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A jet plane traveling 1890 km/h (525 m/s) pulls out of a dive by moving in an arc of radius 5.20 km. What is the plane’s acceleration in g’s?

Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.
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Quick Answer: 

$5.41\textrm{ g's}$

Giancoli 7th Edition, Chapter 5, Problem 2


Chapter 5, Problem 2 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. This jet plane pulls out of a dive in an arc of 5.2 kilometers which is 5200 meters. And has a speed of 525 meters per second. So we will calculate the centripetal acceleration and then convert it into number of g's. So we have the speed is 525 meters per second and we square that divided by the radius of the curve that it's going through, 5200 meters, and we get 53 meters per second squared. And the number of g's will be that centripetal acceleration times 1 g for every 9.8 meters per second squared. And you can consider this meters per second squared as a single unit and it's on the numerator there and it's in the denominator there and as a package they cancel; leaving us with g's and we have 5.41 g's.