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Homework Help: Motion problem

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A body experiences acceleration "a" given by the expression [itex]a=At-Bt^2[/itex] where A and B are constants and t is time. If at time t=0, the body has zero displacement and velocity, at what next value of time does the body again have zero displacement?


    2. Relevant equations

    a is in m/s^2
    v is in m/s
    d is in m


    3. The attempt at a solution
    when t=0
    displacement
    [itex]at^2 = d = 0 = At^3 - Bt^4[/itex]
    velocity
    [itex]at = v = 0 = At^2 - Bt^3[/itex]
     
  2. jcsd
  3. Jan 28, 2012 #2

    Doc Al

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    Staff: Mentor

    Please write the full expressions for v(t) and x(t).
     
  4. Jan 28, 2012 #3
    v(t) is velocity
    dv/dt = At - Bt^2
    dv = dt(At - Bt^2)

    [itex]v(t) = \frac{At^{2}}{2} - \frac{Bt^{3}}{3} + Constant[/itex]

    x(t) is displacement
    dx(t)/dt = (At^2)/2 - (Bt^3)/3 + Constant
    dx(t) = ((At^2)/2 - (Bt^3)/3 + Constant)dt

    [itex] x(t) = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + Constant [/itex]
     
  5. Jan 28, 2012 #4

    Doc Al

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    Staff: Mentor

    Good. What must each "Constant" equal?
     
  6. Jan 28, 2012 #5
    initial velocity
    [itex]v(t) = \frac{At^{2}}{2} - \frac{Bt^{3}}{3} + v_{o}[/itex]

    initial displacement
    [itex] x(t) = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + x_{o} [/itex]

    at t = 0;
    For v(t):
    [itex]v(0) = v_{o}[/itex]

    since [itex]v_{o}[/itex] = 0 at time t = 0
    [itex]v(0) = 0[/itex]

    For x(t):
    [itex]x(0) = x_{o}[/itex]

    since [itex]x_{o} = 0[/itex] at time t = 0
    [itex]x(0) = 0[/itex]


    when :
    x(t) = 0
    [itex] x(t) = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + x_{o} [/itex]

    [itex] 0 = \frac{At^{3}}{6} - \frac{Bt^{4}}{12} + 0 [/itex]

    [itex] \frac{At^{3}}{6} = \frac{Bt^{4}}{12}[/itex]

    [itex] \frac{A}{6} = \frac{Bt}{12}[/itex]

    [itex] \frac{2A}{B} = t[/itex]
     
  7. Jan 28, 2012 #6

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
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