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Vector notation and force and force Help- test this morning

  1. Oct 24, 2007 #1

    ~christina~

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    [SOLVED] vector notation and force and force Help- test this morning..

    1. The problem statement, all variables and given/known data
    A time dependent force F(t)= 20t N/s acts along the possitive y axis on a 4.00kg object. The object starts at the origin with the initial velocity

    [tex]\vec{}v(0)= (1.00m/s)) \hat{}i-(1.00m/s) \hat{}j[/tex].

    a) find the velocity of the object as a function of time.
    b) find the position of the object as a function of time
    c) what is the speed of the object at time t?

    2. Relevant equations
    ?? the ones given?



    3. The attempt at a solution

    I have no idea where to start..it seems simple enough..


    can someone please please please help me out?
    by this morning to be exact..(I have a test)
     
  2. jcsd
  3. Oct 24, 2007 #2

    Doc Al

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    You have the force and the mass, thus you can figure out the acceleration. Given the acceleration as a function of time, how would you find the velocity as a function of time? And then the position? (Hint: Only one component is accelerated. Hint2: A bit of calculus is required.)
     
  4. Oct 24, 2007 #3

    ~christina~

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    Hm..
    so ..acceleration...

    F(t)= 20 t N/s
    a= ?
    m= 4.0kg

    F= ma (not sure how t would get into that unless I plug in 20 t N/s... (I think the

    notation is confusing me)

    (20t N/s)/ 4.0kg= a

    a= 5 t m/s ?? (not sure about the notation once again...)

    I guess if this is correct that I would just do integration to find the v and then the x

    however why was the initial velocity given in the form of vector notation ?

    I was thinking of using that....

    thanks
     
  5. Oct 24, 2007 #4

    Doc Al

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    The acceleration is 5t m/s^2. But in what direction? They tell you: "acts along the positive y axis". That means the acceleration should be written as [itex]5 t \hat{j}[/itex] m/s^2.

    Yes.

    Because direction matters!
     
  6. Oct 24, 2007 #5

    ~christina~

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    so I don't actually use that equation?

    anyways...

    a(t)= 5tj m/s^2

    v(t)= 2.5tj^2 + t

    x(t)= .83 tj^3 + 1/2 t^2 + t

    Is this how it's supposed to look?

    (Once again the units are getting to me...couldn't figure out the units for the v but technically it should be m/s and x should be in m)

    Am I supposed to have anymore j's in the equations??
     
  7. Oct 24, 2007 #6

    Doc Al

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    Seems like the j notation is messing you up a bit.
    a(t)y = 5t (direction: +y axis (a.k.a j); units: m/s^2)
    Integrate a(t) once to get v(t):
    v(t)y = 2.5t^2 + C (y-component of velocity)

    Use the given initial velocity to find the integration constant C:
    v(t)y = 2.5t^2 - 1 (y-component only)

    The complete v(t) is:
    v(t) = (1) i + (2.5t^2 -1) j


    Integrate v(t) to get the y-component of position:
    y(t) = (2.5/3)t^3 -t + C

    Use the given info to find the integration constant:
    y(t) = (2.5/3)t^3 -t

    The x-axis motion is just constant speed:
    x(t) = (1)t = t

    In vector notation, the position as a function of time would be:
    (t) i + ((2.5/3)t^3 -t) j
     
  8. Oct 24, 2007 #7

    ~christina~

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    good grief...I wasn't even close... it was more complicated than I thought it would be

    for part c where I have to find the speed of the object at time t

    wouldn't I take the

    v(t) = (1) i + (2.5t^2 -1) j

    and then find the V from the x and y component's i and j ?

    rad ( 1^2 + (2.5t^2 -1)^2) = ....

    technically that's what I would think but that looks well...complicated...

    Is that correct?

    Thanks alot
     
  9. Oct 24, 2007 #8

    Doc Al

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    Yes, that's correct.
     
  10. Oct 24, 2007 #9

    ~christina~

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    Thank you Doc Al
     
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