2 + (y )2 ]3/2 35 (b) y (t) = 0 when t = 25 √ 125 m 2, y = 14 4 66. (a) a = aT T + aN N, aT = a= (b) cos θ = a2 + a2 = T N d2 s = −7.5 ft/s2 , aN = κ dt2 (7.5)2 + 1322 3000 ds dt 2 = 1 1322 (132)2 = ft/s2 , ρ 3000 2 ≈ 9.49 ft/s2 aT 7.5 a·T = ≈− ≈ −0.79, θ ≈ 2.48 radians ≈ 142◦ aT a 9.49 67. s0 = 0 so x = (v0 cos α)t, y = (v0 sin α)t − gt2 /2 (a) dy/dt = v0 sin α − gt so dy/dt = 0 when t = (v0 sin α)/g , ymax = (v0 sin α)2 /(2g ) 2 2 (b) y = 0 when t = 0 or (2v0 sin α)/g , so x = R = (2v0 sin α cos α)/g = (v0 sin 2α)/g when ◦ ◦ t = (2v0 sin α)/g ; R is maximum when 2α = 90 , α = 45 , and the maximum value of R 2 is v0 /g . 2 2 2 2 68. The range is (v0 sin 2α)/g and the maximum range is v0 /g so (v0 sin 2α)/g = (3/4)v0 /g , −1 −1 ◦ ◦ ◦ sin 2α = 3/4, α = (1/2) sin (3/4) ≈ 24.3 or α = (1/2)[180 − sin (3/4)] ≈ 65.7 . √ 69. v0 = 80, α = 30◦ , s0 = 5 so x = 40 3t, y = 5 + 40t − 16t2 √ √ (a) y = 0 when t = (−40 ± (40)2 − 4(−16)(5))/(−32) = (5 ± 30)/4, reject (5 − 30)/4 to get √ t = (5 + 30)/4 ≈ 2.62 s. √ (b) x ≈ 40 3(2.62) ≈ 181.5 ft. 1 70. (a) v0 = v , s0 = h so x = (v cos α)t, y = h + (v sin α)t − gt2 . If x = R, then (v cos α)t = R, 2 1 R but y = 0 for this value of t so h + (v sin α)[R/(v cos α)] − g [R/(v cos α)]2 = 0, t= v cos α 2 h + (tan α)R − g (sec2 α)R2 /(2v 2 ) = 0, g (sec2 α)R2 − 2v 2 (tan α)R − 2v 2 h = 0. 519 Chapter 14 dR dR dR − 2v 2 sec2 αR − 2v 2 tan α = 0; if = 0 and α = α0 dα dα dα 2 2 2 2 2 wh...
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