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Homework Help: Particle problem

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle moves along the x axis. Its position is given by the equation x=1.5+2.50t-3.9t^2 with x in meters and t in seconds.
    a. Determine its position when it changes direction
    b. what is its velocity when it returns to the position it had at t=o

    2. Relevant equations



    3. The attempt at a solution

    a. I would need to take the derivative and set it equal to 0, right?

    f'(x)=2.50-7.8t=0
    t=(-2.5/-7.8)

    I then need to plug t back into the original equation, right?

    Would the position when it changes direction be 1.80?

    b. What is the velocity when it returns to the position it had at t=0?

    2.50-7.8(0)=2.5

    Thank you very much
     
  2. jcsd
  3. Feb 28, 2008 #2
    Or would this work?

    x=1.50-2.50t-3.90t^2

    2.5-2(3.90)t

    0=2.5-7.8t

    t=.321

    x=1.5+2.5(.321)-3.9(.321)^2
    x=1.9m

    vf^2=vi^2+2a(xf-xi)
    =2.50^2+2(7.8)(0+1.9)
    =6.0

    Thank you very much
     
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