Lightning and Airplanes: Solving a Distance Problem

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AI Thread Summary
The discussion revolves around a homework problem involving the distance from a person to a lightning strike and the average speed of an airplane. The person calculates the distance to the lightning strike as 5,610 feet based on the time it takes for the sound to reach them, using the speed of sound at 1,100 ft/s. They assume this distance is the same for the airplane when it passes overhead 11 seconds later. The average speed of the airplane is calculated to be 510 ft/s by dividing the distance by the time taken. Concerns are raised about the significant figures in the reported answer.
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Can someone confirm the solution? My elder sister who is in college had this problem for homework, and asked me weather I could take a look at it. For some reason however my answers are incorrect, and I was wondering weather anyone could point me in the correct direction.

Homework Statement



a. A person witnesses a lightning strike pass near an airplane. Approximately 5.1 seconds after the lightning strike, the person hears the thunder. The airplane passes overhead 11 seconds following the lightning strike. Assuming the speed of sound to be 1100 ft/s, what is the distance of the person from the airplane precisely when the lightning strikes?

b. What is the average speed the plane is traveling at?

Homework Equations



average speed = distance/time
distance = time * average speed

The Attempt at a Solution



The distance from the person to the lightning strike can be found by:
d = 1100*5.1
= 5610

For simplification, assuming that the distance of the airplane from the person is equivalent
to the distance of the lightning strike from the person, the distance of the airplane from the person is approximately 5, 610 ft.

The solution to the second part of the problem can be found by dividing 5610, the distance of the airplane from the person, by 11, the time it takes for the airplane to reach the person, giving you an average speed of 510 ft/s.
 
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Looks right to me. Only problem I can think of is too many significant figures in the reported answer 5610 ft.
 
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