Pre-Calc

An airplane’s velocity with respect to the air is 630 mph, and its heading is 𝑁 30° 𝐸. The wind, at the altitude of the plane, is directly from the southeast at 45° and has a velocity of 45 mph. Draw a figure that gives a visual representation of the situation. What is the true direction of the plane, and what is its speed with respect to the ground?

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  1. just convert the vectors to x-y components, add them, and you have the resultant.

    Can you get that far? If not, where do you get stuck?

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  2. I did get to that part, I drew the picture, but I think I kept using the wrong angles, I'm not really sure

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  3. 630 @ 𝑁 30° 𝐸 = <315.00,545.58>
    wind from the SE is in the direction NW, so 45 @ NW = <-31.82,31.82>

    Now, if the plane's ground speed is <x,y> we have

    <x,y> + <-31.82,31.82> = <315.00,545.58>

    x = 346.82
    y = 513.76

    That translates into a speed of 619.87 in the direction (from the x-axis) of 55.98°. That makes a heading of 145.98°

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  4. sorry; the heading is 34.02°

    (It's 90-A), not 90+A

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