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Q. On April 15, an airplane takes off at 4:40 P.M. from Belem,Brazil bound for...

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user8235304 | Student, Grade 11 | Valedictorian

Posted August 30, 2013 at 5:39 PM via web

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Q. On April 15, an airplane takes off at 4:40 P.M. from Belem,Brazil bound for Villamil,Ecuador(in the Galapagos).The plane lands at 8:40 P.M. Villamil local time.The sun sets at 6:15 P.M. in Belem(local time) and 7:06 P.M. in Villamil(local time).At what time during the flight do the airplane passengers see the sun set?

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aruv | High School Teacher | Valedictorian

Posted August 30, 2013 at 6:42 PM (Answer #1)

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On April 15, an airplane takes off at 4:40 P.M. from Belem,Brazil bound for Villamil,Ecuador(in the Galapagos).The plane lands at 8:40 P.M. Villamil local time.The sun sets at 6:15 P.M. in Belem(local time) and 7:06 P.M. in Villamil(local time).At what time during the flight do the airplane passengers see the sun set?

Let distance between Belem and Villamil = x km

distance from Belem to  Place where sun sets = y km

Time difference in sun sets between Belem and Villamil= 51/60 hr.

Time of flight base time Belem=4-51/60=189/60 hr.

Speed of plane= x/(189/60)

=(60x)/189

Then

y=((60x)/189 )(95/60)

=(95x)/189

(y/x)=95/189

Thus time variation between place where sun set and Belem = (51/60)(95/189)=.43 hr =26 min.

Thus the time where sun stes see passenger= sun set at Belem +time variation between place where sun set and Belem 

=6:15+:26

=6:41 PM

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