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Position Function and Speed

  1. Mar 16, 2013 #1
    1. The problem statement, all variables and given/known data

    The position function of a particle is given by r(t) = <2t, 3, -2t+1> where t is the time in seconds. When is the speed a minimum? What is the minimum speed?

    2. Relevant equations

    v(t) = r'(t)

    3. The attempt at a solution

    The derivative of r'(t) = 2 + 0 + (-2) = 0. The t is cancelled out so how are we supposed to find t and the minimum speed? I am so confused. Help please?
     
    Last edited by a moderator: Mar 17, 2013
  2. jcsd
  3. Mar 16, 2013 #2

    jbunniii

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    No, the derivative should be a vector. If ##r(t) = \langle a(t), b(t), c(t)\rangle##, then ##r'(t) = \langle a'(t), b'(t), c'(t) \rangle##. You don't sum ##a'(t)##, ##b'(t)##, and ##c'(t)##.
     
  4. Mar 16, 2013 #3
    Ok, so I would just get <2,3,-2>. Then what would I do?
     
  5. Mar 16, 2013 #4

    Ray Vickson

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    Speed = magnitude of velocity.
     
  6. Mar 16, 2013 #5

    jbunniii

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    The second component should not be 3.
     
  7. Mar 16, 2013 #6
    Yea sorry it's <2,0,-2>
     
  8. Mar 16, 2013 #7
    So the speed is 0? How do we find the time then? That's what I am stuck on.
     
  9. Mar 16, 2013 #8

    Fredrik

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    No, it's not. He meant the magnitude (i.e. the norm) of the vector, not the magnitude of the sum of its components.
     
  10. Mar 16, 2013 #9
    I am really confused now. Could you give me some guidance?

    So the derivative is velocity right? The magnitude of the vector would be √((2)^2 + (-2)^2)? Then that would be speed which is √8?

    Is this right?

    Thanks
     
  11. Mar 16, 2013 #10

    Fredrik

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    Yes, that's right.
     
  12. Mar 16, 2013 #11
    Thanks, so for time what would I do since when you find the derivative t is no longer there?
     
  13. Mar 16, 2013 #12

    jbunniii

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    You have shown that the speed is constant, so it achieves both a minimum and a maximum at all times.
     
  14. Mar 17, 2013 #13
    oh! okay that makes sense! Thank you everybody :)
     
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