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18.4: Mwendo katika Vipimo viwili na vitatu

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    Angalia Uelewa Wako

    4.1. (a) Kuchukua derivative kuhusiana na muda wa kazi ya msimamo, tuna\(\vec{v}\) (t) = 9.0t 2\(\hat{i}\) na\(\vec{v}\) (3.0s) = 81.0\(\hat{i}\) m/s. (b) Kwa kuwa kazi ya kasi ni isiyo ya kawaida, tunashutumu kasi ya wastani si sawa na kasi ya papo hapo. Tunaangalia hii na kupata $$\ vec {v} _ {avg} =\ frac {\ vec {r} (t_ {2}) -\ vec {r} (t_ {1})} {t_ {2} - t_ {1}} =\ frac {\ vec {r} (4.0\; s) -\ vec {r} (2.0\ s)} {4.0\; s - 2.0\; s} =\ frac {(188.0\;\ kofia {i} - 20.0\;\ kofia {i})\; m} {2.0\; s} = 84.0\;\ kofia {i}\; m/s, $$ ambayo ni tofauti na\(\vec{v}\) (3.0 s) = 81.0\(\hat{i}\) m/s.

    4.2. Vector ya kuongeza kasi ni mara kwa mara na haibadilika kwa wakati. Ikiwa a, b, na c sio sifuri, basi kazi ya kasi lazima iwe mstari kwa wakati. Tuna\(\vec{v}\) (t) =\(\int \vec{a}\) dt =\(\int\) (a\(\hat{i}\) + b\(\hat{j}\) + c\(\hat{k}\)) dt = (a\(\hat{i}\) + b\(\hat{j}\) + c\(\hat{k}\)) t m/s, tangu kuchukua derivative ya kazi ya kasi inazalisha\(\vec{a}\) (t). Ikiwa sehemu yoyote ya kuongeza kasi ni sifuri, basi sehemu hiyo ya kasi itakuwa mara kwa mara.

    4.3. (a) Chagua juu ya mwamba ambako mwamba unatupwa kutoka asili ya mfumo wa kuratibu. Ingawa ni kiholela, sisi kawaida kuchagua wakati t = 0 yanahusiana na asili. (b) equation inayoelezea mwendo usawa ni x = x 0 + v x t Kwa x 0 = 0, equation hii inakuwa x = v x t. (c) Equation 4.27 kupitia Equation 4.29 na Equation 4.46 kuelezea mwendo wima, lakini tangu y 0 = 0 na v 0y = 0, milinganyo hii hurahisisha sana kuwa y =\(\frac{1}{2}\) (v 0y + v y) t =\(\frac{1}{2}\) v y t, v y = -gt, y = -\(\frac{1}{2}\) gt 2, na v y 2 = -2gy. (d) Tunatumia equations ya kinematic kupata vipengele x na y vya kasi wakati wa athari. Kutumia v y 2 = -2gy na kubainisha hatua ya athari ni -100.0 m, tunapata sehemu y ya kasi katika athari ni v y = 44.3 m/s Tunapewa sehemu x, v x = 15.0 m/s, ili tuweze kuhesabu kasi ya jumla kwa athari: v = 46.8 m/s na\(\theta\) = 71.3° chini ya usawa.

    4.4. Golf risasi katika 30°.

    4.5. 134.0 cm/s

    4.6. Kuandika michango kwa equation vector, tuna B = mashua, R = mto, na E = Dunia. Equation ya vector inakuwa\(\vec{v}_{BE}\) =\(\vec{v}_{BR}\) +\(\vec{v}_{RE}\). Tuna haki pembetatu jiometri inavyoonekana katika takwimu hapa chini. Kutatua kwa\(\vec{v}_{BE}\), tuna $$v_ {BE} =\ sqrt {v_ {BR} ^ {2} + v_ {RE} ^ {2}} =\ sqrt {4.5^ {2} + 3.0^ {2}} $ $$$v_ {BE} = 5.4\; m/s,\ quad\ theta =\ tan^ {-1}\ kushoto (\ dfrac {3.0} 4.5}\ haki) = 33.7^ {o}\ ldotp\]

    Vectors V ndogo B W, V ndogo W E na V ndogo B E huunda pembetatu sahihi. Mashua huonyeshwa kwenye kipeo ambapo mikia ya V ndogo B W na V ndogo B E hukutana. Vector V ndogo B W anasema juu. V ndogo W E anasema na haki. V ndogo B E anasema juu na kulia, kwa pembeni kwa wima. V ndogo B E ni vector jumla ya v ndogo B W na V ndogo W E.

    Maswali ya dhana

    1. Mstari sawa

    3. Mteremko lazima uwe sifuri kwa sababu vector ya kasi ni tangent kwa grafu ya kazi ya nafasi.

    5. Hapana, mwendo katika maelekezo ya perpendicular ni huru.

    7. a. Hapana; b. chini ya kilele cha trajectory na upeo katika uzinduzi na athari; c. hapana, kasi ni vector; d. ndiyo, ambapo ardhi

    9. Wote wawili hupiga ardhi kwa wakati mmoja.

    11. Ndiyo

    13. Ikiwa atapitisha mpira kwa mchezaji mwingine, anahitaji kuweka macho yake kwenye sura ya kumbukumbu ambayo wachezaji wengine kwenye timu wanapatikana.

    15.

    Kielelezo a: trajectory ya kofia ni sawa chini. Kielelezo b: trajectory ya kofia ni parabolic, curving chini na kushoto.

    Matatizo

    17. \(\vec{r}\)= 1.0\(\hat{i}\) -4.0\(\hat{j}\) + 6.0\(\hat{k}\)

    19. \(\Delta \vec{r}_{Total}\)= 472.0 m\(\hat{i}\) + 80.3 m\(\hat{j}\)

    21. Jumla ya uhamisho = -6.4 km\(\hat{i}\) + 9.4 km\(\hat{j}\)

    23. a.\(\vec{v}\) (t) = 8.0t\(\hat{i}\) + 6.0t 2\(\hat{k}\),\(\vec{v}\) (0) = 0,\(\vec{v}\) (1.0) = 8.0\(\hat{i}\) + 6.0\(\hat{k}\) m/s

    b.\(\vec{v}_{avg}\) = 4.0\(\hat{i}\) + 2.0\(\hat{k}\) m/s

    25. \(\Delta \vec{r}_{1}\)= 20.00 m\(\hat{j}\),\(\Delta \vec{r}_{2}\) = (2.000 x 10 4 m) (cos 30°\(\hat{i}\) + dhambi 30°\(\hat{j}\)),\(\Delta \vec{r}\) = 1.700 x 10 4 m\(\hat{i}\) + 1.002 x 10 4 m\(\hat{j}\)

    27. a.\(\vec{v}\) (t) = (4.0t\(\hat{i}\) + 3.0t\(\hat{j}\)) m/s,\(\vec{r}\) (t) = (2.0t 2\(\hat{i}\) +\(\frac{3}{2}\) t 2\(\hat{j}\)) m

    b. x (t) = 2.0t 2m, y (t) =\(\frac{3}{2}\) t 2 m, t 2 =\(\frac{x}{2} \Rightarrow\) y =\(\frac{3}{4}\) x

    Grafu ya kazi ya mstari y inalingana na robo 3 x. grafu ni mstari wa moja kwa moja, mzuri wa mteremko kupitia asili.

    29. a.\(\vec{v}\) (t) = (6.0t\(\hat{i}\) - 21.0t 2\(\hat{j}\) + 10.0t -3\(\hat{k}\)) m/s

    b.\(\vec{a}\) (t) = (6.0\(\hat{i}\) - 42.0t\(\hat{j}\) - 30t -4\(\hat{k}\)) m/s 2

    c.\(\vec{v}\) (2.0s) = (12.0\(\hat{i}\) - 84.0\(\hat{j}\) + 1.25\(\hat{k}\)) m/s

    d.\(\vec{v}\) (1.0 s) = (6.0\(\hat{i}\) - 21.0\(\hat{j}\) + 10.0\(\hat{k}\)) m/s, |\(\vec{v}\) (1.0 s) | = 24.0 m/s;\(\vec{v}\) (3.0 s) = (18.0\(\hat{i}\) - 189.0\(\hat{j}\) + 0.37\(\hat{k}\)) m/s, |\(\vec{v}\) (3.0 s) | = 190.0 m/s

    e.\(\vec{r}\) (t) = (3.0t 2\(\hat{i}\) - 7.0t 3\(\hat{j}\) - 5.0t -2\(\hat{k}\)) cm,\(\vec{v}_{avg}\) = (9.0\(\hat{i}\) - 49.0 - 6.3\(\hat{j}\)\(\hat{k}\)) m/s

    31. a.\(\vec{v}\) (t) = -dhambi (1.0t)\(\hat{i}\) + gharama (1.0t)\(\hat{j}\) +\(\hat{k}\)

    b.\(\vec{a}\) (t) = -cos (1.0t)\(\hat{i}\) - dhambi (1.0t)\(\hat{j}\)

    33. a. t = 0.55 s

    b. x = 110 m

    35. a. t = 0.24s, d = 0.28 m

    b Wao lengo juu.

    Mfano wa mtu kutupa dart. Dart hutolewa kwa usawa umbali wa mita 2.4 kutoka bodi ya dart, kiwango na jicho la ng'ombe la bodi ya dart, kwa kasi ya mita 10 kwa pili.

    37. a. t = 12.8 s, x = 5619 m

    b. v y = 125.0 m/s, v x = 439.0 m/s, |\(\vec{v}\) | = 456.0 m/s

    39. a. v y = v 0y - gt, t = 10s, v y = 0, v 0y = 98.0 m/s, v 0 = 196.0 m/s

    b. h = 490.0 m

    c. v 0x = 169.7 m/s, x = 3394.0 m

    d. x = 2545.5 m, y = 367.5 m,\(\vec{s}\) = 2545.5 m\(\hat{i}\) + 367.5 m\(\hat{j}\)

    41. -100 m = (-2.0 m/s) t - (4.9 m/s 2) t 2, t = 4.3 s, x = 86.0 m

    43. R Mwezi = 48 m

    45. a. v 0y = 24 m/s, v y 2 = v 0y 2 - 2gy\(\Rightarrow\) h = 23.4 m

    b. t = 3 s, v 0x = 18 m/s, x = 54 m

    c. y = -100 m, y 0 = 0, y - y 0 = v 0y, t -\(\frac{1}{2}\) gt 2 - 100 = 24t - 4.9t 2\(\Rightarrow\) t = 7.58 s

    d. x = 136.44 m

    e.$$\ kuanza {mgawanyiko} t & = 2.0\; s, y = 28.4\; m, x = 36\; m\\ t & = 4.0\; s, y = 17.6\; m, x = 22.4\; m\\ t & = 6.0\; s, y = -32.4\; m = 108\; m\ mwisho {mgawanyiko}\]

    47. v 0y = 12.9 m/s, y - y 0 = v 0y t -\(\frac{1}{2}\) gt 2 - 20.0 = 12.9t - 4.9t 2

    t = 3.7 s, v 0x = 15.3 m/s\(\Rightarrow\) x = 56.7 m

    Hivyo risasi ya golfer ardhi 13.3 m mfupi ya kijani.

    49. a. R = 60.8 m

    b R = 137.8 m

    51. a. v y 2 = v 0y 2 - 2gy\(\Rightarrow\) y = 2.9 m/s

    y = 3.3 m/s

    y =\(\frac{v_{0y}^{2}}{2g}\)\(\frac{(v_{0} \sin \theta)^{2}}{2g} \Rightarrow \sin \theta\) = 0.91\(\Rightarrow\)\(\theta\) = 65.5°

    53. R = 18.5 m

    55. y = (tan\(\theta_{0}\)) x -\(\Big[ \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \Big]\) x 2\(\Rightarrow\) v 0 = 16.4 m/s

    57. R\(\frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \Rightarrow \theta_{0}\) = 15.0°

    59. Inachukua mpokeaji pana 1.1 s ili kufikia m 10 ya mwisho ya kukimbia kwake.

    T tof\(\frac{2(v_{0} \sin \theta)}{g} \Rightarrow \sin \theta\) = 0.27\(\Rightarrow \theta\) = 15.6°

    61. a C = 40 m/s 2

    63. C =\(\frac{v^{2}}{r} \Rightarrow\) v 2 = r, C = 78.4, v = 8.85 m/s

    T = 5.68 s, ambayo ni 0.176 rev/s = 10.6 rev/min

    65. Venus ni kilomita milioni 108.2 kutoka Jua na ina kipindi cha orbital cha 0.6152 y.

    r = 1.082 x 10 11 m, T = 1.94 x 10 7 s

    v = 3.5 x 10 4 m/s, C = 1.135 x 10 ÷ 2 m/s 2

    67. 360 rev/min = 6 rev/s

    v = 3.8 m/s, C = 144. m/s 2

    69. a. O( t) = (4.0\(\hat{i}\) + 3.0\(\hat{j}\) + 5.0\(\hat{k}\)) t m

    b.\(\vec{r}_{PS}\) =\(\vec{r}_{PS'}\) +\(\vec{r}_{S'S}\),\(\vec{r}\) (t) =\(\vec{r′}\) (t) + (4.0\(\hat{i}\) + 3.0\(\hat{j}\) + 5.0\(\hat{k}\)) t m

    c.\(\vec{v}\) (t) =\(\vec{v′}\) (t) + (4.0 + 3.0\(\hat{i}\)\(\hat{j}\) + 5.0\(\hat{k}\)) m/s

    d. kasi ni sawa.

    71. \(\vec{v}_{PC}\)= (2.0\(\hat{i}\) + 5.0\(\hat{j}\) + 4.0\(\hat{k}\)) m/s

    73. a. hewa =, S = seagull, G = ardhi

    \(\vec{v}_{SA}\)= 9.0 m/s, kasi ya seagull kwa heshima na hewa bado

    \(\vec{v}_{AG}\)=? ,\(\vec{v}_{SG}\) = 5 m/s,\(\vec{v}_{SG} = \vec{v}_{SA} + \vec{v}_{AG} \Rightarrow \vec{v}_{AG} = \vec{v}_{SG} − \vec{v}_{SA}\)

    \(\vec{v}_{AG}\)= -4.0 m/s

    b.\(\vec{v}_{SG} = \vec{v}_{SA} + \vec{v}_{AG} \Rightarrow \vec{v}_{SG}\) = -13.0 m/s

    \(\frac{−6000\; m}{−13.0\; m/s}\)= 7 min 42 s

    75. Chukua mwelekeo mzuri kuwa mwelekeo huo ambao mto unapita, ambao ni mashariki. S = Pwani/Dunia, W = maji, na B = mashua.

    a.\(\vec{v}_{BS}\) = 11 km/h, t = 8.2 min

    b.\(\vec{v}_{BS}\) = -5 km/h, t = 18 min

    c.\(\vec{v}_{BS} = \vec{v}_{BW} + \vec{v}_{WS}, \theta\) = 22° magharibi mwa kaskazini

    Vectors V ndogo B W, V ndogo W S na V ndogo B S huunda pembetatu sahihi, na V ndogo B W kama hypotenuse. V ndogo B S anasema juu. V ndogo W S anasema na haki. V ndogo B W anasema juu na kushoto, kwa pembe ya theta kwa wima. V ndogo B S ni vector jumla ya v ndogo B W na V ndogo W S.

    d. |\(\vec{v}_{BS}\) | = 7.4 km/h, t = 6.5 min

    e.\(\vec{v}_{BS}\) = 8.54 km/h, lakini tu sehemu ya kasi moja kwa moja katika mto hutumiwa kupata muda

    Vectors V ndogo B W, V ndogo W S na V ndogo B S huunda pembetatu sahihi, na V ndogo B S kama hypotenuse. V ndogo B W anasema juu. V ndogo W S anasema na haki. V ndogo B S anasema juu na kulia, kwa pembe ya theta kwa wima. V ndogo B S ni vector jumla ya v ndogo B W na V ndogo W S.

    t = 6.0 min

    Mto = 0.3 km

    77. \(\vec{v}_{AG} = \vec{v}_{AC} + \vec{v}_{CG}\)

    |\(\vec{v}_{AC}\) | = 25 km/h, |\(\vec{v}_{CG}\) | = 15 km/h, |\(\vec{v}_{AG}\) | = 29.15 km/h,\(\vec{v}_{AG} = \vec{v}_{AC} + \vec{v}_{CG}\)

    Pembe kati\(\vec{v}_{AC}\) na\(\vec{v}_{AG}\) ni 31°, hivyo mwelekeo wa upepo ni 14° kaskazini ya mashariki.

    Vectors V ndogo A C, V ndogo C G na V ndogo A G kuunda pembetatu. V ndogo A C na V ndogo C G ni katika pembe za kulia. V ndogo A G ni vector jumla ya v ndogo C na V ndogo C G.

    Matatizo ya ziada

    79. C = 39.6 m/s 2

    81. 90.0 km/h = 25.0 m/s, 9.0 km/h = 2.5 m/s, 60.0 km/h = 16.7 m/s

    a T = -2.5 m/s 2, C = 1.86 m/s 2, a = 3.1 m/s 2

    83. Radi ya mzunguko wa mapinduzi katika latitude\(\lambda\) ni R E cos\(\lambda\). Kasi ya mwili ni\(\frac{2 \pi r}{T}\). C =\(\frac{4 \pi^{2} R_{E} \cos \lambda}{T^{2}}\) kwa\(\lambda\) = 40°, C = 0.26% g

    85. a T = 3.00 m/s 2

    v (5 s) = 15.00 m/s, C = 150.00 m/s 2,\(\theta\) = 88.8° kuhusiana na tangent kwa mduara wa mapinduzi iliyoelekezwa ndani.

    |\(\vec{a}\) | = 150.03 m/s 2

    87. \(\vec{a}\)(t) =\(\omega^{2}\) -Gharama\(\omega\) kwa\(\hat{i}\) -\(\omega^{2}\) Ishara\(\omega\) t\(\hat{j}\)

    C = 5.0 m\(\omega^{2}\),\(\omega\) = 0.89 rad/s

    \(\vec{v}\)(t) = -2.24 m/s\(\hat{i}\) - 3.87 m/s\(\hat{j}\)

    89. \(\vec{r}_{1}\)= 1.5\(\hat{j}\) + 4.0\(\hat{k}\),\(\vec{r}_{2} = \Delta \vec{r} + \vec{r}_{1}\) = 2.5\(\hat{i}\) + 4.7\(\hat{j}\) + 2.8\(\hat{k}\)

    91. v x (t) = 265.0 m/s, v y (t) = 20.0 m/s,\(\vec{v}\) (5.0 s) = (265.0\(\hat{i}\) + 20.0\(\hat{j}\)) m/s

    93. R = 1.07 m

    95. v 0 = 20.1 m/s

    97. v = 3072.5 m/s, C = 0.223 m/s 2

    Changamoto Matatizo

    99. a. -400.0 m = v 0y t - 4.9t 2, 359.0 m = v 0x t, t =\(\frac{359.0}{v_{0x}}\) - 400.0 = 359.0\(\frac{v_{0y}}{v_{0x}}\) - 4.9\(\left(\dfrac{359.0}{v_{0x}}\right)^{2}\)

    -400.0 = 359.0 tan 40 -\(\frac{631,516.9}{v_{0x}^{2}} \Rightarrow\) v 0x 2 = 900.6, v 0x = 30.0 m/s, v 0y = v 0x tan 40 = 25.2 m/s, v = 39.2 m/s

    b. t = 12.0 s

    101. a.\(\vec{r}_{TC}\) = (-32 + 80t)\(\hat{i}\) + 50t\(\hat{j}\), |\(\vec{r}_{TC}\) | 2 = (-32 + 80t) 2 + (50t) 2

    2r\(\frac{dr}{dt}\) = 2 (-32 + 80t) + 100t,\(\frac{dr}{dt} = \frac{2(−32 + 80t) + 100t}{2r}\) = 0

    260t = 64\(\Rightarrow\) t = 15 min

    b. |\(\vec{r}_{TC}\) | = 17 km