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Motion in Space: Velocity and Acceleration

  1. Oct 4, 2014 #1
    1. The problem statement, all variables and given/known data
    The position function of objects A and B describe different motion along the same path for t >= 0.
    A: [itex] r(t) = cos(t)i + sin(t)j [/itex]
    B: [itex] r(t) = cos(3t)i + sin(3t)j [/itex]
    a) Sketch the path followed by A and B
    b) Find the velocity and acceleration of A and B and discuss the distance
    c) Express the acceleration of A and B in terms of the tangential and normal components and discuss the difference

    2. Relevant equations
    T(t) = velocity/speed (unit tangent vector)
    N(t) = T'(t)/ absolute value of T'(t) (unit normal vector)
    [itex] a_T = v' [/itex]
    [itex] a_N = kv^{2} [/itex]
    [itex] a = v'T + kV^{2}N [/itex] (k is curvature)
    3. The attempt at a solution
    Ok, this was a problem for my multivariable calculus class but if the mods feel that it belongs in a physics subforum then by all means move it there :).
    a) I know that both functions are circles with radius r = 1, traced out in a counterclockwise direction
    b) [itex] v(t)_A = \int r(t) dt = -sin(t)i + cos(t)j [/itex]
    [itex] v(t)_B = \int r(t) dt = -3sin(3t)i + 3cos(3t)j [/itex]
    [itex] a(t)_A = \int v(t) dt = -cos(t)i - sin(t)j [/itex]
    [itex] a(t)_B = \int v(t) dt = -9cos(3t)i - 9sin(3t)j [/itex]
    The difference between the two is that the 'B' function is traced out much faster than the 'A' function
    c) For A:
    [itex] T(t) = -sin(t)i + cos(t)j [/itex]
    [itex] N(t) = -cos(t)i - sin(t)j [/itex]
    The magnitude of T, N is equal to 1, as is the curvature k
    Tangential acceleration is:
    [itex] a_T = -cos(t)i - sin(t)j [/itex]
    Normal acceleration is:
    [itex] a_N = sin^2(t)i + cos^2(t)j [/itex]
    For B:
    [itex] T(t) = -sin(3t)i + cos(3t)j [/itex]
    [itex] N(t) = -cos(3t)i - sin(3t)j [/itex]
    The magnitude of T, N is equal to 3 and the curvature k is equal to 1
    [itex] a_T = -9cos(3t)i - 9sin(3t)j [/itex]
    [itex] a_N = 9sin^2(3t)i + 9cos^2(3t)j [/itex]
    The tangential component shows the rate of change of velocity and the normal component shows the rate of change of direction. In this case the tangential components of the two functions are equal. However B has a larger normal component, therefore it changes direction more quickly. This problem was worth 20 points; however, I only received 16... I am allowed to make a correction for full credit so I was hoping somebody could tell me where I went wrong :)
  2. jcsd
  3. Oct 4, 2014 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    What's the definition of velocity, in terms of position? What's the definition of acceleration, in terms of velocity?

    What you wrote for the answers to the velocity and acceleration are OK, but the definitions you used are incorrect.
  4. Oct 5, 2014 #3
    Thank you for pointing that out, I'll make the changes in a moment...know that when I was finding velocity and accel I did differentiate. I did not integrate
  5. Oct 5, 2014 #4
    I can't seem to edit my original post, perhaps the mods would be so kind as to change my mistake :)
  6. Oct 5, 2014 #5
    Also, besides the definition mistake was there anything else you noticed that was wrong? I have double checked but I believe that I answered everything correctly
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