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Thread: displacement problem

  1. #1
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    displacement problem

    an airplane travels 200 miles due west then 150 miles 60 degrees north of west What is the resultant displacement
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    Quote Originally Posted by leinadwerdna View Post
    an airplane travels 200 miles due west then 150 miles 60 degrees north of west What is the resultant displacement
    You need to assign a reference for your displacement vectors and then sum them. Most intuitive for me would be to think of the directions as if they were on a map: west being negative x-axis, east being positive x-axis, north being positive y-axis, and south being negative y-axis. From there, you see the first vector is -200 and the second vector is (150 \angle 180^\circ - 60^\circ) or (150 \angle 120^\circ)

    Drawing pictures helps with vector problems such as this one. Do do vector addition, you need to break each vector into rectangular components, and add them.
    x: -200 + 150  \cos (120^\circ)
    y: 150\sin (120^\circ)

    Your answer then may look like this:
    (-200 + 150  \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \
    Where \hat i \ and \hat j \ are unit vectors designating only direction and not magnitude.
    If you wanted the answer expressed in phasor form, then do this:
    (\sqrt{(-200 + 150  \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150  \cos (120^\circ)}))
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    Quote Originally Posted by leinadwerdna View Post
    an airplane travels 200 miles due west then 150 miles 60 degrees north of west What is the resultant displacement
    you can use the law of cosines to find the magnitude of the displacement ...

    d = \sqrt{200^2 + 150^2 - 2(200)(150)\cos(120^\circ)}

    then the law of sines to find the direction of the displacement relative to west ...

    \frac{\sin{\theta}}{150} = \frac{\sin(120^\circ)}{d}<br />

    \theta = \arcsin\left[\frac{150\sin(120^\circ)}{d}\right]
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    for some reason the answer key says the displacement is facing north of east i thought it would be north of west
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    Quote Originally Posted by leinadwerdna View Post
    for some reason the answer key says the displacement is facing north of east i thought it would be north of west
    answer keys aren't always correct
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    Quote Originally Posted by 1005 View Post
    You need to assign a reference for your displacement vectors and then sum them. Most intuitive for me would be to think of the directions as if they were on a map: west being negative x-axis, east being positive x-axis, north being positive y-axis, and south being negative y-axis. From there, you see the first vector is -200 and the second vector is (150 \angle 180^\circ - 60^\circ) or (150 \angle 120^\circ)

    Drawing pictures helps with vector problems such as this one. Do do vector addition, you need to break each vector into rectangular components, and add them.
    x: -200 + 150  \cos (120^\circ)
    y: 150\sin (120^\circ)

    Your answer then may look like this:
    (-200 + 150  \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \
    Where \hat i \ and \hat j \ are unit vectors designating only direction and not magnitude.
    If you wanted the answer expressed in phasor form, then do this:
    (\sqrt{(-200 + 150  \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150  \cos (120^\circ)}))
    if u do it this way the angle of the displacement vector is -25 degrees rather than 25 why is that
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