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Galilean form of the law of transformation of velocities

  1. Aug 8, 2013 #1
    May you help me with finding the angle, and what is "line of sight"?(the final answer is 15°)
    Thank you in advance
     

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  2. jcsd
  3. Aug 8, 2013 #2

    Doc Al

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    By "line of sight" they mean the direction in which the police car is moving. (Where they would be looking straight ahead.) So imagine an axis extending forward from the police car.

    To find the angle, figure out the velocity of the motorist in the frame of the police car.
     
  4. Aug 8, 2013 #3
    I get -39°...not 15°...
     
    Last edited: Aug 8, 2013
  5. Aug 8, 2013 #4

    Doc Al

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    Show how you got that.
     
  6. Aug 8, 2013 #5
    α=arctan(-62/76)
     
  7. Aug 8, 2013 #6

    Doc Al

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    Ah, I was wrong about what they meant by "line of sight". Line of sight is the line directed from the police car to the motorist. Find the angle with respect to that line and you'll get the answer given.
     
  8. Aug 8, 2013 #7
    I'm confused, now I got -54° and what does the minus sign mean in that situation, may you give some more details about the solution...
     
  9. Aug 8, 2013 #8

    Doc Al

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    Don't worry about signs. Instead, draw yourself a picture. Show the line of sight (which is based on distances) and then add the direction of the velocity, which you already found. Find the angle between those two.
     
  10. Aug 8, 2013 #9
    Are signs meaningless?
     
  11. Aug 8, 2013 #10

    Doc Al

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    No, but rather than rely on the mechanical use of a formula you would be better off simply drawing the relevant angles. Then you wouldn't have questions about signs.
     
  12. Aug 8, 2013 #11
    I have drawn the angles but how are they related
     

    Attached Files:

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  13. Aug 8, 2013 #12

    Doc Al

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    The first diagram is drawn correctly: The hypotenuse represents the line of sight.

    The second diagram needs to be redrawn to represent the sum of those velocity vectors. That vector will represent the relative velocity.
     
  14. Aug 8, 2013 #13
    76 need to be reversed ...How do those drawings show me that I need to subtract those angles?
     
    Last edited: Aug 8, 2013
  15. Aug 8, 2013 #14

    Doc Al

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    When adding vectors graphically you must have the tail of one start at the head of the other.

    You want the angle that the velocity makes with the line of sight. Draw them both on the same diagram and find the angle between them.
     
  16. Aug 8, 2013 #15
    But they have different dimensions...and why to subtract?
     

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    Last edited: Aug 8, 2013
  17. Aug 8, 2013 #16

    Doc Al

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    (1) We are just comparing directions; dimensions are irrelevant.
    (2) Your diagram is still not right. You want: tail - head - tail - head. You have: head - tail - tail - head.
    (3) Find the angle each direction vector makes with the horizontal; then you can find the angle they make with each other by subtracting.
     
  18. Aug 8, 2013 #17
    Do I need to rotate the triangle of velocity? The angle in this case -15°, is that right?
    I'm grateful to you for your help!
     

    Attached Files:

    • 111.PNG
      111.PNG
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  19. Aug 8, 2013 #18

    Doc Al

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    It's not a question of rotating the triangle, but drawing it correctly. Note that the velocity vector will point towards the police car.

    Yes, but don't worry about the minus sign. I'm sure they just want the magnitude.

    You are most welcome.
     
  20. Aug 8, 2013 #19
    Sorry for asking again...Do you mean that velocity vector of 76 points toward the police while motorist have a velocity vector towards him of 62?
     
    Last edited: Aug 8, 2013
  21. Aug 8, 2013 #20

    Doc Al

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    The velocity vector of the motorist with respect to the Police will point somewhat towards the Police. Not exactly of course--it will make an angle of 15° to the line of sight. In your diagram in post #17 the red arrow, which I presume represents the velocity of the motorist with respect to the police, points in the wrong direction.
     
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