2. Husan A nuqtadan B nuqtaga borish uchun 6 km masofaga \( \frac{2}{3} \) soat yugurdi. Keyin \( 1 \frac{1}{3} \) soat yugurib A ga qaytib keldl. Husan qanday o'rtacha tezlik bilan yugurgan? (3) Adiba soat \( 8: 30 \) da velosipedda \( X \) xiyobondan yo 1 Ig chiqib, \( Y \) xiyobonga soat 10:30 da yetib bordi va to'xtamasdan \( Z \) xiyobonga jo'nab, soat \( 11: 30 \) da yetib bordi. Velosipedda X xiyobondan \( Z \) xiyobonga borishdagi o'rtacha tezligi \( 10 \mathrm{~km} / \mathrm{h} \) bo'lsa, Adiba jami gancha masofani bosib o'tgan?
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Husan A nuqtadan B nuqtaga 6 km masofani \( \frac{2}{3} \) soatda yugurdi. Ularning o'rtacha tezligini hisoblash uchun, tezlik = masofa / vaqt formulasi ishlatiladi. Bu holda, Husan o'rtacha \( \frac{6 \, \text{km}}{\frac{2}{3} \, \text{soat}} = 6 \times \frac{3}{2} = 9 \, \text{km/soat} \) tezlik bilan yugurgan. Adibaning harakatini hisoblaymiz. U \( X \) xiyobondan \( Y \) xiyobongacha 2 soat (8:30 dan 10:30 gacha) ketdi, so'ngra \( Y \) dan \( Z \) xiyobonga 1 soat (10:30 dan 11:30 gacha) ketdi. Agar \( Z \) xiyobonga yo'lning o'rtacha tezligi \( 10 \, \text{km/soat} \) bo'lsa, masofani hisoblash uchun vaqtni tezlikka ko'paytiramiz: \( 10 \, \text{km/soat} \times 1 \, \text{soat} = 10 \, \text{km} \). Shunday qilib, Adiba jami 6 km + 10 km = 16 km masofani boshlab o'tgan.