13.4E: Mazoezi ya Sehemu ya 13.4

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Edwin “Jed” Herman & Gilbert Strang
OpenStax

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1) Kutokana\(\vecs r(t)=(3t^2−2)\,\hat{\mathbf{i}}+(2t−\sin t)\,\hat{\mathbf{j}}\),

a. kupata kasi ya chembe kusonga pamoja Curve hii.

b. kupata kasi ya chembe kusonga pamoja Curve hii.

Jibu: a.\(\vecs v(t)=6t\,\hat{\mathbf{i}}+(2−\cos t)\,\hat{\mathbf{i}}\)
b.\(\vecs a(t)=6\,\hat{\mathbf{i}}+\sin t\,\hat{\mathbf{i}}\)

Katika maswali 2 - 5, kutokana na kazi ya msimamo, pata kasi, kasi, na kasi kulingana na parameter\(t\).

2)\(\vecs r(t)=e^{−t}\,\hat{\mathbf{i}}+t^2\,\hat{\mathbf{j}}+\tan t\,\hat{\mathbf{k}}\)

3)\(\vecs r(t)=⟨3\cos t,\,3\sin t,\,t^2⟩\)

Jibu: \(\vecs v(t)=-3\sin t\,\hat{\mathbf{i}}+3\cos t\,\hat{\mathbf{j}}+2t\,\hat{\mathbf{k}}\)
\(\vecs a(t)=-3\cos t\,\hat{\mathbf{i}}-3\sin t\,\hat{\mathbf{j}}+2\,\hat{\mathbf{k}}\)
\(\text{Speed}(t) = \|\vecs v(t)\| = \sqrt{9 + 4t^2}\)

4)\(\vecs r(t)=t^5\,\hat{\mathbf{i}}+(3t^2+2t- 5)\,\hat{\mathbf{j}}+(3t-1)\,\hat{\mathbf{k}}\)

5)\(\vecs r(t)=2\cos t\,\hat{\mathbf{j}}+3\sin t\,\hat{\mathbf{k}}\). Grafu imeonyeshwa hapa:

Jibu: \(\vecs v(t)=-2\sin t\,\hat{\mathbf{j}}+3\cos t\,\hat{\mathbf{k}}\)
\(\vecs a(t)=-2\cos t\,\hat{\mathbf{j}}-3\sin t\,\hat{\mathbf{k}}\)
\(\text{Speed}(t) = \|\vecs v(t)\| = \sqrt{4\sin^2 t+9\cos^2 t}=\sqrt{4+5\cos^2 t}\)

Katika maswali 6 - 8, pata kasi, kasi, na kasi ya chembe na kazi iliyopewa nafasi.

6)\(\vecs r(t)=⟨t^2−1,t⟩\)

7)\(\vecs r(t)=⟨e^t,e^{−t}⟩\)

Jibu: \(\vecs v(t)=⟨e^t,−e^{−t}⟩\),
\(\vecs a(t)=⟨e^t, e^{−t}⟩,\)
\( \|\vecs v(t)\| = \sqrt{e^{2t}+e^{−2t}}\)

8)\(\vecs r(t)=⟨\sin t,t,\cos t⟩\). Grafu imeonyeshwa hapa:

9) Kazi ya nafasi ya kitu hutolewa na\(\vecs r(t)=⟨t^2,5t,t^2−16t⟩\). Wakati gani kasi ni kiwango cha chini?

Jibu: \(t = 4\)

10) Hebu\(\vecs r(t)=r\cosh(ωt)\,\hat{\mathbf{i}}+r\sinh(ωt)\,\hat{\mathbf{j}}\). Kupata kasi na kuongeza kasi wadudu na kuonyesha kwamba kuongeza kasi ni sawia na\(\vecs r(t)\).

11) Fikiria mwendo wa uhakika juu ya mzunguko wa mduara unaoendelea. Kama mzunguko unavyozunguka, huzalisha cycloid\(\vecs r(t)=(ωt−\sin(ωt))\,\hat{\mathbf{i}}+(1−\cos(ωt))\,\hat{\mathbf{j}}\), wapi\(\omega\) kasi ya angular ya mduara na\(b\) ni radius ya mduara:

Kupata equations kwa kasi, kuongeza kasi, na kasi ya chembe wakati wowote.

Jibu: \(\vecs v(t)=(ω−ω\cos(ωt))\,\hat{\mathbf{i}}+(ω\sin(ωt))\,\hat{\mathbf{j}}\)
\(\vecs a(t)=(ω^2\sin(ωt))\,\hat{\mathbf{i}}+(ω^2\cos(ωt))\,\hat{\mathbf{j}}\)
\ (\ kuanza {align*}\ Nakala {kasi} (t) &=\ sqrt {(ωωω\ cos (ωt)) ^2 + (ω\ dhambi (ωt)) ^2}\\
&=\ sqrt {ω^2 - 2ω^2\ cos (ωt) + ω^2\ sin^2 (ωt)}\\
&=\ sqrt {2ω^2 (1 -\ cos (ωt))}\ mwisho {align*}\)

12) Mtu kwenye glider hutegemea inazunguka zaidi kutokana na hewa inayoongezeka kwa kasi kwenye njia ya kuwa na vector ya msimamo\(\vecs r(t)=(3\cos t)\,\hat{\mathbf{i}}+(3\sin t)\,\hat{\mathbf{j}}+t^2\,\hat{\mathbf{k}}\). Njia hiyo inafanana na ile ya helix ingawa si helix. Grafu imeonyeshwa hapa:

Pata kiasi zifuatazo:

a. kasi na kuongeza kasi wadudu

b. kasi ya glider wakati wowote

Jibu: \(\|\vecs v(t)\|=\sqrt{9+4t^2}\)

c. nyakati, kama ipo, ambapo kasi ya glider ni orthogonal kwa kasi yake

13) Kutokana na kwamba\(\vecs r(t)=⟨e^{−5t}\sin t,\, e^{−5t}\cos t,\, 4e^{−5t}⟩\) ni vector nafasi ya chembe kusonga, kupata kiasi zifuatazo:

a. kasi ya chembe

Jibu: \(\vecs v(t)=⟨e^{−5t}(\cos t−5\sin t),\, −e^{−5t}(\sin t+5\cos t),\, −20e^{−5t}⟩\)

b. kasi ya chembe

c. kuongeza kasi ya chembe

Jibu: \(\vecs a(t)=⟨e^{−5t}(−\sin t−5\cos t)−5e^{−5t}(\cos t−5\sin t), \; −e^{−5t}(\cos t−5\sin t)+5e^{−5t}(\sin t+5\cos t),\; 100e^{−5t}⟩\)

14) Pata kasi ya kiwango cha juu ya mzunguko wa tairi ya gari ya radius\(1\) ft wakati gari linasafiri kwa\(55\) mph.

15) Kupata nafasi vector-thamani kazi\(\vecs r(t)\), kutokana na kwamba\(\vecs a(t)=\hat{\mathbf{i}}+e^t \,\hat{\mathbf{j}}, \quad \vecs v(0)=2\,\hat{\mathbf{j}}\), na\(\vecs r(0)=2\,\hat{\mathbf{i}}\).

16) Kupata\(\vecs r(t)\) kutokana na kwamba\(\vecs a(t)=−32\,\hat{\mathbf{j}}, \vecs v(0)=600\sqrt{3} \,\hat{\mathbf{i}}+600\,\hat{\mathbf{j}}\), na\(\vecs r(0)=\vecs 0\).

17) Kuongeza kasi ya kitu hutolewa na\(\vecs a(t)=t\,\hat{\mathbf{j}}+t\,\hat{\mathbf{k}}\). Kasi katika\(t=1\) sec ni\(\vecs v(1)=5\,\hat{\mathbf{j}}\) na nafasi ya kitu katika\(t=1\) sec ni\(\vecs r(1)=0\,\hat{\mathbf{i}}+0\,\hat{\mathbf{j}}+0\,\hat{\mathbf{k}}\). Pata nafasi ya kitu wakati wowote.

Jibu: \(\vecs r(t)=0\,\hat{\mathbf{i}}+\left(\frac{1}{6}t^3+4.5t−\frac{14}{3}\right)\,\hat{\mathbf{j}}+\left(\frac{1}{6}t^3−\frac{1}{2}t+\frac{1}{3}\right)\,\hat{\mathbf{k}}\)

Projectile Motion

18) Projectile hupigwa hewani kutoka ngazi ya chini na kasi ya awali ya\(500\) m/sec kwa pembe ya 60° na usawa.

a Wakati gani projectile hufikia urefu wa juu?

Jibu: \(44.185\)sekunde

b Urefu wa kiwango cha juu cha projectile ni nini?

c Wakati gani kiwango cha juu cha projectile kinachopatikana?

Jibu: \(t=88.37\)sekunde

d Je, ni kiwango cha juu gani?

e Ni wakati gani wa kukimbia wa projectile?

Jibu: \(t=88.37\)sekunde

19) Projectile inafukuzwa kwa kimo cha\(1.5\) m juu ya ardhi na kasi ya awali ya\(100\) m/sec na kwa pembe ya 30° juu ya usawa. Tumia habari hii kujibu maswali yafuatayo:

a Kuamua urefu wa juu wa projectile.

b Kuamua aina mbalimbali za projectile.

Jibu: Mbalimbali ni takriban\(886.29\) m.

20) mpira golf ni hit katika mwelekeo usawa mbali makali ya juu ya jengo ambayo ni 100 ft mrefu. Jinsi ya kufunga lazima mpira ilizinduliwa nchi\(450\) ft mbali?

21) Projectile inafukuzwa kutoka ngazi ya chini kwa pembe ya 8° na usawa. Projectile ni kuwa na meta mbalimbali\(50\) Pata kasi ya chini (kasi) muhimu ili kufikia aina hii.

Jibu: \(v=42.16\)m/sec

22) Thibitisha kwamba kitu kinachohamia kwenye mstari wa moja kwa moja kwa kasi ya mara kwa mara kina kasi ya sifuri.

Kutafuta Vipengele vya Sheria za Kuharakisha na Kepler

23) Pata vipengele vya kawaida na vya kawaida vya kuongeza kasi kwa\(\vecs r(t)=t^2\,\hat{\mathbf{i}}+2t \,\hat{\mathbf{j}}\) wakati\(t=1\).

Jibu: \(a_\vecs{T}=\sqrt{2}, \quad a_\vecs{N}=\sqrt{2}\)

Katika maswali 24 - 30, pata vipengele vya kawaida na vya kawaida vya kuongeza kasi.

24)\(\vecs r(t)=⟨\cos(2t),\,\sin(2t),1⟩\)

25)\(\vecs r(t)=⟨e^t \cos t,\,e^t\sin t,\,e^t⟩\). Grafu imeonyeshwa hapa:

Jibu: \(a_\vecs{T}=\sqrt{3}e^t, \quad a_\vecs{N}=\sqrt{2}e^t\)

26)\(\vecs r(t)=⟨\frac{2}{3}(1+t)^{3/2}, \,\frac{2}{3}(1-t)^{3/2},\,\sqrt{2}t⟩\)

27)\(\vecs r(t)=\left\langle 2t,\,t^2,\,\dfrac{t^3}{3}\right\rangle\)

Jibu: \(a_\vecs{T}=2t, \quad a_\vecs{N}=2\)

28)\(\vecs r(t)=t^2\,\hat{\mathbf{i}}+t^2\,\hat{\mathbf{j}}+t^3\,\hat{\mathbf{k}}\)

29)\(\vecs r(t)=⟨6t,\,3t^2,\,2t^3⟩\)

Jibu: \(a_\vecs{T}=\dfrac{6t +12t^3}{\sqrt{1+t^2+t^4}}, \quad a_\vecs{N}=6\sqrt{\dfrac{1+4t^2+t^4}{1+t^2+t^4}}\)

30)\(\vecs r(t)=3\cos(2πt)\,\hat{\mathbf{i}}+3\sin(2πt)\,\hat{\mathbf{j}}\)

Jibu: \(a_\vecs{T}=0, \quad a_\vecs{N}=12\pi^2\)

31) Kupata vipengele tangential na ya kawaida ya kuongeza kasi kwa\(\vecs r(t)=a\cos(ωt)\,\hat{\mathbf{i}}+b\sin(ωt)\,\hat{\mathbf{j}}\) saa\(t=0\).

Jibu: \(a_\vecs{T}=0, \quad a_\vecs{N}=aω^2\)

32) Tuseme kwamba kazi ya msimamo kwa kitu katika vipimo vitatu inatolewa na equation\(\vecs r(t)=t\cos(t)\,\hat{\mathbf{i}}+t\sin(t)\,\hat{\mathbf{j}}+3t\,\hat{\mathbf{k}}\).

Onyesha kwamba chembe huenda kwenye koni ya mviringo.

b. kupata angle kati ya kasi na kuongeza kasi wadudu wakati\(t=1.5\).

c Kupata vipengele tangential na ya kawaida ya kuongeza kasi wakati\(t=1.5\).

Jibu: c.\(a_\vecs{T}=0.43\,\text{m/sec}^2, \quad a_\vecs{N}=2.46\,\text{m/sec}^2\)

33) Nguvu kwenye chembe hutolewa na\(\vecs f(t)=(\cos t)\,\hat{\mathbf{i}}+(\sin t)\,\hat{\mathbf{j}}\). Chembe iko katika hatua\((c,0)\) katika\(t=0\). Kasi ya awali ya chembe hutolewa na\(\vecs v(0)=v_0\,\hat{\mathbf{j}}\). Pata njia ya chembe ya wingi\(m\). (Kumbuka,\(\vecs F=m\vecs a\).)

Jibu: \(\vecs r(t)=\left(\dfrac{-\cos t}{m}+c+\frac{1}{m}\right)\,\hat{\mathbf{i}}+\left(\dfrac{−\sin t}{m}+\left(v_0+\frac{1}{m}\right)t\right)\,\hat{\mathbf{j}}\)

34) magari ambayo ina uzito\(2700\) lb inafanya kurejea kwenye barabara gorofa wakati wa kusafiri katika\(56\) ft/sec. Ikiwa radius ya kugeuka ni\(70\) ft, ni nguvu gani ya msuguano inayohitajika ili kuweka gari kutoka skidding?

35) Kutumia sheria za Kepler, inaweza kuonyeshwa kuwa\(v_0=\sqrt{\dfrac{2GM}{r_0}}\) ni kasi ya chini inayohitajika wakati\(\theta=0\) ili kitu kitatoroka kutoka kwenye kuvuta kwa nguvu kuu inayotokana na wingi\(M\). Tumia matokeo haya ili kupata kasi ya chini wakati\(\theta=0\) kwa capsule ya nafasi kutoroka kutoka kwenye mvuto wa mvuto wa Dunia ikiwa probe iko kwenye urefu wa\(300\) kilomita juu ya uso wa Dunia.

Jibu: \(10.94\)km/sec

36) Kupata muda katika miaka inachukua sayari kibete Pluto kufanya obiti moja kuhusu Jua kutokana na kwamba\(a=39.5\) A.U.

Wachangiaji

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