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Kinematics problem

  1. Aug 30, 2008 #1
    A particle moves along a straight line with accel. a = 2[tex]\sqrt{v}[/tex] where v = [tex]\frac{dx}{dt}[/tex] is the velocity. At t = 2s, the particle is located at x= (64/3)m with velocity v = 16m/s. Find the location and acceleration at time t = 3s (hint: integrate)

    Now, if the differential equation is dv/dt = 2[tex]\sqrt{v}[/tex]. I don´t know how to solve such an equatoin (because of the square root). But then I thought v must be constant for the integration to work, but that would mean that the acceleration would be zero and that makes no sense. I need help.
  2. jcsd
  3. Aug 30, 2008 #2
    never mind, I figured it out as soon as I clicked submit post.

    dv over the sqrt(v) = 2dt

    No w8, this makes no sense, if you integrate you get 2sqrt(v) = 2t + c
    Last edited: Aug 30, 2008
  4. Aug 31, 2008 #3
    Square both sides of that expression and integrate again. You have enough data to find the constants.
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