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Homework Help: How to find kinetic energy of a particle as a function of time?

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    The position of a 5kg particle moving in the x-y plane varies as a function of time according to the expression

    r = [(3t + 1)m]i + [(-4t^2 +5t + 2)m]j

    (r is a vector I just didn't know how to make the arrow over head)


    Find the kinetic energy of the particle as a function of time


    2. Relevant equations

    k = 1/2 mx+b



    3. The attempt at a solution

    d/dx r = [(3t + 1)m]i + [(-4t^2 +5t + 2)m]j

    v = 3i + (-8t +5)j

    v =(3i^2 + (-8t +5)j^2)^(1/2)?????????????

    ......................................

    3t+1 = 0
    t= 1/3 for x

    -4t^2 +5t + 2

    t= -.44 or t= 1.69 for y????????????

    Very confused don't know what vector r is. Is it already a function of time or what. I'm very confused and don't understand how to solve please help
     
  2. jcsd
  3. Apr 24, 2010 #2

    rock.freak667

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    KE=1/2 mv2

    and v2 is the same as v.v
     
  4. Apr 24, 2010 #3
    so ...

    v = 3i * (-8t +5)j

    v = (-24t +15)^1/2????????????? and if so were do I go from there?
     
  5. Apr 24, 2010 #4

    rock.freak667

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    If v = 3i + (-8t +5)j, then compute v.v, the dot product of v and v.
     
  6. Apr 24, 2010 #5
    I'm not sure if i'm correct in my assumption that v = 3i + (-8t +5)j also I don't see how the dot product could be used since we have the variable t ?
     
  7. Apr 24, 2010 #6

    ideasrule

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    What was that "m" in your initial equation for r? If that shouldn't have been there, your expression for v is correct.

    Who cares about the t. Just do v dot v the usual way and see what happens. Do you get something with t in it? That's how v^2 varies with t.
     
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