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desmiyoungmoore
25.06.2019 •
Mathematics
Two automobiles left simultaneously from cities a and b heading towards each other and met in 5 hours. the speed of the automobile that left city a was 10 km/hour less than the speed of the other automobile. if the first automobile had left city a 4 1 2 hours earlier than the other automobile left city b, then the two would have met 150 km away from b. find the distance between a and b.
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Ответ:
450 km
Step-by-step explanation:
Let's say Va is the speed of the car from city A, Ta is the time it spent traveling, and Da is the distance it traveled.
Similarly, Vb is the speed of the car from city B, Tb is the time it spent traveling, and Db is the distance it traveled.
Given:
Va = Vb - 10
Ta₁ = Tb₁ = 5
Ta₂ = Tb₂ + 4.5
Db₂ = 150
Find:
D = Da₁ + Db₁ = Da₂ + Db₂
Distance = rate × time
In the first scenario:
Da₁ = Va Ta₁
Da₁ = (Vb - 10) (5)
Da₁ = 5Vb - 50
Db₁ = Vb Tb₁
Db₁ = Vb (5)
Db₁ = 5Vb
So:
D = Da₁ + Db₁
D = 10Vb - 50
In the second scenario:
Da₂ = Va Ta₂
Da₂ = (Vb - 10) (Tb₂ + 4.5)
Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45
Db₂ = Vb Tb₂
150 = Vb Tb₂
Substituting:
Da₂ = 150 + 4.5Vb - 10Tb₂ - 45
Da₂ = 105 + 4.5Vb - 10Tb₂
Da₂ = 105 + 4.5Vb - 10 (150 / Vb)
Da₂ = 105 + 4.5Vb - (1500 / Vb)
So:
D = Da₂ + Db₂
D = 105 + 4.5Vb - (1500 / Vb) + 150
D = 255 + 4.5Vb - (1500 / Vb)
Setting this equal to the equation we found for D from the first scenario:
10Vb - 50 = 255 + 4.5Vb - (1500 / Vb)
5.5Vb - 305 = -1500 / Vb
5.5Vb² - 305Vb = -1500
5.5Vb² - 305Vb + 1500 = 0
11Vb² - 610Vb + 3000 = 0
(Vb - 50) (11Vb - 60) = 0
Vb = 50, 5.45
Since Vb > 10, Vb = 50 km/hr.
So the distance between the cities is:
D = 10Vb - 50
D = 10(50) - 50
D = 450 km
Ответ: