Question

The takeoff speed for an Airbus A320 jetliner is 82 m/s . Velocity data measured during takeoff are as follows: t(s) vx(m/s) 0 0 10 23 20 46 30 69

What is the jetliner's acceleration during takeoff, in g's?

At what time do the wheels leave the ground?

For safety reasons, in case of an aborted takeoff, the length of the runway must be three times the takeoff distance. What is the minimum length runway this aircraft can use?

Answer 1

Ans

a) Using the equation of costant acceleration

v = u + a*t

From the data, u = 0 m/s at t = 0 s and at t = 10 s, v = 23 m/s

Therefore a = (v - u)/t = (23 - 0)/10 = 2.30 m/s²

The acceleration of the jetliner = 2.30 m/s²

Writing it in terms of g, a = (2.30/g)*g = 0.235g (multiplied and divided a by g)

The acceleration of jetliner in terms of g = 0.235g

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b) The wheels leave the ground when v = 82 m/s

Again using the same equation to find time t when v = 82 m/s

t = (v - u)/a = (82 - 0)/2.30 = 35.65 s

The wheels leave the ground at t = 35.65 s

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c) First we find the takeoff distance using s = u*t + (1/2)*a*t²

To find s when plane is moving at v = 82m/s i.e after time t = 35.65 s and with a = 2.30 m/s²

u = 0 m/s, s = 0 + 0.5*2.30*(35.65)^2 = 1461.56 m

The takeoff distance = 1461.56 m

So the minimum length of the runway that can be used by the aircraft is equal to 3*takeoff distance

i.e 3*s = 3*1461.56 = 4384.68 m