A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 232 km and a direc

Question

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 232 km and a direction 30.0o north of east. The displacement vector B for the second segment has a magnitude of 168 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle θ with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle θ.

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Ngọc Diệp 1 week 2021-07-21T12:42:20+00:00 1 Answers 0 views 0

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    2021-07-21T12:43:58+00:00

    Answer:

    a) 121 km

    b) 74°

    Explanation:

    To start with, we assume that there exist two components, the East and the North. We’d be representing the East, by “e” and the North, by “n”

    Now we start with the

    First vector:

    east1 = 232 cos 30 = 201

    north1 = 232 sin 30 = 116

    Now, that of the second vector will be

    east 2 = – 168

    north2 = 0

    Next, we add the two together and get

    East components

    201 – 168 = 33 east

    North components

    116 + 0 = 116

    Therefore, the magnitude has to be

    magnitude = √(33² + 116²)

    Magnitude = √14545

    Magnitude = 121

    tanθ = 116/33

    Tanθ = 3.51

    θ = tan^-1 3.51

    θ = 74° North East

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