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Derive Velocity Equation

  1. Jan 31, 2007 #1
    Given a(t) = -Cv^2(t) where C is a constant, express v(t) and x(t) as explicit functions of time t. Assume v(0) = v0

    So I tried integrating both sides of the equation to get:
    v(t)+C' = -C integral v^2(t) dt

    but then I how am I supposed to integrate v^2(t) dt...

    Also, I thought I could do it another way:

    a(t) = -Cv^2(t)
    v(t)/t = -Cv^2(t)
    -1/(Ct) = v(t)

    -1/(Ct) = v(t)
    -1/(Ct) = x(t)/t
    -1/C = x(t)

    Well the distance can't be constant so I guess that doesn't work either. I don't even know why this 2nd method doesn't work... It looks fine to me. Any help would be appreciated.
  2. jcsd
  3. Jan 31, 2007 #2


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    Staff: Mentor

    You are correct in using integration to go from a(t) to an expression for v(t).

    Are you just asking how to do the indefinite integral [tex] \int v^2 (t) dt [/tex] ?
  4. Jan 31, 2007 #3
    Yeah, I don't get how to integrate that. Wouldn't it be something like v^3(t)/(3a(t)) but where does the a(t) come from? It's not a constant so you just can't add it in right?
  5. Jan 31, 2007 #4
    notice that:
    how can you solve this differential equation?

    hint: separation of variable!
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