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14 August, 14:32

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0 seconds what is the direction of the velocity of the plane?

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  1. 14 August, 14:39
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    38.5° north of east

    Explanation:

    Directly northwest is 135° from east. If we say east is + x and north is + y, then the components of the acceleration are:

    ax = 2.88 cos 135° ≈ - 2.036 m/s²

    ay = 2.88 sin 135° ≈ 2.036 m/s²

    And the initial velocity is:

    v₀x = 115 m/s

    v₀y = 0 m/s

    After, 25.0 seconds, the final velocity components are:

    vx = at + v₀x

    vx = (-2.036) (25.0) + 115

    vx ≈ 64.1 m/s

    vy = at + v₀y

    vy = (2.036) (25.0) + 0

    vy ≈ 50.9 m/s

    The plane's direction is:

    θ = atan (vy / vx)

    θ = atan (50.9 / 64.1)

    θ ≈ 38.5°

    The direction of the plane's velocity is approximately 38.5° north of east.
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