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Position of a particle

  1. Oct 13, 2007 #1
    The position of a particle as a function of time is given by r = (6.50 ihat + 3.10 jhat)t^2 , where t is in seconds.

    What is the particle's distance from the origin at t_1 = 2.6 seconds?
    What is the particle's distance from the origin at t_2 = 6.2 seconds?

    What is the particle's speed at t = 2.6 seconds?
    What is the particle's speed at t = 6.2 seconds?

    I know that at t = 0 seconds the particle is 0 meters away and has a speed of 0 m/.

    How do I go about solving for the other questions?
     
  2. jcsd
  3. Oct 14, 2007 #2
    Well I figured out that the magnitude of the vector r = (6.5+3.1)t^2 meters is 7.20.

    Therefore I solved t = 2.6s.... 7.2 x 2.6^2 = 48.7 meters and t = 6.2 s.... 7.2 x 6.2^2 = 276.8 meters.

    But now I am not sure of how to solve for the particle's speed?

    Any help would be great.

    Thanks
     
  4. Oct 15, 2007 #3

    CompuChip

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    You forgot to write a [itex]t^2[/itex] after the number, but I think you did it correct as you get the correct answer.

    In one dimension, if the position of the particle is given by x(t), the velocity is given by x'(t). The same is true for (in this case) two dimensions: the derivative of the vector r(t) will give you the velocity vector v(t) = r'(t).
     
  5. Oct 15, 2007 #4
    So r'(t) = v(t)

    I am not getting the right answer so maybe I am doing something wrong when taking the derivative..

    r'(t) = 2(6.5+3.1)t or 19.2t Correct???
     
  6. Oct 16, 2007 #5

    CompuChip

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    No, the derivative is again a vector. It's not a number. If [itex]r(t) = (x(t), y(t), z(t)[/itex] in three dimensions, where x, y and z are just functions of t (the components of the vector), then [itex]r'(t) = (x'(t), y'(t), z'(t))[/itex] - the components are the derivatives of the original vector.
    In this case [itex]r(t) = \left( 6.50 t^2 \hat \i + 3.10 t^2 \hat \j \right)[/itex]. The derivative is then
    [tex]r'(t) = \left( \frac{d}{dt} 6.50 t^2 \hat \i + \frac{d}{dt} 3.10 t^2 \hat \j \right)
    = \left( 2 \times 6.50 t \hat \i + 2 \times 3.10 t \hat \j \right),
    [/tex]
    which you can write as
    [tex]r'(t) = (6.50 \hat \i + 3.10 \hat \j) 2 t[/tex].

    I made the attached (very ugly!) image to show what I mean. The v(t) vector is supposed to be tangent to the black path of the object.
     

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  7. Sep 16, 2011 #6
    TonkaQD4, did you get the correct answer using that^ equation? If so what is it because I can't seem to get it.
     
  8. Sep 19, 2011 #7

    CompuChip

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    Hi Becca.

    Can you
    a) not revive an old thread (the last post was almost 4 years ago)?
    b) show some of your work? The answer is quite straightforward, so if you don't get it, it's most likely an algebraic error somewhere.
     
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