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Mechanics 1 vector help

  1. Aug 30, 2007 #1
    Hi.

    Could someone please help me with the following question? I've no idea how to begin

    Two ships P and Q are moving along straight lines with constant velocities. Initially P is at a point O and the position vector of Q relative to O is (6i + 12j) km, where i and j are unit vectors directed due east and due north respectively. The ship P is moving with velocity 10j km/h and Q is moving with velocity (-8i + 6j) km/h. At time t hours the position vectors of P and Q relative to O are p km and q km respectively.

    (a) Find p and q in terms of t.


    Thank you.

    Cathy
     
  2. jcsd
  3. Aug 30, 2007 #2

    learningphysics

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    Does it say p and q are the magnitudes of the position vectors?
     
  4. Aug 30, 2007 #3

    berkeman

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    You have initial position vectors and velocities. Express the relative motion in rectangular coordinates.
     
  5. Aug 30, 2007 #4
    No. It just says position vectors.

    Cathy
     
  6. Aug 30, 2007 #5
    What is the initial position vector for P?

    Cathy
     
  7. Aug 30, 2007 #6

    learningphysics

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    Oh... I think I was just misunderstanding... So p is the final position of ship P and q is the final position of ship Q (all relative to 0)...

    you know that [tex]\vec{s} = \vec{s_{0}} + \vec{v}t[/tex]

    where [tex]\vec{s}[/tex] is the final position. Use this formula to find the final position of P and Q... treat P and Q as separate problems... one really has nothing to do with the other...

    So what is [tex]\vec{s_{0}}[/tex] and [tex]\vec{v}[/tex] for P?

    What is [tex]\vec{s_{0}}[/tex] and [tex]\vec{v}[/tex] for Q?
     
  8. Aug 30, 2007 #7

    learningphysics

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    It is just 0. Or 0i + 0j.
     
  9. Aug 30, 2007 #8
    Oh, thanks very much. I understand now!

    Cathy
     
  10. Aug 30, 2007 #9

    learningphysics

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    No prob.
     
  11. Aug 30, 2007 #10
    Oh yikes - I'm now stuck on these two parts!

    Could someone please give me some hints? I would really appreciate any help!

    Calculate the distance of Q from P when t = 3.

    Is t measured in seconds, minutes or hours?

    Calculate the value of t when Q is due north of P.

    Thank you.

    Cathy
     
  12. Aug 30, 2007 #11

    learningphysics

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    t is in hours according to the question description... did you calculate the formulas for p and q?

    Do you have any ideas of how to approach the problem? Hint: Use your position formulas p and q
     
    Last edited: Aug 30, 2007
  13. Aug 30, 2007 #12
    Oh yeah, of course it is!

    Yup, I got p = 10jt and q = (6 - 8t)i + (6t + 12)j.

    Would it be possible to work out the displacement and then use v = d/t?

    Cathy
     
  14. Aug 30, 2007 #13

    learningphysics

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    The formulas look good. Yeah, you need the displacement from P to Q. But you don't need velocity. You need distance... so get the distance from your displacement.
     
  15. Aug 30, 2007 #14
    Thanks so much.

    I got that the distance is 18 km.

    Am not sure how to do the next part, though.

    Could you please help?

    Thank you.

    Cathy
     
  16. Aug 30, 2007 #15

    learningphysics

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    The distance looks good to me. :)

    For the next part. Use your displacement formula from P to Q which is in terms of t...

    Hint: you want to find the time when the i component of the displacement 0. when that happens, either Q is north or south of P. you should check the number you get to make sure Q is north and not south of P.
     
  17. Aug 30, 2007 #16
    Thanks for your help.

    I got that t = 0.75 hours.

    I'm not sure how to check the number to check that it is north and not south of P. Could you please explain this?

    Cathy
     
  18. Aug 30, 2007 #17

    learningphysics

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    Cool. That's what I get.

    In your displacement formula from P to Q, when you substitute in t = 0.75, you should get a positive coefficient for j... that means that Q is north of P... if it was negative Q would be south of P.

    I get the displacement from P to Q at t=0.75 as q - p = 9j, so since 9>0 Q is north of P...
     
  19. Aug 30, 2007 #18
    Okay. I get it now!

    Thanks again for all your help.

    I really appreciate it! :-)

    Cathy
     
  20. Aug 30, 2007 #19

    learningphysics

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    You're welcome. :smile:
     
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