Between two

There are regular flights between the two airports 690 km apart. A plane leaves the first airport at 6:30 with an average speed of 60 km/h greater than the second plane, which leaves the second airport at 7:00. The planes always pass each other at 9:00. What is the speed of both planes? How far from the first airport do they always meet?

Correct answer:

v₁ =  180 km/h
v₂ =  120 km/h
s₁ =  450 km

Step-by-step explanation:

$s = 690 \ \text{km} \ \\ \ \\ v_{1} \cdot \ (9.00-6.5) + v_{2} \cdot \ (9.00-7.00) = s \ \\ v_{1} = 60+v_{2} \ \\ v_{1} \cdot \ (9.00-6.5) + v_{2} \cdot \ (9.00-7.00) = 690 \ \\ v_{1} = 60+v_{2} \ \\ \ \\ 2.5v_{1}+2v_{2} = 690 \ \\ v_{1}-v_{2} = 60 \ \\ \ \\ Row 2 - \dfrac{ 1 }{ 2.5 } \cdot \ Row 1 → Row 2 \ \\ 2.5v_{1}+2v_{2} = 690 \ \\ -1.8v_{2} = -216 \ \\ \ \\ v_{2} = \dfrac{ -216 }{ -1.8 } = 120 \ \\ v_{1} = \dfrac{ 690-2v_{2} }{ 2.5 } = \dfrac{ 690-2 \cdot \ 120 }{ 2.5 } = 180 \ \\ \ \\ v_{1} = 180 \ \\ v_{2} = 120$

$s_1=v_1\cdot (9.00-6.5)=180\cdot (9.00-6.5)=450\text{km}$

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