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An airplane originally at rest on a runway accelerates uniformly at 6.0 meters per second2 for 12 seconds. During this 12-second intervai, the airplane travels a distance of approximately

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The distance travelled by the airplane, is 432 meters.

Given the following data:

  • Initial velocity = 0 m/s (since the airplane starts from rest).
  • Acceleration = 6.0 [tex]m/s^2[/tex]
  • Time = 12 seconds.

To find the distance travelled by the airplane, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2} at^2[/tex]

Where:

  • S is the distance travelled.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the parameters into the formula, we have;

[tex]S = 0(12) + \frac{1}{2}(6)(12^2)\\\\S = 3(144)[/tex]

[tex]S = 3[/tex] × [tex]144[/tex]

Distance, S = 432 meters.

Therefore, the distance travelled by the airplane, is 432 meters.

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