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Finding the location of a kinematic particle in motion based on time.

  1. May 27, 2014 #1
    Hello there, I have a problem that I've been trying to solve for some time now and I find myself struggling.

    I am trying to develop a method of calculating the position of a particle using kinematic motion in a 2D Cartesian coordinates system given a set of starting values based on how much time has passed. In my case the particle is being used to represent a vehicle such as a car or otherwise that moves only forward and backward with no sideways motion and uses rotation to turn.

    Just for further clarification I'm going to point out that given the above description, the rotation of the particle is important as it defines what is "forward" and "backward" for the particle and as such changes the line of motion.

    The given values I have to work with are: [itex]v_i[/itex] Initial Velocity, [itex]a_f[/itex] Forward Acceleration, [itex]ω[/itex] Angular Velocity, [itex]\alpha[/itex] Angular Acceleration, and [itex]t[/itex] Time.

    The starting coordinates and rotation of the particle are all assumed to be starting at zero.
    [itex]x = 0[/itex], [itex]y = 0[/itex], [itex]\theta = 0[/itex]

    If the particle is traveling in a straight line with no angular velocity or acceleration, calculating the position of the particle is easy using the kinematic formula for displacement.
    [itex]d = v_i\times t + 1/2\times a_f\times t^2[/itex]

    And if the particle is traveling in a perfect circle with no acceleration it's not so bad because I can find the circumference, radius, and origin of the circle just looking at the forward and angular velocities and find its place on the circle based on that.

    But if the particle has a forward acceleration or angular acceleration while turning, it is no longer a circle but a spiral path of motion. This is where I am failing to find a solution.

    Any assistance anyone could lend would be greatly appreciated!
    Thank you in advance, I will be back to check this post tomorrow.
     
  2. jcsd
  3. May 27, 2014 #2
    Don't know if I get what you ask, but if you know

    [tex]\vec{a}(t) = a_x(t)\vec{i} + a_y(t)\vec{j}[/tex]

    (i.e. if you know the acceleration vector at each time)

    and you also know

    [tex]\vec{r}(t_0)[/tex] and [tex]\vec{v}(t_0)[/tex]

    (i.e. the position vector and velocity vector at one given time instant t_0 )

    then you know [tex]\vec{r}(t)[/tex] and [tex]\vec{v}(t)[/tex] at any time t.


    Because:


    [tex]\vec{v}(t) = \vec{v}(t_0) + \int_{t_0}^t \vec{a}(s)ds[/tex]


    and


    [tex]\vec{r}(t) = \vec{r}(t_0) + \vec{v}(t_0) (t-t_0) + \int_{t_0}^t\left(\int_{t_0}^s \vec{a}(w)dw\right)ds[/tex]
     
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