1D Kinematics problem with a plane taking off

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SUMMARY

The discussion centers on a 1D kinematics problem involving an Airbus A380 that requires a velocity of 280 km/h for takeoff, with a maximum acceleration of 0.95 m/s². Calculations confirm that the aircraft can successfully take off from a 3805-meter runway, achieving a velocity of 85.05 m/s. However, the 2400-meter runway is insufficient, yielding only 67.545 m/s, which is below the required velocity. The conversation highlights the importance of understanding maximum acceleration versus actual acceleration in determining runway suitability.

PREREQUISITES
  • Understanding of kinematic equations, specifically: \(x = x_0 + v_0 t + \frac{1}{2} a t^2\) and \(v = v_0 + at\)
  • Basic knowledge of unit conversions, particularly between km/h and m/s
  • Familiarity with concepts of maximum acceleration in physics
  • Ability to analyze and interpret motion problems in one dimension
NEXT STEPS
  • Study the implications of maximum versus average acceleration in kinematics
  • Learn about the physics of aircraft takeoff and the factors influencing runway length requirements
  • Explore real-world applications of kinematic equations in aviation scenarios
  • Investigate the role of safety margins in runway length calculations for different aircraft
USEFUL FOR

Aerospace engineers, physics students, and anyone involved in aviation safety and runway design will benefit from this discussion.

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Homework Statement


An Airbus 380 needs to reach the velocity of 280 kmh^-1 before it takes off. The maximum acceleration the plain reaches in the runway is 0.95 ms^-2. Verify THAT the plane can use an airport with this runways.
Runway 1: 3805 meters (SSW-NNE)
Runway 2: 2400 meters (S-N)

Homework Equations


$$x=x_0+v_0 t + 1/2 a t^2 $$
$$ v= v_0 + at $$

The Attempt at a Solution


At first I thought this was a pretty easy problem.
Using equations of motion:
$$x=1/2 a t^2 $$
$$ v= at $$

For the first runway, solving
$$3805=1/2 \times 0.95 \times t^2$$
$$t=89.5 s$$

$$v= 0.95 \times 89.5 $$
$$v=85.05 ms^-1$$

Which confirms it can use this runway (since the velocity is bigger than the necessary velocity).

However for the second runway, I get

$$2400=1/2 \times 0.95 \times t^2$$
$$t=71.1 s$$

$$v= 0.95 \times 71.1 $$
$$v=67.545 ms^-1$$

Which is inferior to the necessary velocity. But that contradicts the problem statement that it can indeed use the airport.

I might be making an incorrect assumption. I suspect it might be related to the acceleration. In fact they say that's the maximum acceleration not the acceleration. So the might be a time dependence on the acceleration?
But how do I determine it?

Thanks!
 
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Does a plane have to be able to use every runway at an airport in order for it to use that airport? The problem statement doesn't give any direct guidance on this matter, so I think that you are free to impose your own (reasonable) assumption. Just state up front in your answer what assumption you choose.
 

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