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Acceleration of an object

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data


    Acceleration of an object is described by a function of x, a(x) = bx.
    (b is a constant 2 sec^-2)

    1. If the speed at x=1m is zero, what is the speed at x=3m?

    2. How long does the travel from x = 1m to x = 3m takes?


    2. Relevant equations

    chain rule : dr/dt = dr/ds*ds/dt


    3. The attempt at a solution

    I have thought of spring in a simple harmonic motion.
    But I could not take any further steps... Somebody help me please!
     
  2. jcsd
  3. Sep 21, 2008 #2

    CompuChip

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    The first formula you give says that
    x''(t) = b x(t).

    This is a differential equation. Do you know of any functions who are their own derivative (maybe up to a constant numerical factor?)
     
  4. Sep 21, 2008 #3
    I thought of sine and cosine fuction in a simple harmonic motion.
    But how can I relate them to the chain rule?
     
  5. Sep 21, 2008 #4

    CompuChip

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    Sine and cosine sounds good.
    I don't know what you want to do with the chain rule, but you can just use
    x(t) = A cos(t) + B sin(t)
    then differentiate to get the velocity x'(t) and acceleration y''(t) and use that to determine the constants A and B.
     
  6. Sep 21, 2008 #5
    What do you mean by y''(t)? Did you mean x(t)" ??

    I also don't know what the chain rule has to do with this problem but the instruction was to utilize this rule when solving the problem =<
    Could you please think a little bit more about the relevance of the chain rule to this problem?

    Also,
    Is there any reason that you included both cosine and sine fuction in the fuction of x(t)?
    Originally I intended to use x(t) = Asin([tex]\omega[/tex]t + [tex]\phi[/tex])....
    If there is any further advices or recommendations, please let me know

    Thanks!!
     
  7. Sep 22, 2008 #6

    CompuChip

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    Yes, that was a typo. I apologise.

    Hmm, maybe you have to use the chain rule with something like
    [tex]\frac{da}{dx} \frac{dx}{dt} [/tex]

    Well, my reason was the following (rather technical point): x'' = a x is a second order equation (because there are two derivatives), so there must be two independent solutions. In this case, they can be easily found to be sin(x) and cos(x). The general solution to the equation is then A sin(x) + B cos(x), with A and B two constants which have to be determined from initial conditions (for example, at t = 0, x must equal 23 and x' must equal 0). You need two conditions to fix A and B.
    In the expression you gave, there are also two undetermined constants, [itex]\omega[/itex] and [itex]\phi[/itex]. If you take [itex]\sin(\omega t + \phi)[/itex] you can use the two conditions to fix these constants.
    It doesn't really matter which of the two you take, in the end they are equivalent. In fact, a cosine is nothing more than a sine which is shifted a bit horizontally so you can always write cos(t) in the form [itex]\sin(t + \phi)[/itex] by choosing [itex]\phi[/itex] properly.

    I don't know how much you know about differential equations and related topics, but I hope that makes sense. I'll stick to [itex]\sin(\omega t + \phi)[/itex] :smile:
     
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