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A Boeing 747 aircraft has a landing speed of 72 m/s, and upon landing it is able to come to a stop in 2,000 m. Assuming constant deceleration, how long would it take for a Boeing 747 to come to a full stop?
  • a)
    14 seconds
  • b)
    28 seconds
  • c)
    56 seconds
  • d)
    72 seconds
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A Boeing 747 aircraft has a landing speed of 72 m/s, and upon landing ...
We can relate time and displacement with constant deceleration just the same as we can with constant acceleration, with the formula d= Δt (va).
We know displacement must equal 2,000m, therefore we can plug in our values to solve for Δt.
2000 = Δt 1/2(0+72), therefore Δt = 55.55 or roughly 56 seconds.
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Most Upvoted Answer
A Boeing 747 aircraft has a landing speed of 72 m/s, and upon landing ...
Given data:
- Landing speed = 72 m/s
- Stopping distance = 2000 m

Calculating deceleration:
- Initial velocity (u) = 72 m/s
- Final velocity (v) = 0 m/s
- Distance (s) = 2000 m
- Using the equation of motion: v^2 = u^2 + 2as
- 0 = (72)^2 + 2*a*2000
- Solving for acceleration (a): a = -72^2 / (2*2000) = -1.296 m/s^2

Calculating time taken to stop:
- Using the equation of motion: v = u + at
- 0 = 72 + (-1.296)*t
- Solving for time (t): t = 72 / 1.296 ≈ 55.56 seconds
Therefore, it would take approximately 56 seconds for a Boeing 747 to come to a full stop after landing. So, the correct answer is option 'C' - 56 seconds.
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