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Dynamics Question

  1. Jan 7, 2006 #1
    My hw problems looks like this:

    The acceleration of a particle is directly proportional to the square of the time t. When t=0, the particle is at x=36 ft. Knowing that at t=9 s, x=144 ft. and v=27 ft/s, express x and v in terms of t.

    I'm not exactly sure what directly proportional means. Can someone help me out?

    Also I have another question I'm stuck on:

    The acceleration of a particle is defined by the relation a= 12x-28, where a and x are expressed in m/s2 and meters, respectively. Knowing that v=8 m/s when x=0, determine (a) the maximum value of x, (b) the velocity when the particle has traveled a total distance of 3m.
  2. jcsd
  3. Jan 7, 2006 #2

    Tom Mattson

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    MathWorld is your friend.


    Note that [itex]a=\frac{d^2x}{dt^2}[/itex] and solve the resulting differential equation for [itex]x[/itex]. You should be able to answer the questions using that solution.
  4. Jan 7, 2006 #3
    Sorry, I'm kinda slow. I'm still not sure what they mean in the question with directly proportional. Does it mean that a = t squared?
  5. Jan 7, 2006 #4
    a = 12x - 28 \Rightarrow \frac{{d^2 x}}{{dt^2 }} - 12x = - 28

    You have a second order ODE. The ICs and the equation both suggest that you can consider this equation as one in which the independent variable (t) is missing.

    p\left( x \right) = \frac{{dx}}{{dt}} \Rightarrow \frac{{d^2 x}}{{dt^2 }} = \frac{d}{{dt}}\left( {p\left( x \right)} \right) = \frac{{dp}}{{dx}}\frac{{dx}}{{dt}} = p\frac{{dp}}{{dx}}

    The DE becomes:

    p\frac{{dp}}{{dx}} - 12x = - 28
    [/tex] which is separable after a little rearrangement.

    You should be able to solve it from there.

    (if you run into problems just reply to this thread - I haven't done the whole question so I don't know if the problem is easy or not)

    As for your other question I would interpret it as meaning [tex]a = kt^2 [/tex].

    Edit: Fixed statement above regarding the type of the DE.
    Last edited: Jan 8, 2006
  6. Jan 8, 2006 #5


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    Benny is right. It just means that [itex]a = kt^2 [/itex], where k is a constant. The use of the word 'directly' is rather redundant.
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