Two aircraft

Two planes fly to the airport. At some point, the first airplane is away from the airport 98 km and the second 138 km. The first aircraft flies at an average speed of 420 km/h, the second average speed is 360 km/h, while the tracks of both planes are perpendicular to one another. What is the distance of the aircraft in 9 minutes?

Result

x =  91 km

Solution:

t=9/60=320=0.15 h a=98−420⋅ t=98−420⋅ 0.15=35 km b=138−360⋅ t=138−360⋅ 0.15=84 km x=a2+b2=352+842=91 kmt=9/60=\dfrac{ 3 }{ 20 }=0.15 \ \text{h} \ \\ a=98-420 \cdot \ t=98-420 \cdot \ 0.15=35 \ \text{km} \ \\ b=138-360 \cdot \ t=138-360 \cdot \ 0.15=84 \ \text{km} \ \\ x=\sqrt{ a^{ 2 }+b^{ 2 } }=\sqrt{ 35^{ 2 }+84^{ 2 } }=91 \ \text{km}

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