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An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.50 $\mathrm{km}$ . If the airplane rounds half the circle in $1.50 \times 10^{2} \mathrm{s},$ determine the magnitude of its (a) displacement and (b) average velocity during that time. (c) What is the airplane's average speed during the same time interval?

(a) displacement $= 7 \mathrm { km }$

(b) $v = 46.67 \mathrm { ms } ^ { - 1 }$

(c) $v = 733.04 \mathrm { ms } ^ { - 1 }$

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Rutgers, The State University of New Jersey

Numerade Educator

Hope College

University of Sheffield

silver Party. We can say that after moving halfway around the circle, the magnitude that plane's displacement would simply be equal to a diameter or two times the radius. So we can say that the displacement This would be doubt, our vector. The displacement vector would equal two times the radius. And so this would be equaling seven 0.0 kilometers or, we can simply say, 7.0 times 10 to the third meters. And so, for part B, the magnitude of the airplanes average velocity we can save the average velocity would be equaling. Delta are divided about out to tea. This is equaling 7.0 times, 10 to the third meters, divided by 150 seconds. And so this is equaling 46.7 meters per second. This would be our answer for part B R answer for part A. And then for part C. We know that the airplanes average speed is equaling to the path length divided by the elapsed time. So here it won't be Delta are over tea. We can say that the, um, average speed would actually be instead of the diameter because you're moving halfway around the circle. It'll be half of the circumference, so this would be equaling 1/2 times. Two pi r These air conference so 1/2 times a circumference divided by again delta t. So this would be pie multiplied by 3.50 times 10 to the third meters, divided by 150 seconds. And this is giving us 73.3 meters per second for the average speed. That is the end of the solution. Thank you for watching.