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16 January, 21:07

An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B with a take-off velocity of 270 km/h, determine (a) the acceleration a, (b) distance AB.

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  1. 17 January, 00:23
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    a) The acceleration of the airplane is 2.5 m/s².

    b) The distance AB is 1125 m.

    Explanation:

    Hi there!

    a) The equation of velocity of an object moving in a straight line with constant acceleration is the following:

    v = v0 + a · t

    Where:

    v = velocity of the object at time "t".

    v0 = initial velocity.

    a = acceleration.

    t = time

    We have the following information:

    The airplane starts with zero velocity (v0 = 0) and its velocity after 30 s is 270 km/h (converted into m/s: 270 km/h · 1000 m/1 km · 1 h / 3600 s = 75 m/s). Then, we can solve the equation to obtain the acceleration:

    75 m/s = a · 30 s

    75 m/s / 30 s = a

    a = 2.5 m/s²

    The acceleration of the airplane is 2.5 m/s².

    b) The distance AB can be calculated using the equation of position of an object moving in a straight line with constant acceleration:

    x = x0 + v0 · t + 1/2 · a · t²

    Where:

    x = position of the object at time "t".

    x0 = initial position.

    v0 = initial velocity.

    t = time.

    a = acceleration.

    If we place the origin of the frame of reference at A, then, x0 = 0. Since the airplane is initially at rest, v0 = 0. So, the equation gets reduced to this:

    x = 1/2 · a · t²

    Let's find the position of the airplane after 30 s:

    x = 1/2 · 2.5 m/s² · (30 s) ²

    x = 1125 m

    The position of B is 1125 m away from A (the origin), then, the distance AB is 1125 m.
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