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27 October, 01:51

A plane is flying east at 135 m/s. The wind accelerates it at 2.18 m/s^2 directly northeast. After 18.0s, what is the magnitude of the displacement of the plane?

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  1. 27 October, 05:34
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    Directly northeast is 45˚ north of east. Let’s determine the north and east components of the acceleration.

    North = 2.18 * sin 45

    East = 2.18 * cos 45

    Both of these are approximately 1.54 m/s^2. Let’s use the following to determine the east component of the plane’s displacement.

    d = vi * t + ½ * a * t^2

    d = 135 * 18 + ½ * 2.18 * cos 45 * 18^2

    d = 2430 + 353.16 * cos 45

    This is approximately 2680 meters.

    North = ½ * 2.18 * sin 45 * 18^2 = 353.16 * sin 45

    This is approximately 250 meters. To determine the angle north of east, use the following equation.

    Tan θ = North : East

    Tan θ = 353.16 * sin 45 : (2430 + 353.16 * cos 45)

    This is approximately 5.3˚ north of east.
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