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Equations (calc)

  1. Jan 8, 2004 #1
    A particle moves along the x-axis in such a way that its acceleration at time t for t>= 0 is given by a(t)=4cos(2t). At time t=0, the velocity of the particle is v(0)=1 and its position is x(0)=0.
    a. Write an equation for the velocity of v(t) of the particle.
    b. Write an equation for the position x(t) of the particle.
    c. For what values of t, 0<=t<=pi, is the particle at rest

    -------------------------------------
    a = dv/dt
    4*cos(2t) = dv/dt
    v = Integral[4*cos(2t) dt] + C
    v = 2 * sin(2t) + C
    t = 0 -> v = 1.
    1 = 2 * sin(2*0) + C
    C = 1
    v(t) = 2 * sin(2t) + 1

    How would I go about x(t)?

    And for c. 2sin(2t)+1=0 sin2t=-1/2
    arcsin2t=(1/2)
    t=-.261799?
     
  2. jcsd
  3. Jan 9, 2004 #2
    [tex]\frac{d^2x}{dt^2}=4\cos(2t)[/tex]
    [tex]\frac{dx}{dt}=\int 4\cos(2t)dt+c = 2sin2t+c[/tex]
    [tex]x=\int {2sin2t+c}dt+k[/tex]
     
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