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Homework Help: Dynamics with complex equations

  1. Sep 16, 2010 #1
    A particle starts moving from the position
    r= -2i+j

    with its velocity given as a function of time, t, as

    What is the distance of the particle from the origin at time t = 10 s?

    The i's a j's both have hats if that makes any difference.

    My attempt:
    distance = velocity x time
    so multiply the v equation by t and then subsitute 10s into it.
    This will give the distance travelled in both the x and y direction add these values to the values given in the first equation.
    So (5t^2)*t + t*2j

    then sub in 10 and add the i and j values to their respective values in the first equation to find the final displacement.

    I'm unsure if this is correct or not.
  2. jcsd
  3. Sep 16, 2010 #2


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    No, "displacement", not "distance" equals velocity x time, and even that is only true in the case of constant velocity.

    Instead, you will want to integrate the velocity to get the position as a function of time, and then determine the distance from the origin at a given time by using the usual formula for the distance between 2 points.

  4. Sep 16, 2010 #3
    is this not exactly what you're telling me to do?
    (5t^2)*t + t*2j
  5. Sep 16, 2010 #4


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    No, do you know how to integrate?
  6. Sep 16, 2010 #5
    Sorry didn't read what you first said properly.
    So once the integral is found, and the equation becomes something which represents the displacement of the object, can't I just sub 10 into this equation and then add -2i and 1j to the resulting i and j values?
  7. Sep 17, 2010 #6


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    Sort of, your integration (like any integration) will have a constant on the end. In this case, the constant will be a vector and equal to the initial position of the particle:

    [tex]\int_0^t \textbf{v}(t')dt' = \int_0^t \frac{d\textbf{r}(t')}{dt'}dt' = \textbf{r}(t)-\textbf{r}(0)[/tex]
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