Airports 4605

Regular planes fly between the two airports 690 km away. The aircraft takes off from the first airport at 6:30 am at an average speed of 60 km/h faster than the aircraft taking off at 7.00 am from the second airport. The planes always meet at 9.00 am. At what distance from the first airport?

Correct answer:

s1 =  450 km

Step-by-step explanation:

s=690 t1=9.006.5=2.5 t2=9.007.00=2 d=60 s1+s2=s v1 t1 + v2 t2 = s (v2+60) t1+v2 t2 = s v2=(s60 t1)/(t1+t2)=(69060 2.5)/(2.5+2)=120 v1=v2+60=120+60=180 s1=v1 t1=180 2.5=450=450 km s2=v2 t2=120 2=240

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