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Acceleration as a function of displacement

  1. Apr 20, 2007 #1
    1. The problem statement, all variables and given/known data

    All right, the problem is that I have a motor that's acceleration at a rate of a = 6 theta rad/s^2. For the bigger problem that this is part of, I need to find out the velocity function. How do I do this if a isn't given in terms of t?

    2. Relevant equations

    omega = d theta/dt a = d omega / dt

    3. The attempt at a solution

    All I can think of is that

    a = 6 theta
    integrate both sides for t

    v = 6 theta t
    theta = 3 theta t^2? that doesn't make sense!

    and theta is really theta(t), so how can I integrate that function if I don't know what i is? :confused:

    help me calculus gurus!
  2. jcsd
  3. Apr 20, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    I Think you have to use the result [tex]a=x''=\frac{d}{dx} (\frac{1}{2}v^2)[/tex].

    That result can be obtained as such - [tex]\frac{d}{dx} ((1/2) v^2) = \frac{d}{dv} ((1/2) v^2) \cdot \frac{dv}{dx}=v\frac{dv}{dx}[/tex] by the chain rule. Another application of the chain rule: [tex]v\frac{dv}{dx} = \frac{dx}{dt}\cdot\frac{dv}{dx} = \frac{dv}{dt}=a[/tex]
  4. Apr 20, 2007 #3
    actually as it turned out:

    a(t) = 6 theta

    better put as

    a(t) = 6 theta(t)

    a(t) is just the second derative of theta(t), so

    theta''(t) = 6 theta(t)

    tossing away the 6 mentally for a second, what function equals its own second derative? e does! so let's play with that and see what we get

    if theta(t) = e^t, then theta''(t) = e^t. Okay, but we need to bring that 6 back in. So, theta (t) = e^sqrt6 t, theta'(t) = v(t) = sqrt6 e^sqrt6 t, and v''(t) = a(t) = 6 e^sqrt6 t.

    To check it:a(t) = 6 e^sqrt6 t. e^sqrt6 t = theta(t) substitute and we get back to the original a(t) = 6 theta.

    yay :) using this I got the right answer for my problem!

    But I'm on another problem (last one, promise) with similar premises.

    omega(t) = 5 (theta(t))^2 is what I'm given.

    So theta'(t) = 5 (theta(t))^2

    Toss away the 5, we can deal with it later...but what function equals its derative when you square it? :confused:
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