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 = v t = 20 × 5 = 100 x = vt = 20 \times 5 = 100 x = v t = 20 × 5 = 100 m
Masala 2 ⭐⭐
Velosiped 0 dan 15 m/s gacha 10 s da tezlashdi. Tezlanishni toping.
Yechim a = v − v 0 t = 15 − 0 10 = 1.5 a = \frac{v - v_0}{t} = \frac{15 - 0}{10} = 1.5 a = t v − v 0 = 10 15 − 0 = 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 = v 0 + a t = 0 + 2 ( 6 ) = 12 v = v_0 + at = 0 + 2(6) = 12 v = v 0 + a t = 0 + 2 ( 6 ) = 12 m/s
x = v 0 t + 1 2 a t 2 = 0 + 1 2 ( 2 ) ( 36 ) = 36 x = v_0 t + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(36) = 36 x = v 0 t + 2 1 a t 2 = 0 + 2 1 ( 2 ) ( 36 ) = 36 m
Masala 4 ⭐⭐
Tosh 50 m balandlikdan erkin tushirildi. Yerga tushguncha qancha vaqt o'tadi? (g = 10 g = 10 g = 10 m/s²)
Yechim h = 1 2 g t 2 h = \frac{1}{2}gt^2 h = 2 1 g t 2
50 = 1 2 ( 10 ) t 2 50 = \frac{1}{2}(10)t^2 50 = 2 1 ( 10 ) t 2
t 2 = 10 t^2 = 10 t 2 = 10
t = 10 ≈ 3.16 t = \sqrt{10} \approx 3.16 t = 10 ≈ 3.16 s
Masala 5 ⭐⭐
Avtomobil 30 m/s tezlikda ketayotib, 4 m/s² tezlanish bilan tormozladi. To'xtaguncha qancha masofa bosadi?
Yechim v 2 = v 0 2 + 2 a x v^2 = v_0^2 + 2ax v 2 = v 0 2 + 2 a x
0 = 900 + 2 ( − 4 ) x 0 = 900 + 2(-4)x 0 = 900 + 2 ( − 4 ) x
x = 900 8 = 112.5 x = \frac{900}{8} = 112.5 x = 8 900 = 112.5 m
Masala 6 ⭐⭐
To'p 45° burchak ostida 20 m/s tezlik bilan otildi. Maksimal balandlik? (g = 10 g = 10 g = 10 m/s²)
Yechim h m a x = v 0 2 sin 2 θ 2 g = 400 × 0.5 20 = 10 h_{max} = \frac{v_0^2 \sin^2\theta}{2g} = \frac{400 \times 0.5}{20} = 10 h ma x = 2 g v 0 2 s i n 2 θ = 20 400 × 0.5 = 10 m
Masala 7 ⭐⭐
Disk 100 rpm tezlikda aylanmoqda. Burchak tezlikni rad/s da toping.
Yechim ω = 2 π × 100 60 = 200 π 60 = 10 π 3 ≈ 10.47 \omega = \frac{2\pi \times 100}{60} = \frac{200\pi}{60} = \frac{10\pi}{3} \approx 10.47 ω = 60 2 π × 100 = 60 200 π = 3 10 π ≈ 10.47 rad/s
Masala 8 ⭐⭐
G'ildirak radiusi 0.3 m, burchak tezligi 20 rad/s. Chiziqli tezlik?
Yechim v = r ω = 0.3 × 20 = 6 v = r\omega = 0.3 \times 20 = 6 v = r ω = 0.3 × 20 = 6 m/s
Masala 9 ⭐⭐
Samolyot 200 m/s tezlik bilan 500 m radiusli aylana bo'ylab uchmoqda. Markazga intilma tezlanish?
Yechim a n = v 2 r = 40000 500 = 80 a_n = \frac{v^2}{r} = \frac{40000}{500} = 80 a n = r v 2 = 500 40000 = 80 m/s²
Masala 10 ⭐⭐
Proektil 30° burchak ostida 50 m/s bilan otildi. Gorizontal va vertikal tezlik komponentlari?
Yechim v x = v 0 cos 30 ° = 50 × 0.866 = 43.3 v_x = v_0 \cos 30° = 50 \times 0.866 = 43.3 v x = v 0 cos 30° = 50 × 0.866 = 43.3 m/s
v y = v 0 sin 30 ° = 50 × 0.5 = 25 v_y = v_0 \sin 30° = 50 \times 0.5 = 25 v y = v 0 sin 30° = 50 × 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 Δ r ⃗ = ( 100 − 0 , 50 − 0 , 30 − 10 ) = ( 100 , 50 , 20 ) \Delta \vec{r} = (100-0, 50-0, 30-10) = (100, 50, 20) Δ r = ( 100 − 0 , 50 − 0 , 30 − 10 ) = ( 100 , 50 , 20 ) m
v ⃗ a v g = Δ r ⃗ Δ t = ( 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) v a v g = Δ t Δ r = 20 ( 100 , 50 , 20 ) = ( 5 , 2.5 , 1 ) m/s
∣ v ⃗ a v g ∣ = 25 + 6.25 + 1 = 32.25 ≈ 5.68 |\vec{v}_{avg}| = \sqrt{25 + 6.25 + 1} = \sqrt{32.25} \approx 5.68 ∣ v a v g ∣ = 25 + 6.25 + 1 = 32.25 ≈ 5.68 m/s
Masala 12 ⭐⭐⭐
Robot g'ildiragi 0.1 m radiusli, motor 500 rpm beradi. Robotning chiziqli tezligi?
Yechim ω = 2 π × 500 60 = 52.36 \omega = \frac{2\pi \times 500}{60} = 52.36 ω = 60 2 π × 500 = 52.36 rad/s
v = r ω = 0.1 × 52.36 = 5.24 v = r\omega = 0.1 \times 52.36 = 5.24 v = r ω = 0.1 × 52.36 = 5.24 m/s
Masala 13 ⭐⭐⭐
To'p 60° burchak ostida 40 m/s bilan otildi. Uchish masofasi va vaqti? (g = 10 g = 10 g = 10 m/s²)
Yechim R = v 0 2 sin 2 θ g = 1600 × sin 120 ° 10 = 1600 × 0.866 10 = 138.6 R = \frac{v_0^2 \sin 2\theta}{g} = \frac{1600 \times \sin 120°}{10} = \frac{1600 \times 0.866}{10} = 138.6 R = g v 0 2 s i n 2 θ = 10 1600 × s i n 120° = 10 1600 × 0.866 = 138.6 m
T = 2 v 0 sin θ g = 2 × 40 × 0.866 10 = 6.93 T = \frac{2v_0 \sin\theta}{g} = \frac{2 \times 40 \times 0.866}{10} = 6.93 T = g 2 v 0 s i n θ = 10 2 × 40 × 0.866 = 6.93 s
Masala 14 ⭐⭐⭐
Jism r ⃗ ( t ) = ( 3 t 2 , 4 t , 5 ) \vec{r}(t) = (3t^2, 4t, 5) r ( t ) = ( 3 t 2 , 4 t , 5 ) m qonun bo'yicha harakat qilmoqda. t = 2 t = 2 t = 2 s dagi tezlik va tezlanish?
Yechim v ⃗ = d r ⃗ d t = ( 6 t , 4 , 0 ) \vec{v} = \frac{d\vec{r}}{dt} = (6t, 4, 0) v = d t d r = ( 6 t , 4 , 0 )
t = 2 t = 2 t = 2 : v ⃗ ( 2 ) = ( 12 , 4 , 0 ) \vec{v}(2) = (12, 4, 0) v ( 2 ) = ( 12 , 4 , 0 ) m/s
∣ v ⃗ ∣ = 144 + 16 = 160 ≈ 12.65 |\vec{v}| = \sqrt{144 + 16} = \sqrt{160} \approx 12.65 ∣ v ∣ = 144 + 16 = 160 ≈ 12.65 m/s
a ⃗ = d v ⃗ d t = ( 6 , 0 , 0 ) \vec{a} = \frac{d\vec{v}}{dt} = (6, 0, 0) a = d t d v = ( 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 = 60 2 + 80 2 = 3600 + 6400 = 10000 = 100 d = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 d = 6 0 2 + 8 0 2 = 3600 + 6400 = 10000 = 100 km
Masala 16 ⭐⭐⭐
Raketa vertikal ko'tarilmoqda. Tezlanish a = 20 a = 20 a = 20 m/s². 10 s da qanday balandlikka ko'tariladi va tezligi qancha?
Yechim h = 1 2 a t 2 = 1 2 ( 20 ) ( 100 ) = 1000 h = \frac{1}{2}at^2 = \frac{1}{2}(20)(100) = 1000 h = 2 1 a t 2 = 2 1 ( 20 ) ( 100 ) = 1000 m
v = a t = 20 × 10 = 200 v = at = 20 \times 10 = 200 v = a t = 20 × 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 = v R + v L 2 = 1.5 + 1 2 = 1.25 v = \frac{v_R + v_L}{2} = \frac{1.5 + 1}{2} = 1.25 v = 2 v R + v L = 2 1.5 + 1 = 1.25 m/s
ω = v R − v L L = 1.5 − 1 0.5 = 1 \omega = \frac{v_R - v_L}{L} = \frac{1.5 - 1}{0.5} = 1 ω = L v R − v L = 0.5 1.5 − 1 = 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 ω 0 = 0 \omega_0 = 0 ω 0 = 0 , ω = 2 π × 10000 60 = 1047.2 \omega = \frac{2\pi \times 10000}{60} = 1047.2 ω = 60 2 π × 10000 = 1047.2 rad/s
α = ω − ω 0 t = 1047.2 2 = 523.6 \alpha = \frac{\omega - \omega_0}{t} = \frac{1047.2}{2} = 523.6 α = t ω − ω 0 = 2 1047.2 = 523.6 rad/s²
θ = ω 0 t + 1 2 α t 2 = 0 + 1 2 ( 523.6 ) ( 4 ) = 1047.2 \theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(523.6)(4) = 1047.2 θ = ω 0 t + 2 1 α t 2 = 0 + 2 1 ( 523.6 ) ( 4 ) = 1047.2 rad
Aylanishlar: n = θ 2 π = 1047.2 2 π = 166.7 n = \frac{\theta}{2\pi} = \frac{1047.2}{2\pi} = 166.7 n = 2 π θ = 2 π 1047.2 = 166.7 aylanish
Masala 19 ⭐⭐⭐
Jism 5 m radiusli aylana bo'ylab harakat qilmoqda. Tezligi v = 2 t v = 2t v = 2 t m/s. t = 3 t = 3 t = 3 s da tangensial va normal tezlanish?
Yechim t = 3 t = 3 t = 3 s da: v = 6 v = 6 v = 6 m/s
Tangensial: a t = d v d t = 2 a_t = \frac{dv}{dt} = 2 a t = d t d v = 2 m/s²
Normal: a n = v 2 r = 36 5 = 7.2 a_n = \frac{v^2}{r} = \frac{36}{5} = 7.2 a n = r v 2 = 5 36 = 7.2 m/s²
Umumiy: ∣ a ∣ = 4 + 51.84 = 55.84 ≈ 7.47 |a| = \sqrt{4 + 51.84} = \sqrt{55.84} \approx 7.47 ∣ a ∣ = 4 + 51.84 = 55.84 ≈ 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 v ⃗ A = ( 0 , 30 ) \vec{v}_A = (0, 30) v A = ( 0 , 30 ) km/h (shimolga)
v ⃗ B = ( 40 , 0 ) \vec{v}_B = (40, 0) v B = ( 40 , 0 ) km/h (sharqqa)
v ⃗ A / B = v ⃗ A − v ⃗ B = ( − 40 , 30 ) \vec{v}_{A/B} = \vec{v}_A - \vec{v}_B = (-40, 30) v A / B = v A − v B = ( − 40 , 30 ) km/h
∣ v ⃗ A / B ∣ = 1600 + 900 = 50 |\vec{v}_{A/B}| = \sqrt{1600 + 900} = 50 ∣ v A / B ∣ = 1600 + 900 = 50 km/h
Yo'nalish: θ = arctan ( 30 − 40 ) = 143.1 ° \theta = \arctan(\frac{30}{-40}) = 143.1° θ = arctan ( − 40 30 ) = 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:
R 0 = v 0 2 sin 90 ° g = 10000 10 = 1000 R_0 = \frac{v_0^2 \sin 90°}{g} = \frac{10000}{10} = 1000 R 0 = g v 0 2 s i n 90° = 10 10000 = 1000 m
Shamol bilan (gorizontal tezlik kamayadi):
v x = v 0 cos 45 ° − 10 = 70.7 − 10 = 60.7 v_x = v_0 \cos 45° - 10 = 70.7 - 10 = 60.7 v x = v 0 cos 45° − 10 = 70.7 − 10 = 60.7 m/s
Uchish vaqti (vertikal komponent o'zgarmaydi):
T = 2 v 0 sin 45 ° g = 2 × 70.7 10 = 14.14 T = \frac{2 v_0 \sin 45°}{g} = \frac{2 \times 70.7}{10} = 14.14 T = g 2 v 0 s i n 45° = 10 2 × 70.7 = 14.14 s
R = v x × T = 60.7 × 14.14 = 858.3 R = v_x \times T = 60.7 \times 14.14 = 858.3 R = v x × T = 60.7 × 14.14 = 858.3 m
Farq: 1000 − 858.3 = 141.7 1000 - 858.3 = 141.7 1000 − 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: g e f f = g + a = 10 + 3 = 13 g_{eff} = g + a = 10 + 3 = 13 g e f f = g + a = 10 + 3 = 13 m/s²
Sezilgan og'irlik: W = m g e f f = 70 × 13 = 910 W = mg_{eff} = 70 \times 13 = 910 W = m g e f f = 70 × 13 = 910 N
Oddiy og'irlik: W 0 = 70 × 10 = 700 W_0 = 70 \times 10 = 700 W 0 = 70 × 10 = 700 N
30% ko'proq og'irlik seziladi.
Murakkab Masalalar (23-30)
Masala 23 ⭐⭐⭐⭐
Dron r ⃗ ( t ) = ( 10 cos ( 0.5 t ) , 10 sin ( 0.5 t ) , 2 t ) \vec{r}(t) = (10\cos(0.5t), 10\sin(0.5t), 2t) r ( t ) = ( 10 cos ( 0.5 t ) , 10 sin ( 0.5 t ) , 2 t ) m traektoriya bo'ylab uchmoqda. t = 4 t = 4 t = 4 s da tezlik va tezlanish vektorlarini toping.
Yechim v ⃗ = d r ⃗ d t = ( − 5 sin ( 0.5 t ) , 5 cos ( 0.5 t ) , 2 ) \vec{v} = \frac{d\vec{r}}{dt} = (-5\sin(0.5t), 5\cos(0.5t), 2) v = d t d r = ( − 5 sin ( 0.5 t ) , 5 cos ( 0.5 t ) , 2 )
t = 4 t = 4 t = 4 : 0.5 × 4 = 2 0.5 \times 4 = 2 0.5 × 4 = 2 rad
v ⃗ ( 4 ) = ( − 5 sin 2 , 5 cos 2 , 2 ) = ( − 4.55 , − 2.08 , 2 ) \vec{v}(4) = (-5\sin 2, 5\cos 2, 2) = (-4.55, -2.08, 2) 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 ∣ v ∣ = 20.7 + 4.33 + 4 = 5.39 m/s
a ⃗ = d v ⃗ d t = ( − 2.5 cos ( 0.5 t ) , − 2.5 sin ( 0.5 t ) , 0 ) \vec{a} = \frac{d\vec{v}}{dt} = (-2.5\cos(0.5t), -2.5\sin(0.5t), 0) a = d t d v = ( − 2.5 cos ( 0.5 t ) , − 2.5 sin ( 0.5 t ) , 0 )
a ⃗ ( 4 ) = ( − 2.5 cos 2 , − 2.5 sin 2 , 0 ) = ( 1.04 , − 2.27 , 0 ) \vec{a}(4) = (-2.5\cos 2, -2.5\sin 2, 0) = (1.04, -2.27, 0) 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: L 1 = 0.5 L_1 = 0.5 L 1 = 0.5 m, L 2 = 0.3 L_2 = 0.3 L 2 = 0.3 m. Burchaklar: θ 1 = 45 ° \theta_1 = 45° θ 1 = 45° , θ 2 = 30 ° \theta_2 = 30° θ 2 = 30° . Burchak tezliklar: θ ˙ 1 = 2 \dot{\theta}_1 = 2 θ ˙ 1 = 2 rad/s, θ ˙ 2 = 1 \dot{\theta}_2 = 1 θ ˙ 2 = 1 rad/s. End effector tezligini toping.
Yechim Forward kinematika:
x = L 1 cos θ 1 + L 2 cos ( θ 1 + θ 2 ) x = L_1\cos\theta_1 + L_2\cos(\theta_1+\theta_2) x = L 1 cos θ 1 + L 2 cos ( θ 1 + θ 2 )
y = L 1 sin θ 1 + L 2 sin ( θ 1 + θ 2 ) y = L_1\sin\theta_1 + L_2\sin(\theta_1+\theta_2) y = L 1 sin θ 1 + L 2 sin ( θ 1 + θ 2 )
Tezlik (Jacobian usuli):
x ˙ = − L 1 sin θ 1 θ ˙ 1 − L 2 sin ( θ 1 + θ 2 ) ( 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) x ˙ = − L 1 sin θ 1 θ ˙ 1 − L 2 sin ( θ 1 + θ 2 ) ( θ ˙ 1 + θ ˙ 2 )
y ˙ = L 1 cos θ 1 θ ˙ 1 + L 2 cos ( θ 1 + θ 2 ) ( 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) y ˙ = L 1 cos θ 1 θ ˙ 1 + L 2 cos ( θ 1 + θ 2 ) ( θ ˙ 1 + θ ˙ 2 )
θ 1 = 45 ° \theta_1 = 45° θ 1 = 45° , θ 1 + θ 2 = 75 ° \theta_1+\theta_2 = 75° θ 1 + θ 2 = 75°
x ˙ = − 0.5 ( 0.707 ) ( 2 ) − 0.3 ( 0.966 ) ( 3 ) = − 0.707 − 0.869 = − 1.576 \dot{x} = -0.5(0.707)(2) - 0.3(0.966)(3) = -0.707 - 0.869 = -1.576 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 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 ∣ v ∣ = 2.48 + 0.88 = 1.83 m/s
Masala 25 ⭐⭐⭐⭐
Raketa ko'tarilishda. Massasi m ( t ) = 1000 − 10 t m(t) = 1000 - 10t m ( t ) = 1000 − 10 t kg (yonilg'i sarfi). Itarish kuchi F = 15000 F = 15000 F = 15000 N. t = 50 t = 50 t = 50 s dagi tezlanish? (Havo qarshiligini hisobga olmang)
Yechim t = 50 t = 50 t = 50 s da massa: m = 1000 − 500 = 500 m = 1000 - 500 = 500 m = 1000 − 500 = 500 kg
Natijaviy kuch: F n e t = F − m g = 15000 − 500 ( 10 ) = 10000 F_{net} = F - mg = 15000 - 500(10) = 10000 F n e t = F − m g = 15000 − 500 ( 10 ) = 10000 N
Tezlanish: a = F n e t m = 10000 500 = 20 a = \frac{F_{net}}{m} = \frac{10000}{500} = 20 a = m F n e t = 500 10000 = 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: r ⃗ B ( t ) = ( 10 t , 5 t , 20 + 0.5 t ) \vec{r}_B(t) = (10t, 5t, 20 + 0.5t) r B ( t ) = ( 10 t , 5 t , 20 + 0.5 t )
Orasidagi vektor:
d ⃗ ( t ) = r ⃗ A − r ⃗ B = ( 100 − 10 t , 50 − 5 t , 10 − 0.5 t ) \vec{d}(t) = \vec{r}_A - \vec{r}_B = (100-10t, 50-5t, 10-0.5t) d ( t ) = r A − r B = ( 100 − 10 t , 50 − 5 t , 10 − 0.5 t )
Masofa kvadrati:
d 2 = ( 100 − 10 t ) 2 + ( 50 − 5 t ) 2 + ( 10 − 0.5 t ) 2 d^2 = (100-10t)^2 + (50-5t)^2 + (10-0.5t)^2 d 2 = ( 100 − 10 t ) 2 + ( 50 − 5 t ) 2 + ( 10 − 0.5 t ) 2
Minimum uchun: d ( d 2 ) d t = 0 \frac{d(d^2)}{dt} = 0 d t d ( d 2 ) = 0
− 20 ( 100 − 10 t ) − 10 ( 50 − 5 t ) − 1 ( 10 − 0.5 t ) = 0 -20(100-10t) - 10(50-5t) - 1(10-0.5t) = 0 − 20 ( 100 − 10 t ) − 10 ( 50 − 5 t ) − 1 ( 10 − 0.5 t ) = 0
− 2000 + 200 t − 500 + 50 t − 10 + 0.5 t = 0 -2000 + 200t - 500 + 50t - 10 + 0.5t = 0 − 2000 + 200 t − 500 + 50 t − 10 + 0.5 t = 0
250.5 t = 2510 250.5t = 2510 250.5 t = 2510
t = 10.02 t = 10.02 t = 10.02 s
Masala 27 ⭐⭐⭐⭐
G'ildirak radiusi 0.2 m, burchak tezlanish α = 5 \alpha = 5 α = 5 rad/s². Boshlang'ich ω 0 = 10 \omega_0 = 10 ω 0 = 10 rad/s. 4 s da g'ildirak chetidagi nuqtaning umumiy tezlanishi?
Yechim t = 4 t = 4 t = 4 s da burchak tezlik:
ω = ω 0 + α t = 10 + 5 ( 4 ) = 30 \omega = \omega_0 + \alpha t = 10 + 5(4) = 30 ω = ω 0 + α t = 10 + 5 ( 4 ) = 30 rad/s
Tangensial tezlanish:
a t = r α = 0.2 × 5 = 1 a_t = r\alpha = 0.2 \times 5 = 1 a t = r α = 0.2 × 5 = 1 m/s²
Normal tezlanish:
a n = r ω 2 = 0.2 × 900 = 180 a_n = r\omega^2 = 0.2 \times 900 = 180 a n = r ω 2 = 0.2 × 900 = 180 m/s²
Umumiy:
∣ a ∣ = 1 2 + 180 2 = 32401 ≈ 180 |a| = \sqrt{1^2 + 180^2} = \sqrt{32401} \approx 180 ∣ a ∣ = 1 2 + 18 0 2 = 32401 ≈ 180 m/s²
Masala 28 ⭐⭐⭐⭐
Yo'ldosh Yer atrofida 400 km balandlikda aylanmoqda. Orbital tezlik va davr? (Yer radiusi 6371 km, g = 9.81 g = 9.81 g = 9.81 m/s²)
Yechim Orbital radius: r = 6371 + 400 = 6771 r = 6371 + 400 = 6771 r = 6371 + 400 = 6771 km = 6.771 × 10 6 6.771 \times 10^6 6.771 × 1 0 6 m
Orbital tezlik: v = g R 2 r = 9.81 × ( 6.371 × 10 6 ) 2 6.771 × 10 6 v = \sqrt{\frac{gR^2}{r}} = \sqrt{\frac{9.81 \times (6.371 \times 10^6)^2}{6.771 \times 10^6}} v = r g R 2 = 6.771 × 1 0 6 9.81 × ( 6.371 × 1 0 6 ) 2
v = 398.2 × 10 12 6.771 × 10 6 = 58.8 × 10 6 = 7670 v = \sqrt{\frac{398.2 \times 10^{12}}{6.771 \times 10^6}} = \sqrt{58.8 \times 10^6} = 7670 v = 6.771 × 1 0 6 398.2 × 1 0 12 = 58.8 × 1 0 6 = 7670 m/s ≈ 7.67 km/s
Davr: T = 2 π r v = 2 π × 6.771 × 10 6 7670 = 5545 T = \frac{2\pi r}{v} = \frac{2\pi \times 6.771 \times 10^6}{7670} = 5545 T = v 2 π r = 7670 2 π × 6.771 × 1 0 6 = 5545 s ≈ 92.4 min
Masala 29 ⭐⭐⭐⭐
Avtomobil egri yo'lda harakat qilmoqda. Egrilik radiusi r = 100 r = 100 r = 100 m. Tezlik v = 20 v = 20 v = 20 m/s va 2 m/s² bilan ortmoqda. Rul burchagi qanday o'zgarmoqda?
Yechim Tangensial: a t = 2 a_t = 2 a t = 2 m/s² (tezlik ortishi)
Normal: a n = v 2 r = 400 100 = 4 a_n = \frac{v^2}{r} = \frac{400}{100} = 4 a n = r v 2 = 100 400 = 4 m/s²
Umumiy tezlanish burchagi (yo'nalishdan):
ϕ = arctan ( a n a t ) = arctan ( 4 2 ) = arctan ( 2 ) = 63.4 ° \phi = \arctan(\frac{a_n}{a_t}) = \arctan(\frac{4}{2}) = \arctan(2) = 63.4° ϕ = arctan ( a t a n ) = arctan ( 2 4 ) = arctan ( 2 ) = 63.4°
Masala 30 ⭐⭐⭐⭐
Quadcopter GPS'dan pozitsiya olmoqda (10 Hz). Ikki o'lchov: P 1 = ( 100.0 , 50.0 , 30.0 ) P_1 = (100.0, 50.0, 30.0) P 1 = ( 100.0 , 50.0 , 30.0 ) m, P 2 = ( 100.5 , 50.2 , 30.1 ) P_2 = (100.5, 50.2, 30.1) P 2 = ( 100.5 , 50.2 , 30.1 ) m. Tezlik va yo'nalish (heading)?
Yechim Δ t = 0.1 \Delta t = 0.1 Δ t = 0.1 s (10 Hz)
Δ r ⃗ = ( 0.5 , 0.2 , 0.1 ) \Delta \vec{r} = (0.5, 0.2, 0.1) Δ r = ( 0.5 , 0.2 , 0.1 ) m
Tezlik: v ⃗ = Δ r ⃗ Δ t = ( 5 , 2 , 1 ) \vec{v} = \frac{\Delta \vec{r}}{\Delta t} = (5, 2, 1) v = Δ t Δ r = ( 5 , 2 , 1 ) m/s
∣ v ⃗ ∣ = 25 + 4 + 1 = 30 ≈ 5.48 |\vec{v}| = \sqrt{25 + 4 + 1} = \sqrt{30} \approx 5.48 ∣ v ∣ = 25 + 4 + 1 = 30 ≈ 5.48 m/s
Heading (gorizontal):
ψ = arctan ( v y v x ) = arctan ( 2 5 ) = 21.8 ° \psi = \arctan(\frac{v_y}{v_x}) = \arctan(\frac{2}{5}) = 21.8° ψ = arctan ( v x v y ) = arctan ( 5 2 ) = 21.8° (shimoli-sharqqa)
Elevatsiya:
θ = arctan ( v z v x 2 + v y 2 ) = arctan ( 1 29 ) = 10.5 ° \theta = \arctan(\frac{v_z}{\sqrt{v_x^2+v_y^2}}) = \arctan(\frac{1}{\sqrt{29}}) = 10.5° θ = arctan ( v x 2 + v y 2 v z ) = arctan ( 29 1 ) = 10.5°
✅ Tekshirish Ro'yxati
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