A 2940 kg airplane is 38 m long and has a wing span of 27 m. The airport taxiway

Question

A 2940 kg airplane is 38 m long and has a wing span of 27 m. The airport taxiways are
38 m wide and the runway is 54 m wide and 910 m long. The airplane taxies at a
constant velocity of 5 m/s. The airplane can take off when it reaches a velocity of 57 m/s.

a)On the way from the airport terminal to the runway, the plane crosses a taxiway.
How long was the airplane in the taxiway?

b)The plane starts from rest and uses the entire length of the runway during takeoff.
How much time was needed for the airplane to take off?

c)What was the airplane's acceleration during take off?

Explanation / Answer

Hi, It seems some part(text) of the question is missing. However, letz solve the problem with given and understood data. Given the length of the plane = 38m. And the taxiways is 38m wide. Given that plane has crossed the taxiway and hence the total distance it has to cover from the point its front just enters the taxiway to the time its tail end just leaves the taxiway is sum of the above two = 38 + 38 = 76m. Since the plane is just reaching the runway from the terminal, it will travelling at a constant speed of 5m/s. Hence the plane will take a time of t = s/v = 76/5 = 15.2 sec. i.e., the plane was for 15.2 sec in the taxiway. b) & c) Length of the runway = 910m. So the maximum affective length that is available for the plane on the runway = 910 - 38 = 872m.     it starts from rest, u = 0 and can take off at a velocity v = 57m/s. Hence from v^2 - u^2 = 2 * a * s we have, 57^2 - 0 = 2 * a * 872       => a = 1.86 m/s^2. and hence from v = u + at, we have t = (57 - 0) / 1.86 = 30.65 sec. i.e., the airplane will take 30.65sec to takeoff and has an acceleration of 1.86 m/s^2 during takeoff. Hope this helps you. Hope this helps you.

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