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What is the relative velocity of the airplane to the surface

  1. Jan 15, 2006 #1
    ok i have tried to draw this problem several times and i get different answers every time...my answers range from 804-880km/h and i cant get the degree down either...please help....

    an airplane pilot tries to fly directly east of the velocity of 800.0km/h if a wind comes from the southwest at 80.0km/h what is the relative velocity of the airplane to the surface of the earth...how do u do this mathematically???? please help!!
     
  2. jcsd
  3. Jan 15, 2006 #2
    Break the vectors into components and add them
     
  4. Jan 15, 2006 #3

    Ouabache

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    Welcome balletprincess34 to the Physics Forums !! :smile:
    You have an interesting question.. You have correctly completed the first step in asking a question on HW forums, describing your problem.. The next step, before we can help you, is to explain what you have done so far and where you are getting stuck.. (Besides the range of answers you found, how did you get there?)

    don't worry, we have plenty of helpful & knowlegable folks here who will be happy to steer you in a successful direction. Andrew has already given you some useful information. We can always elaborate on that if you don't understand..

    Be sure to read this sticky at the top of this subpage to useful advice in using this forum.
     
  5. Jan 15, 2006 #4
    well for the most part i just drew the vectors....however depending on my scale....the answers varied like i would get one measurement but the incorrect angle value....how can u just solve this problem mathematically...i know that u have to break the 80.0km/hr in its vertical and horizontal components and i got 56.569 for both components...but what do i do now?
     
  6. Jan 15, 2006 #5

    tony873004

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    56.569 is your horizontal component for the wind and
    56.679 is your vertical component for the wind.

    What are your horizontal and vertical components for the airplane?
     
  7. Jan 15, 2006 #6
    well if the airplane is flying directly east wouldnt that mean that the entire component is horizontal and the vertical component is equal to zero?
     
  8. Jan 15, 2006 #7

    tony873004

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    Yes. So the airplane's horizontal component is 800 and its vertical component is 0. Now you can add them to the wind's components.
     
  9. Jan 15, 2006 #8
    ok so i got 858km/hr...now how can i solve for the angle?
     
  10. Jan 15, 2006 #9

    tony873004

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    You weren't asked to solve for the angle. But if you want to know it, use Trig. You are asked for the airplane's speed relative to the ground. Construct a triangle with the horizontal component and vertical component as 2 legs. Do you know how to find the hypotenuse of this triangle?
     
  11. Jan 15, 2006 #10

    tony873004

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    Also, 858 km/hr is the horizontal component. What about the vertical component?
     
  12. Jan 15, 2006 #11

    tony873004

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    Actually, how did you get 858? It's close but not quite right.
     
  13. Jan 15, 2006 #12
    wait i just used pythagorean thm to solve for the 858km/hr and thanx fer the angle help....i get the problem now..i just always to forget to add the components first....thanx again
     
  14. Jan 15, 2006 #13

    tony873004

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    Sorry, you were a step ahead of me. 858 is correct. sqrt(856.6^2+56.6^2).
    To get the angle, draw a triangle. You know all 3 sides, so you can use SIN, COS or TAN to get the angle. And since they ask for velocity, not speed, you should include this angle in your answer.
     
  15. Jan 15, 2006 #14
    all right ok so i got 858km/hr at 3.8 degrees N of E...is that what u got? i hope so and thank u sooo much fer all your help!!! :)
     
  16. Jan 15, 2006 #15

    tony873004

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    yes. That's what I got. Making a triangle, the vertical component of 56.6 is opposite the angle. The horizontal component of 856.6 is adjacent the angle. TAN-1(56.6/856.6)=3.8 degrees. Since the wind comes from the southwest, it pushes the eastbound aircraft north of east. Aren't these easy!
     
  17. Jan 18, 2006 #16

    Ouabache

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    Thanks for jumping in Tony, I didn't get back to this thread till just now. BTW how did you get different values for F(wind_horiz) and F(wind_vert)??

    [I get the same value for sin (45 deg) and cos (45 deg)]

    The most confusing part of this question is interpreting horizontal and vertical velocities of wind and the airplane, as both being parallel to the earth's surface.

    What we normally think of as vertical in space is actually perpendicular to both vertical and horizontal velocity axes in this question.

    Once we realize this, the rest of the problem becomes easy.
     
    Last edited: Jan 18, 2006
  18. Jan 18, 2006 #17

    tony873004

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    Typo :rolleyes:
     
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