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21 December, 12:37

What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 265 m?

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  1. 21 December, 12:59
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    a) a = 2.096 m/s² = 2.1 m/s²

    b) Take off time = 15.9 s

    Explanation:

    Using the equations of motion

    x = take off run = 265 m

    Initial velocity, u = 0 m/s (plane starts from rest)

    final velocity = lift off speed = v = 120 km/h = 33.33 m/s

    a = ?

    a) v² = u² + 2ax

    33.33² = 0² + 2*a*265

    a = 1111.1111/530

    a = 2.096 m/s²

    b) v = u + at

    33.33 = 0 + (2.096 * t)

    t = 33.33/2.096

    t = 15.9 s
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