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Vectors question problem

  1. Nov 17, 2014 #1
    Question
    An object is travelling eastward at a speed of 500 ms-1 flies into a 90ms-1 crosswind blowing southward.
    a) calculate the resultant velocity of the plane relative to the object as it flies through the crosswind.

    b) A person is steering the object, calculate the direction the person would have to steer in order for the resultant velocity off the plane to remain eastward.

    My attempt

    a) Magnitude of R=squareroot(902+5002)
    = 508ms-1

    Direction= tan-1(90/500) = 10.2
    90-10.2= 79.8

    Resultant= 508ms-1 at 79.8 degrees to the ground

    b) dont know what to do, would you please be able to help me on this one? I dont really understand what it wants me to do

    Thanks for any help you can give
     
  2. jcsd
  3. Nov 17, 2014 #2

    Bystander

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    Plane? Taxiing? Parked? Sitting in a hangar? Are the "plane" and the "object" one and the same?
     
  4. Nov 17, 2014 #3
    Sorry missed the bit at the bottom its a plane
     
  5. Nov 17, 2014 #4

    BvU

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    Make a drawing for case a)
    Make a drawing for the case b)
     
  6. Nov 17, 2014 #5
    Bump still dont know how to do part b
     
  7. Nov 17, 2014 #6

    Bystander

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    For b) you want the plane/object moving at 500m/s to travel true east while the wind is blowing S at 90 m/s. What velocity N is necessary to offset the wind? What velocity E combined with that gives you a 500 m/s resultant air speed?
     
  8. Nov 17, 2014 #7
    Thank you for the help
    so I want to find a velocity (with angle north) that will give me 500km h-1 east?

    if what I said above is right, I would

    500= sqaureroot(902 +X2) then doing some rearranging and calculating

    X= 491.83 km h-1

    then to work out the direction would do

    500= 90 + 491.83cos(y)
    y= cos-1(410/491.83)
    y= 33.53 degrees to the eastward

    therefore the direction and speed that the plane would need to travel is 491.83km h-1 at an angle of 33.53 degrees to the eastwood?

    is that correct

    thank you again for the help
     
  9. Nov 17, 2014 #8

    Bystander

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    My calculator just cratered --- I'll take your word --- it looks close.

    No. BvU suggested you make drawings. Which part of the right triangle is 500 for this part of the problem? Keep in mind that part b) may be different than part a).
     
  10. Nov 17, 2014 #9
    so the 500 is the resultant? making 491.83=500sin(y)?? and this makes y to the north?

    is that correct
     
    Last edited: Nov 17, 2014
  11. Nov 17, 2014 #10

    Bystander

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    Yes. And yes, or cos(y), whichever you prefer. You measure angles from a reference direction in either clockwise (navigation), or counterclockwise (right handed coordinate systems in mathematics) direction; what reference direction are you using, and are you measuring cw or ccw?
     
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