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The position function from given velocity or acceleration function

  1. Mar 20, 2010 #1

    Find the position function from the given velocity or acceleration function.

    a(t) = <e^-3t,t,sint>, v(0)=<4,-2,4>, r(0)=<0,4,-2>


    To find the answer the integral must be taken...

    Integral of a(t) = <-(1/3)e^(-3t), (1/2)t^2, -cos(t)>

    Since taking it with respect to t, it becomes velocity

    (acceleration = x/t/t distance over time 2)

    Integral of acceleration with respect to t is... (xt^-1, x/t)

    Since that is just the velocity function, you do have an initial velocity at v(0)...(time at 0)
    adding these to each part of the integral.

    The velocity function is this: <-(1/3)e^(-3t)+4, (1/2)t^2-2, -cos(t)+4>

    Again, the integral must be taken and it becomes the position function

    x(t) = <(1/9)e^(-3t)+4t, (1/6)t^3-2t, 4t-sin(t)>

    And add constants again r(0)

    therefore the answer r(t) (x(t)) = <(1/9)e^(-3t)+4t+0, (1/6)t^3-2t+4, 4t-sin(t)-2>

    BUT my teacher said the first and the third portions are wrong? I do not understand how.
  2. jcsd
  3. Mar 20, 2010 #2


    Staff: Mentor

    Put the constants in right away.
    v(t) = <-(1/3)e-3t + c1, (1/2)t2 + c2, -cos(t) + c3>
    You're given that v(0) = <4, -2, 4>, so you can solve for the three constants. It's not a simple matter of adding them in as you did.

    Do the same thing when you find r(t).
  4. Mar 20, 2010 #3
    so I just put the constants in and then solve? I'm still confused.
  5. Mar 20, 2010 #4


    Staff: Mentor

    You are given that v(0) = <4, -2, 4>, and from your work you can substitute 0 for t to get v(0) = <-(1/3)e-3*0 + c1, (1/2)02 + c2, -cos(0) + c3>.

    From this you can solve for the constants.
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