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Need help with Kinematics

  1. Nov 12, 2006 #1
    A rocket initially at rest accelerates with constant net acceleration B from t = 0 to t = T1 at which time the fuel is exhausted. Neglect air resistance. If the rocket's net acceleration, B, is equal to 1.0g, find an expression for the total time [itex] T_{max} [/itex] (from liftoff until it hits the ground).

    So [tex] T_{max} = T_{1} + t [/tex]

    [tex] \frac{1}{2}BT_{1}^{2} + BT_{1}t - \frac{1}{2}gt^{2} = 0 [/tex]

    I know that [tex] t = \frac{BT_{1}}{g} [/tex]

    What do I do from here? I got [tex] T_{max} = 2T_{1} = 2 t [/tex]

  2. jcsd
  3. Nov 12, 2006 #2


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    Looks okay, since, when the rocket is left without any fuel, its motion is a free fall with y(t) = yo + vo t - 1/2 g t^2, where yo is the well-known height yo = y(T1) = 1/2 B T1^2 = 1/2 g T1^2, and v0 = BT1. You're on the right track. Now just solve for t, and plug it into Tmax = T1 + t.
  4. Nov 13, 2006 #3
    So is the equation [tex] \frac{1}{2}gt^{2} + gt^{2} - \frac{1}{2}gt^{2} [/tex]?
  5. Nov 13, 2006 #4


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    The equation is [tex] \frac{1}{2}gT_{1}^{2} + gT_{1}t - \frac{1}{2}gt^{2} = 0 [/tex], as you already wrote. Now solve for t.
  6. Nov 13, 2006 #5
    If [tex] B = g [/tex] how do we get [tex] t^{2} - 2T_{1}t - T_{1}^{2} = 0 [/tex]?

    I factored the equation: [tex] g(\frac{1}{2}T_{1}^{2} + T_{1}t - \frac{1}{2}t^{2}) = 0 [/tex]. I guess they used the relation that [tex] t = T_{1} [/tex] and multiplied both sides by 2?
    Last edited: Nov 13, 2006
  7. Nov 13, 2006 #6


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    Again, solve the equation (i.e. find the roots of the parabola) [tex] \frac{1}{2}gT_{1}^{2} + gT_{1}t - \frac{1}{2}gt^{2} = 0 [/tex] for t. It is the only unknown.
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