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As an airplane descends toward an airport, it drops a vertical distance of 24 m and moves forward a horizontal distance of 320 m .what is the distance covered by the plane during this time?

Respuesta :

skyluke89

Answer:

320.9 m

Explanation:

The total distance covered by the plane during this time is the resultant of the displacements of the plane in the two directions.

In fact:

- The plane moved 24 m in the vertical direction: [tex]d_y = 24 m[/tex]

- The plane moved 320 m in the horizontal direction: [tex]d_x = 320 m[/tex]

These two lengths represent the sides of a right triangle, so we can find the hypothenuse, which corresponds to the distance covered by the plane:

[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(24 m)^2+(320 m)^2}=320.9 m[/tex]

snehashish65

The distance covered by the airplane during this time is 320.89 m.

Given data:

The vertical distance is, [tex]s_{y}=24 \;\rm m[/tex].

The horizontal distance is, [tex]s_{x}=320 \;\rm m[/tex].

The actual distance covered by the airplane is equal to the net distance with respect to the vertical and horizontal direction.

Then,

[tex]D_{net}=\sqrt{d^2_{x}+d^2_{y}}[/tex]

Solving as,

[tex]D_{net}=\sqrt{320^2+24^2}\\D_{net}=320.89 \;\rm m[/tex]

Thus, the distance covered by the airplane during this time is 320.89 m.

Learn more about the speed and distance here:

https://brainly.com/question/15172156?referrer=searchResults

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