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Find all time t at which the particle is accelerating in a direction

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle travels along a path C in R^3 which velocity v(t) =<t^2, cos(pi*t), t> at time t. Assume that the particle's initial position is at the point p=(1, 0, -1). Find all times t at which the particle is accelerating in a direction that is perpendicular to the plane 4x+2z= squareroot(squareroot(squareroot(2))).




    2. Relevant equations



    3. The attempt at a solution


    I have calculated that the point f(t) = (1/3*t^3, sin(pi*t)/pi +1, .5t^2 -1) at time t, that the acceleration a(t) = <2t,-pi*sin(pi*t), 1> for time t, and that the normal of the plane is n=<4,0,2> and so the direction vector of the particle should be equal to <4,0,2> but i don't know what to do after that?

    thanks!
     
  2. jcsd
  3. Sep 28, 2009 #2

    Dick

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    You don't need f(t). What you do need is for a(t) to be parallel to n. That means that a(t) and n are multiples of each other. For one thing the ratio of the x and z components of each vector must be the same. What does that tell you about t?
     
  4. Sep 28, 2009 #3
    Okay, so then that would mean that t would have to be any real number that satisfies any multiple of <4,0,2> .. which would only be when t is 1, since we have a constant 1 in the acceleration formula. Am i on the right track?

    Thanks so much!:)
     
  5. Sep 28, 2009 #4

    Dick

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    Right. Then you also need to check that at t=1, that the y component of a(t) is zero.
     
  6. Sep 28, 2009 #5
    Ah! Okay, i understand! :) yay! thanks!
     
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