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Homework Help: Kinematics of Particles

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle moves in a plane. The particle (x,y) position on the plane is given by:

    [tex]x = -34t^4 − 28t^3 + 7[/tex]
    [tex]y = 25t^2 + 13t^3 + 5[/tex]

    Therefore the particle's displacement [from the (x,y) origin) at time t = 68 s is

    [tex](736000000m) i + (4200000m)j[/tex]

    (a) What is the particle's velocity at 68 s ?

    (b) What is the particle's acceleration at 68 s ?

    3. The attempt at a solution

    (a) for this part the correct answer has to be

    [tex]v_p = (43200000m) i + (184000m)j[/tex]

    But I can't see how they have arrived at this answer! The velocity is [tex]\frac{\Delta x}{\Delta t}[/tex]

    [tex]\Delta t = t_f-t_i=68-0 =68[/tex]

    [tex]\Delta x = x_f-x_i = 736000000 - 7 = 375999993[/tex]

    So, [tex]\frac{375999993}{68}=10823529.31[/tex].

    This is not the right answer for the i component. What is the problem?
     
  2. jcsd
  3. Mar 22, 2010 #2

    rock.freak667

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    just use v(t)= (dx/dt)i+(dy/dt)j
     
  4. Mar 22, 2010 #3
    That's exactly what I did! And I don't know why I get the wrong answer...
     
  5. Mar 22, 2010 #4

    rock.freak667

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    Can you show exactly how you did it? You may have made a mistake in the algebra somewhere.
     
  6. Mar 22, 2010 #5
    Here is my working for the i component:

    The particle's displacement is [tex](736000000m) i + (4200000m)j[/tex]

    [tex]\Delta t = t_f-t_i=68-0 =68[/tex]

    Since for t=0 the first equation, [tex]x=-34t^4-28t^3+7[/tex], will be 7.

    [tex]\Delta x = x_f-x_i = 736000000 - 7 = 375999993[/tex]

    So, [tex]v=\frac{\Delta x}{\Delta t}=\frac{375999993}{68}=10823529.31[/tex].
     
  7. Mar 22, 2010 #6

    rock.freak667

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    In your answers are you working with a specified degree of accuracy?

    Also if x=-34t4-28t3+7, how is your i component positive?

    Alos what is dx/dt and dy/dt equal to in terms of t?
     
  8. Mar 22, 2010 #7
    I think 2% is the accuracy tolerance.

    oops, I'm sorry, the answer has to be -43200000 i. :redface:

    [tex]\frac{375999993}{68}=10823529.31[/tex]

    If this not true then I have no idea...
     
  9. Mar 22, 2010 #8

    rock.freak667

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    If x=t5, then dx/dt is 5t4


    So if x=-34t4-28t3+7 and y =25t2+13t3+5

    what is dx/dt and dy/dt equal to?
     
  10. Mar 23, 2010 #9
    Thanks a lot it worked! :)

    Finally, the question asks "what is the particle's acceleration at 68 s?". If I divide the particle's velocity at 68 s by time=68 seconds, it doesn't give me the correct answer. So what formula do I need to use?
     
  11. Mar 23, 2010 #10

    rock.freak667

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    You use the same method.


    a(t)=(d2x/dt2)i+(d2y/dt2}j
     
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