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Answer :
The distance covered by an airplane starting from rest, assuming a constant acceleration of a = 4.3 m/s and taking off after 18 seconds is 696.6 meters.
The formula for the distance covered by an object starting from rest and assuming a constant acceleration is:
s = (1/2) * a * t² Where;
- s is the distance covered
- a is the constant acceleration
- t is the time taken
Substituting the given values into the formula above;
s = (1/2) * a * t² = (1/2) * 4.3 m/s² * (18 s)²
s = 696.6 meters
Therefore, the airplane has moved 696.6 meters down the runway after 18 seconds of takeoff.
To learn more about acceleration: https://brainly.com/question/460763
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