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2.1 Kinematika — Masalalar

Jami: 30 ta | Yechim bilan:


Asosiy Masalalar (1-10)

Masala 1 ⭐⭐

Avtomobil 20 m/s tezlik bilan harakatlanmoqda. 5 s da qancha masofa bosadi?

Yechim

x=vt=20×5=100x = vt = 20 \times 5 = 100 m


Masala 2 ⭐⭐

Velosiped 0 dan 15 m/s gacha 10 s da tezlashdi. Tezlanishni toping.

Yechim

a=vv0t=15010=1.5a = \frac{v - v_0}{t} = \frac{15 - 0}{10} = 1.5 m/s²


Masala 3 ⭐⭐

Jism 2 m/s² tezlanish bilan tinch holatdan harakatlandi. 6 s dagi tezligi va bosib o'tgan yo'li?

Yechim

v=v0+at=0+2(6)=12v = v_0 + at = 0 + 2(6) = 12 m/s

x=v0t+12at2=0+12(2)(36)=36x = v_0 t + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(36) = 36 m


Masala 4 ⭐⭐

Tosh 50 m balandlikdan erkin tushirildi. Yerga tushguncha qancha vaqt o'tadi? (g=10g = 10 m/s²)

Yechim

h=12gt2h = \frac{1}{2}gt^2

50=12(10)t250 = \frac{1}{2}(10)t^2

t2=10t^2 = 10

t=103.16t = \sqrt{10} \approx 3.16 s


Masala 5 ⭐⭐

Avtomobil 30 m/s tezlikda ketayotib, 4 m/s² tezlanish bilan tormozladi. To'xtaguncha qancha masofa bosadi?

Yechim

v2=v02+2axv^2 = v_0^2 + 2ax

0=900+2(4)x0 = 900 + 2(-4)x

x=9008=112.5x = \frac{900}{8} = 112.5 m


Masala 6 ⭐⭐

To'p 45° burchak ostida 20 m/s tezlik bilan otildi. Maksimal balandlik? (g=10g = 10 m/s²)

Yechim

hmax=v02sin2θ2g=400×0.520=10h_{max} = \frac{v_0^2 \sin^2\theta}{2g} = \frac{400 \times 0.5}{20} = 10 m


Masala 7 ⭐⭐

Disk 100 rpm tezlikda aylanmoqda. Burchak tezlikni rad/s da toping.

Yechim

ω=2π×10060=200π60=10π310.47\omega = \frac{2\pi \times 100}{60} = \frac{200\pi}{60} = \frac{10\pi}{3} \approx 10.47 rad/s


Masala 8 ⭐⭐

G'ildirak radiusi 0.3 m, burchak tezligi 20 rad/s. Chiziqli tezlik?

Yechim

v=rω=0.3×20=6v = r\omega = 0.3 \times 20 = 6 m/s


Masala 9 ⭐⭐

Samolyot 200 m/s tezlik bilan 500 m radiusli aylana bo'ylab uchmoqda. Markazga intilma tezlanish?

Yechim

an=v2r=40000500=80a_n = \frac{v^2}{r} = \frac{40000}{500} = 80 m/s²


Masala 10 ⭐⭐

Proektil 30° burchak ostida 50 m/s bilan otildi. Gorizontal va vertikal tezlik komponentlari?

Yechim

vx=v0cos30°=50×0.866=43.3v_x = v_0 \cos 30° = 50 \times 0.866 = 43.3 m/s

vy=v0sin30°=50×0.5=25v_y = v_0 \sin 30° = 50 \times 0.5 = 25 m/s


O'rtacha Masalalar (11-22)

Masala 11 ⭐⭐⭐

Dron A(0, 0, 10) dan B(100, 50, 30) m ga 20 s da uchdi. O'rtacha tezlik vektorini toping.

Yechim

Δr=(1000,500,3010)=(100,50,20)\Delta \vec{r} = (100-0, 50-0, 30-10) = (100, 50, 20) m

vavg=ΔrΔt=(100,50,20)20=(5,2.5,1)\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{(100, 50, 20)}{20} = (5, 2.5, 1) m/s

vavg=25+6.25+1=32.255.68|\vec{v}_{avg}| = \sqrt{25 + 6.25 + 1} = \sqrt{32.25} \approx 5.68 m/s


Masala 12 ⭐⭐⭐

Robot g'ildiragi 0.1 m radiusli, motor 500 rpm beradi. Robotning chiziqli tezligi?

Yechim

ω=2π×50060=52.36\omega = \frac{2\pi \times 500}{60} = 52.36 rad/s

v=rω=0.1×52.36=5.24v = r\omega = 0.1 \times 52.36 = 5.24 m/s


Masala 13 ⭐⭐⭐

To'p 60° burchak ostida 40 m/s bilan otildi. Uchish masofasi va vaqti? (g=10g = 10 m/s²)

Yechim

R=v02sin2θg=1600×sin120°10=1600×0.86610=138.6R = \frac{v_0^2 \sin 2\theta}{g} = \frac{1600 \times \sin 120°}{10} = \frac{1600 \times 0.866}{10} = 138.6 m

T=2v0sinθg=2×40×0.86610=6.93T = \frac{2v_0 \sin\theta}{g} = \frac{2 \times 40 \times 0.866}{10} = 6.93 s


Masala 14 ⭐⭐⭐

Jism r(t)=(3t2,4t,5)\vec{r}(t) = (3t^2, 4t, 5) m qonun bo'yicha harakat qilmoqda. t=2t = 2 s dagi tezlik va tezlanish?

Yechim

v=drdt=(6t,4,0)\vec{v} = \frac{d\vec{r}}{dt} = (6t, 4, 0)

t=2t = 2: v(2)=(12,4,0)\vec{v}(2) = (12, 4, 0) m/s

v=144+16=16012.65|\vec{v}| = \sqrt{144 + 16} = \sqrt{160} \approx 12.65 m/s

a=dvdt=(6,0,0)\vec{a} = \frac{d\vec{v}}{dt} = (6, 0, 0) m/s² (doimiy)


Masala 15 ⭐⭐⭐

Ikki mashina A va B bir nuqtadan bir vaqtda harakatlandi. A shimolga 60 km/h, B sharqqa 80 km/h. 1 soatdan keyin ular orasidagi masofa?

Yechim

1 soatda:

  • A: 60 km shimolga
  • B: 80 km sharqqa

Masofa: d=602+802=3600+6400=10000=100d = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 km


Masala 16 ⭐⭐⭐

Raketa vertikal ko'tarilmoqda. Tezlanish a=20a = 20 m/s². 10 s da qanday balandlikka ko'tariladi va tezligi qancha?

Yechim

h=12at2=12(20)(100)=1000h = \frac{1}{2}at^2 = \frac{1}{2}(20)(100) = 1000 m

v=at=20×10=200v = at = 20 \times 10 = 200 m/s


Masala 17 ⭐⭐⭐

Differensial drive robot: chap g'ildirak 1 m/s, o'ng 1.5 m/s, g'ildiraklar orasidagi masofa 0.5 m. Robot tezligi va burchak tezligi?

Yechim

v=vR+vL2=1.5+12=1.25v = \frac{v_R + v_L}{2} = \frac{1.5 + 1}{2} = 1.25 m/s

ω=vRvLL=1.510.5=1\omega = \frac{v_R - v_L}{L} = \frac{1.5 - 1}{0.5} = 1 rad/s

Robot chapga burilmoqda.


Masala 18 ⭐⭐⭐

Quadcopter propeller 0 dan 10000 rpm gacha 2 s da tezlashdi. Burchak tezlanish va bu vaqtda necha marta aylandi?

Yechim

ω0=0\omega_0 = 0, ω=2π×1000060=1047.2\omega = \frac{2\pi \times 10000}{60} = 1047.2 rad/s

α=ωω0t=1047.22=523.6\alpha = \frac{\omega - \omega_0}{t} = \frac{1047.2}{2} = 523.6 rad/s²

θ=ω0t+12αt2=0+12(523.6)(4)=1047.2\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(523.6)(4) = 1047.2 rad

Aylanishlar: n=θ2π=1047.22π=166.7n = \frac{\theta}{2\pi} = \frac{1047.2}{2\pi} = 166.7 aylanish


Masala 19 ⭐⭐⭐

Jism 5 m radiusli aylana bo'ylab harakat qilmoqda. Tezligi v=2tv = 2t m/s. t=3t = 3 s da tangensial va normal tezlanish?

Yechim

t=3t = 3 s da: v=6v = 6 m/s

Tangensial: at=dvdt=2a_t = \frac{dv}{dt} = 2 m/s²

Normal: an=v2r=365=7.2a_n = \frac{v^2}{r} = \frac{36}{5} = 7.2 m/s²

Umumiy: a=4+51.84=55.847.47|a| = \sqrt{4 + 51.84} = \sqrt{55.84} \approx 7.47 m/s²


Masala 20 ⭐⭐⭐

Kema A shimolga 30 km/h, kema B sharqqa 40 km/h harakat qilmoqda. B dan A ning nisbiy tezligi?

Yechim

vA=(0,30)\vec{v}_A = (0, 30) km/h (shimolga) vB=(40,0)\vec{v}_B = (40, 0) km/h (sharqqa)

vA/B=vAvB=(40,30)\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B = (-40, 30) km/h

vA/B=1600+900=50|\vec{v}_{A/B}| = \sqrt{1600 + 900} = 50 km/h

Yo'nalish: θ=arctan(3040)=143.1°\theta = \arctan(\frac{30}{-40}) = 143.1° (shimoli-g'arbga)


Masala 21 ⭐⭐⭐

Snaryad 45° burchak ostida 100 m/s bilan otildi, lekin shamol gorizontal 10 m/s tezlikda esmoqda (qarshi). Uchish masofasi qanchaga kamayadi?

Yechim

Shamolsiz: R0=v02sin90°g=1000010=1000R_0 = \frac{v_0^2 \sin 90°}{g} = \frac{10000}{10} = 1000 m

Shamol bilan (gorizontal tezlik kamayadi): vx=v0cos45°10=70.710=60.7v_x = v_0 \cos 45° - 10 = 70.7 - 10 = 60.7 m/s

Uchish vaqti (vertikal komponent o'zgarmaydi): T=2v0sin45°g=2×70.710=14.14T = \frac{2 v_0 \sin 45°}{g} = \frac{2 \times 70.7}{10} = 14.14 s

R=vx×T=60.7×14.14=858.3R = v_x \times T = 60.7 \times 14.14 = 858.3 m

Farq: 1000858.3=141.71000 - 858.3 = 141.7 m kamaydi


Masala 22 ⭐⭐⭐

Lift 3 m/s² tezlanish bilan yuqoriga ko'tarilmoqda. Unda turgan odam (70 kg) o'z og'irligini qanday sezadi?

Yechim

Effektiv tezlanish: geff=g+a=10+3=13g_{eff} = g + a = 10 + 3 = 13 m/s²

Sezilgan og'irlik: W=mgeff=70×13=910W = mg_{eff} = 70 \times 13 = 910 N

Oddiy og'irlik: W0=70×10=700W_0 = 70 \times 10 = 700 N

30% ko'proq og'irlik seziladi.


Murakkab Masalalar (23-30)

Masala 23 ⭐⭐⭐⭐

Dron r(t)=(10cos(0.5t),10sin(0.5t),2t)\vec{r}(t) = (10\cos(0.5t), 10\sin(0.5t), 2t) m traektoriya bo'ylab uchmoqda. t=4t = 4 s da tezlik va tezlanish vektorlarini toping.

Yechim

v=drdt=(5sin(0.5t),5cos(0.5t),2)\vec{v} = \frac{d\vec{r}}{dt} = (-5\sin(0.5t), 5\cos(0.5t), 2)

t=4t = 4: 0.5×4=20.5 \times 4 = 2 rad

v(4)=(5sin2,5cos2,2)=(4.55,2.08,2)\vec{v}(4) = (-5\sin 2, 5\cos 2, 2) = (-4.55, -2.08, 2) m/s

v=20.7+4.33+4=5.39|\vec{v}| = \sqrt{20.7 + 4.33 + 4} = 5.39 m/s

a=dvdt=(2.5cos(0.5t),2.5sin(0.5t),0)\vec{a} = \frac{d\vec{v}}{dt} = (-2.5\cos(0.5t), -2.5\sin(0.5t), 0)

a(4)=(2.5cos2,2.5sin2,0)=(1.04,2.27,0)\vec{a}(4) = (-2.5\cos 2, -2.5\sin 2, 0) = (1.04, -2.27, 0) m/s²

Bu spiral (helix) traektoriya.


Masala 24 ⭐⭐⭐⭐

Robot qo'li ikkita bo'g'inli: L1=0.5L_1 = 0.5 m, L2=0.3L_2 = 0.3 m. Burchaklar: θ1=45°\theta_1 = 45°, θ2=30°\theta_2 = 30°. Burchak tezliklar: θ˙1=2\dot{\theta}_1 = 2 rad/s, θ˙2=1\dot{\theta}_2 = 1 rad/s. End effector tezligini toping.

Yechim

Forward kinematika: x=L1cosθ1+L2cos(θ1+θ2)x = L_1\cos\theta_1 + L_2\cos(\theta_1+\theta_2) y=L1sinθ1+L2sin(θ1+θ2)y = L_1\sin\theta_1 + L_2\sin(\theta_1+\theta_2)

Tezlik (Jacobian usuli): x˙=L1sinθ1θ˙1L2sin(θ1+θ2)(θ˙1+θ˙2)\dot{x} = -L_1\sin\theta_1 \dot{\theta}_1 - L_2\sin(\theta_1+\theta_2)(\dot{\theta}_1+\dot{\theta}_2) y˙=L1cosθ1θ˙1+L2cos(θ1+θ2)(θ˙1+θ˙2)\dot{y} = L_1\cos\theta_1 \dot{\theta}_1 + L_2\cos(\theta_1+\theta_2)(\dot{\theta}_1+\dot{\theta}_2)

θ1=45°\theta_1 = 45°, θ1+θ2=75°\theta_1+\theta_2 = 75°

x˙=0.5(0.707)(2)0.3(0.966)(3)=0.7070.869=1.576\dot{x} = -0.5(0.707)(2) - 0.3(0.966)(3) = -0.707 - 0.869 = -1.576 m/s

y˙=0.5(0.707)(2)+0.3(0.259)(3)=0.707+0.233=0.940\dot{y} = 0.5(0.707)(2) + 0.3(0.259)(3) = 0.707 + 0.233 = 0.940 m/s

v=2.48+0.88=1.83|\vec{v}| = \sqrt{2.48 + 0.88} = 1.83 m/s


Masala 25 ⭐⭐⭐⭐

Raketa ko'tarilishda. Massasi m(t)=100010tm(t) = 1000 - 10t kg (yonilg'i sarfi). Itarish kuchi F=15000F = 15000 N. t=50t = 50 s dagi tezlanish? (Havo qarshiligini hisobga olmang)

Yechim

t=50t = 50 s da massa: m=1000500=500m = 1000 - 500 = 500 kg

Natijaviy kuch: Fnet=Fmg=15000500(10)=10000F_{net} = F - mg = 15000 - 500(10) = 10000 N

Tezlanish: a=Fnetm=10000500=20a = \frac{F_{net}}{m} = \frac{10000}{500} = 20 m/s²


Masala 26 ⭐⭐⭐⭐

Dron A va B bir-birini kuzatmoqda. A(100, 50, 30) m da turgan, B (0, 0, 20) dan (10, 5, 0.5) m/s tezlik bilan harakatlanmoqda. Qancha vaqtdan keyin eng yaqin bo'ladi?

Yechim

B pozitsiyasi: rB(t)=(10t,5t,20+0.5t)\vec{r}_B(t) = (10t, 5t, 20 + 0.5t)

Orasidagi vektor: d(t)=rArB=(10010t,505t,100.5t)\vec{d}(t) = \vec{r}_A - \vec{r}_B = (100-10t, 50-5t, 10-0.5t)

Masofa kvadrati: d2=(10010t)2+(505t)2+(100.5t)2d^2 = (100-10t)^2 + (50-5t)^2 + (10-0.5t)^2

Minimum uchun: d(d2)dt=0\frac{d(d^2)}{dt} = 0

20(10010t)10(505t)1(100.5t)=0-20(100-10t) - 10(50-5t) - 1(10-0.5t) = 0

2000+200t500+50t10+0.5t=0-2000 + 200t - 500 + 50t - 10 + 0.5t = 0

250.5t=2510250.5t = 2510

t=10.02t = 10.02 s


Masala 27 ⭐⭐⭐⭐

G'ildirak radiusi 0.2 m, burchak tezlanish α=5\alpha = 5 rad/s². Boshlang'ich ω0=10\omega_0 = 10 rad/s. 4 s da g'ildirak chetidagi nuqtaning umumiy tezlanishi?

Yechim

t=4t = 4 s da burchak tezlik: ω=ω0+αt=10+5(4)=30\omega = \omega_0 + \alpha t = 10 + 5(4) = 30 rad/s

Tangensial tezlanish: at=rα=0.2×5=1a_t = r\alpha = 0.2 \times 5 = 1 m/s²

Normal tezlanish: an=rω2=0.2×900=180a_n = r\omega^2 = 0.2 \times 900 = 180 m/s²

Umumiy: a=12+1802=32401180|a| = \sqrt{1^2 + 180^2} = \sqrt{32401} \approx 180 m/s²


Masala 28 ⭐⭐⭐⭐

Yo'ldosh Yer atrofida 400 km balandlikda aylanmoqda. Orbital tezlik va davr? (Yer radiusi 6371 km, g=9.81g = 9.81 m/s²)

Yechim

Orbital radius: r=6371+400=6771r = 6371 + 400 = 6771 km = 6.771×1066.771 \times 10^6 m

Orbital tezlik: v=gR2r=9.81×(6.371×106)26.771×106v = \sqrt{\frac{gR^2}{r}} = \sqrt{\frac{9.81 \times (6.371 \times 10^6)^2}{6.771 \times 10^6}}

v=398.2×10126.771×106=58.8×106=7670v = \sqrt{\frac{398.2 \times 10^{12}}{6.771 \times 10^6}} = \sqrt{58.8 \times 10^6} = 7670 m/s ≈ 7.67 km/s

Davr: T=2πrv=2π×6.771×1067670=5545T = \frac{2\pi r}{v} = \frac{2\pi \times 6.771 \times 10^6}{7670} = 5545 s ≈ 92.4 min


Masala 29 ⭐⭐⭐⭐

Avtomobil egri yo'lda harakat qilmoqda. Egrilik radiusi r=100r = 100 m. Tezlik v=20v = 20 m/s va 2 m/s² bilan ortmoqda. Rul burchagi qanday o'zgarmoqda?

Yechim

Tangensial: at=2a_t = 2 m/s² (tezlik ortishi)

Normal: an=v2r=400100=4a_n = \frac{v^2}{r} = \frac{400}{100} = 4 m/s²

Umumiy tezlanish burchagi (yo'nalishdan): ϕ=arctan(anat)=arctan(42)=arctan(2)=63.4°\phi = \arctan(\frac{a_n}{a_t}) = \arctan(\frac{4}{2}) = \arctan(2) = 63.4°


Masala 30 ⭐⭐⭐⭐

Quadcopter GPS'dan pozitsiya olmoqda (10 Hz). Ikki o'lchov: P1=(100.0,50.0,30.0)P_1 = (100.0, 50.0, 30.0) m, P2=(100.5,50.2,30.1)P_2 = (100.5, 50.2, 30.1) m. Tezlik va yo'nalish (heading)?

Yechim

Δt=0.1\Delta t = 0.1 s (10 Hz)

Δr=(0.5,0.2,0.1)\Delta \vec{r} = (0.5, 0.2, 0.1) m

Tezlik: v=ΔrΔt=(5,2,1)\vec{v} = \frac{\Delta \vec{r}}{\Delta t} = (5, 2, 1) m/s

v=25+4+1=305.48|\vec{v}| = \sqrt{25 + 4 + 1} = \sqrt{30} \approx 5.48 m/s

Heading (gorizontal): ψ=arctan(vyvx)=arctan(25)=21.8°\psi = \arctan(\frac{v_y}{v_x}) = \arctan(\frac{2}{5}) = 21.8° (shimoli-sharqqa)

Elevatsiya: θ=arctan(vzvx2+vy2)=arctan(129)=10.5°\theta = \arctan(\frac{v_z}{\sqrt{v_x^2+v_y^2}}) = \arctan(\frac{1}{\sqrt{29}}) = 10.5°


✅ Tekshirish Ro'yxati

  • 1-10: Asosiy kinematika
  • 11-22: O'rtacha masalalar
  • 23-30: Murakkab masalalar

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