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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!