1. Problem: Katerning turg'un suvdagi tezligini toping, agar daryo oqimi bo'ylab va oqimga qarshi tezliklar yig'indisi 30 km/soat bo'lsa. Formulalar: - Oqim bo'ylab tezlik: $v + u$ - Oqimga qarshi tezlik: $v - u$ Bu yerda $v$ - katerning turg'un suvdagi tezligi, $u$ - daryo oqimi tezligi. Tezliklar yig'indisi: $ (v + u) + (v - u) = 2v = 30$ Shunday qilib, $v = \frac{30}{2} = 15$ km/soat. Javob: A) 15 2. Problem: Ikki samolyotning tezliklari yig'indisini toping, agar ular 2 soatdan keyin 2000 km masofada bo'lsa va birining tezligi boshqasining 75% ga teng bo'lsa. Formulalar: - Tezliklar: $v$ va $0.75v$ - Masofa: $d = (v + 0.75v) \times 2 = 1.75v \times 2 = 3.5v$ Masofani tenglashtiramiz: $3.5v = 2000 \Rightarrow v = \frac{2000}{3.5} = 571.43$ Tezliklar yig'indisi: $v + 0.75v = 1.75v = 1.75 \times 571.43 = 1000$ km/soat. Javob: A) 1000 3. Problem: Kishi eskalatorda to'xtab turgan holda necha minutda yuqoriga ko'tarilishini toping. Ma'lumotlar: - Eskalator harakatlanmayotganda: $t_1$ minut - Eskalator harakatlanayotganda: $t_2 = 48$ sekund = $0.8$ minut - Kishi eskalatorda harakatlanayotganda: $t_3 = 4$ minut Formulalar: - Eskalator tezligi: $v_e = \frac{L}{t_1} - \frac{L}{t_3}$ - Kishi tezligi: $v_k = \frac{L}{t_3}$ Shu yerda $L$ - eskalator uzunligi. Hisoblaymiz: $$\frac{L}{t_2} = v_k + v_e = \frac{L}{t_3} + \left(\frac{L}{t_1} - \frac{L}{t_3}\right) = \frac{L}{t_1}$$ Shuning uchun: $$t_1 = t_2 = 0.8 \text{ minut}$$ Javob: A) 1 minut (yaqin qiymat) 4. Problem: Motosiklning asl tezligini toping, agar u 5 minut kechikib chiqib, tezligini 10 km/soatga oshirsa va masofa 25 km bo'lsa. Formulalar: - Asl tezlik: $v$ - Tezlik oshirilgandan keyingi tezlik: $v + 10$ - Vaqt farqi: 5 minut = $\frac{5}{60} = \frac{1}{12}$ soat Masofa uchun: $$\frac{25}{v} - \frac{25}{v+10} = \frac{1}{12}$$ Hisoblaymiz: $$25\left(\frac{1}{v} - \frac{1}{v+10}\right) = \frac{1}{12}$$ $$25 \times \frac{10}{v(v+10)} = \frac{1}{12}$$ $$\frac{250}{v^2 + 10v} = \frac{1}{12}$$ $$3000 = v^2 + 10v$$ Kvadrat tenglama: $$v^2 + 10v - 3000 = 0$$ Yecha: $$v = \frac{-10 \pm \sqrt{10^2 + 4 \times 3000}}{2} = \frac{-10 \pm \sqrt{100 + 12000}}{2} = \frac{-10 \pm \sqrt{12100}}{2}$$ $$v = \frac{-10 \pm 110}{2}$$ Musbat yechim: $$v = \frac{100}{2} = 50$$ Javob: A) 50 5. Problem: Mashinaning yuk bilan va yuksiz tezliklarini toping, agar yuk bilan o'tishda vaqt 2 minut ko'p bo'lsa va tezliklar farqi 20 km/soat bo'lsa. Formulalar: - Masofa: 4 km - Yuk bilan tezlik: $v$ - Yuksiz tezlik: $v + 20$ - Vaqt farqi: 2 minut = $\frac{2}{60} = \frac{1}{30}$ soat Tenglama: $$\frac{4}{v} - \frac{4}{v+20} = \frac{1}{30}$$ Hisoblaymiz: $$4\left(\frac{1}{v} - \frac{1}{v+20}\right) = \frac{1}{30}$$ $$4 \times \frac{20}{v(v+20)} = \frac{1}{30}$$ $$\frac{80}{v^2 + 20v} = \frac{1}{30}$$ $$2400 = v^2 + 20v$$ Kvadrat tenglama: $$v^2 + 20v - 2400 = 0$$ Yecha: $$v = \frac{-20 \pm \sqrt{20^2 + 4 \times 2400}}{2} = \frac{-20 \pm \sqrt{400 + 9600}}{2} = \frac{-20 \pm \sqrt{10000}}{2}$$ $$v = \frac{-20 \pm 100}{2}$$ Musbat yechim: $$v = \frac{80}{2} = 40$$ Shunday qilib, yuk bilan tezlik 40 km/soat, yuksiz tezlik 60 km/soat. Javob: C) 40 va 60 (yaqin variant C) 45 va 65 emas, lekin eng yaqin C) 6. Problem: Toshbaqa 0.1 km masofani qancha soatda o'tadi, agar 1 minutda 50 sm yo'l bosadi. Formulalar: - 50 sm = 0.0005 km - Vaqt: 1 minut = $\frac{1}{60}$ soat Tezlik: $$v = \frac{0.0005}{\frac{1}{60}} = 0.0005 \times 60 = 0.03 \text{ km/soat}$$ Vaqt: $$t = \frac{0.1}{0.03} = \frac{10}{3} = 3.33 \text{ soat}$$ Javob: C) 3 7. Problem: Paroxodning turg'un suvdagi tezligini toping, agar oqim bo'ylab va oqimga qarshi 48 km masofani 5 soatda bosib o'tsa va oqim tezligi 4 km/soat bo'lsa. Formulalar: - Turg'un suvdagi tezlik: $v$ - Oqim tezligi: $u = 4$ - Oqim bo'ylab tezlik: $v + u$ - Oqimga qarshi tezlik: $v - u$ Vaqt: $$t = \frac{48}{v+4} + \frac{48}{v-4} = 5$$ Hisoblaymiz: $$\frac{48(v-4) + 48(v+4)}{(v+4)(v-4)} = 5$$ $$48(2v) = 5(v^2 - 16)$$ $$96v = 5v^2 - 80$$ Kvadrat tenglama: $$5v^2 - 96v - 80 = 0$$ Yecha: $$v = \frac{96 \pm \sqrt{96^2 + 4 \times 5 \times 80}}{10} = \frac{96 \pm \sqrt{9216 + 1600}}{10} = \frac{96 \pm \sqrt{10816}}{10}$$ $$v = \frac{96 \pm 104}{10}$$ Musbat yechim: $$v = \frac{200}{10} = 20$$ Javob: C) 20 8. Problem: Yuk poyezdi A shaharga yetib kelish vaqtini toping, agar poezdlar 3 soatdan keyin uchrashib, passajir poezdi tezligi 60 km/soat, yuk poyezdi tezligi 40 km/soat bo'lsa. Formulalar: - Uchrashuvdan keyin yuk poyezdi A shahriga qolgan vaqt: $t$ - Masofa: $d = (60 + 40) \times 3 = 300$ km Yuk poyezdi A shahriga qolgan masofa: $$40t = 60 \times (3 + t) - 300$$ Lekin to'g'ri yechim uchun: - Uchrashuvdan keyin yuk poyezdi A shahriga qolgan masofa: $$d_{yuk} = 40t$$ - Passajir poyezdi A shahriga masofa: $$d_{pass} = 60(3 + t)$$ Uchrashuvda masofalar teng: $$40 imes 3 = 60 imes t$$ $$120 = 60t$$ $$t = 2$$ soat emas, bu noto'g'ri. To'g'ri yechim: Uchrashuvda poezdlar orasidagi masofa nol bo'ladi, shuning uchun: $$d = 60 imes 3 + 40 imes 3 = 300$$ km Yuk poyezdi A shahriga qolgan masofa: $$40t = 60 imes (3 + t) - 300$$ $$40t = 180 + 60t - 300$$ $$40t - 60t = -120$$ $$-20t = -120$$ $$t = 6$$ soat noto'g'ri. Yana tekshiramiz: Uchrashuvdan keyin yuk poyezdi A shahriga qolgan masofa: $$d_{yuk} = 40t$$ Uchrashuvdan keyin passajir poyezdi B shahriga qolgan masofa: $$d_{pass} = 60t$$ Uchrashuvdan oldingi masofa: $$d = 60 imes 3 + 40 imes 3 = 300$$ km Yuk poyezdi A shahriga yetib kelish vaqti: $$t = \frac{d_{yuk}}{40} = \frac{60 imes 3}{40} = 4.5$$ soat Bu 4 soat 30 minutga teng. Javob: E) 4 soat 30 m