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Deriving Vector and position vectors from Force vector

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    An object of mass m is at rest at equilibirum at the origin. At t=0, a new force [tex]\vec{F}(t)[/tex] is applied that has components
    [tex]F_{x}[/tex](t) = k1+k2y [tex]F_{y}[/tex](t)=k3t
    where k1, k2, and k3 are constants. Calculate the position r(t) and velocity v(t) vectors as functions of time.
    2. Relevant equations
    We know that Force = ma.
    and that :
    [tex]\int \vec{a}dt = \vec{v}(t)[/tex]
    [tex]\int \vec{v}dt = \vec{r}(t)[/tex]

    3. The attempt at a solution

    I'm not sure if I'm doing this right but
    I did
    [tex]\vec{F}(t)[/tex] =[tex](k1+k2y)\hat{i}[/tex]+[tex](k3t)\hat{j}[/tex]
    I divided the Force vector by the scalar value of m, the mass to get [tex]\vec{a}[/tex]
    [tex]\vec{a}(t)[/tex] = [tex](k1+k2y)/m\hat{i}[/tex]+[tex](k3t)/m\hat{j}[/tex]

    [tex]\vec{v}(t)[/tex]=[tex]\int \vec{a}dt[/tex] = [tex](k1+k2y)t/m \hat{i}[/tex] + [tex](k3t^2)/2m \hat{j}[/tex]

    then integrate the velocity vector to get the position vector, am I doing this right at all?
  2. jcsd
  3. Apr 18, 2008 #2


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    Gold Member

    You reasoning seems good, but you are forgetting some constants when you integrate.
  4. Apr 18, 2008 #3
    Thanks, just noticed that myself as well.
  5. Apr 19, 2008 #4
    note that in Fx(t) = k1 + k2y, the y is not a constant, so the integral of k2y dt is not equal to k2yt
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