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Plane landing, finding the velocity!

  1. Jan 24, 2014 #1
    1. The problem statement, all variables and given/known data
    A plane, B, is approaching a runway along a straight line 15 degrees (angle phi) below horizontal, while the radar antenna, A, is monitoring the distance, r, between A and B, as well as the angle between A and b, theta. The plane has a constant approach speed v0. In addition, when theta=20 degrees, it is known that r(dot)=216 ft/s and theta(dot)=-0.022 rad/s. Determine the corresponding values of v0 and of the distance between the plane and the radar antenna.

    Phi=15 degrees
    Theta=20 degrees
    r(dot)=216 ft/s
    theta(dot)= -0.022 rad/s
    Since v is constant, a=0.

    2. Relevant equations
    a=dv/dt
    v=ds/dt


    3. The attempt at a solution
    I tried solving knowing that a=0. However, I do not know how to utilize the r(dot) and theta(dot).
     
  2. jcsd
  3. Jan 24, 2014 #2

    BvU

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    This is more math than physics. You already have a = 0 because v0 is constant.
    You aren't interested in he trajectory of the plane, only in the components of the vector v0,
    for which you were given the angle w.r.t. horizontal.
    All you have to do is convert the vector v0 to dr/dt and dtheta/dt. Or rather the other way around.
    Make a drawing showing A, B and vector v0.
    I hope the radar is between the plane and the runway....
     
  4. Jan 24, 2014 #3
    The radar is between the runway and the plane. So when I convert v0 to dr/dt and dtheta/dt would it look like this:

    v0=cos(35+15)*dr/dt - cos(90-50)*dtheta/dt
    v0=cos(50)*(216 ft/s) - cos(40)*(-0.022 rad/s)

    if so I don't know r so I cant convert rad/s to ft/s in order to simplify?
     
  5. Jan 24, 2014 #4

    BvU

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    You also know the direction of v0. That should give you a second equation. Did you make the drawing ?
     
  6. Jan 24, 2014 #5

    BvU

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    Reason I ask is because I don't see where the 35 degrees comes in.
     
  7. Jan 24, 2014 #6

    BvU

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    Bedtime for me. make the drawing. Project v0 on the r vector and on the theta direction.
     
  8. Jan 24, 2014 #7
    Yes, v0 has a component in the u_r direction (x component) and in the u_theta direction (y direction). Here's what I did but it is not correct:

    v0 = (216)(cos(35)) + (-0.022)(sin(35)) = 176.949

    obviously this is wrong because the -.022 is in units rad/s and I am not sure how to convert to ft/s unless in know r.
     
  9. Jan 24, 2014 #8
    it should have been cos(20+15) not cos(35+15).. stupid error
     
  10. Jan 24, 2014 #9

    BvU

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    Don't forget to correct your conversion formula: you want ft/s in the second term, not radians/s

    Do these flying folks still use feet/s ? ;-)
     
  11. Jan 24, 2014 #10

    BvU

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    Still don't see where the addition of the angles comes from. The plane was landing, wasn't it ?
     
  12. Jan 24, 2014 #11

    BvU

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    Suppose it was flying straight at the radar. Would you calculate with 20 degrees + 20 degrees ?
     
  13. Jan 24, 2014 #12

    BvU

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    r is an unknown, so you have to leave it in.
     
  14. Jan 24, 2014 #13

    BvU

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    I really have to turn in now. If your drawing doesn't help you out, a different tack is to convert v0 to a horizontal and a vertical component, and then do the same with v(r, theta) = (vr, vtheta) theta = 20 deg, r is unknown.
     
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