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Basic Velocity -> Speed -> Position of particle

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle's trajectory is described by x=((1/2)*(t^3)-2t^2)m and y=((1/2)*(t^2)-2t)m, where t is in s.

    a) What are the particle's position and speed at t=0s and t=4s?

    b) What is the particle's direction of motion, measured as an angle from the x-axis, at t=0s and t=4s?


    2. Relevant equations
    Listed above


    3. The attempt at a solution
    I feel like I am over-thinking the problem. As I understand it the equations provided are velocities in the directions of the x and y axies. So, the integral of the equation for 'x' would give me the speed. Integrating it a second time would give me position. I can work through the integrals but something doesn't feel right about the relationship between the work I'm doing and the material in the chapter.

    integral of x = 1/8 t^4 - 2/3 t^3 and second integral of x = t^5/40 - t^4/6

    integral of y = 1/6 t^3 - t^2 and second integral of y = t^4/24 - t^3/3

    Plugging in the values of 0 and 4:

    The position at 0 seconds = (0,0)
    The speed at 0 seconds = 0 m/s

    The position at 4 seconds = (-17 1/15, -10 2/3)
    The speed at 4 seconds = -10 2/3 m/s

    The second part I can simply use the triangles created to find the angle in relation to x.

    Am I on the right track?
     
  2. jcsd
  3. Sep 11, 2011 #2

    lewando

    User Avatar
    Gold Member

    Workin' in the wrong direction (differentiate, not integrate). dx/dt = v, dv/dt = a
     
  4. Sep 11, 2011 #3
    Bah, of course. Thank you!
     
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