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Location-dependent acceleration

  1. Sep 30, 2004 #1
    I'm having a really hard time figuring this one out. At first glance it seems very trivial, but I think it is not. This is not a homework assignment but rather a calculation I need to do for personal reasons.

    The problem is as follows:
    An object is accelerated along a distance of s. We have a linear relationship between acceleration and the distance travelled. We also know the acceleration at the beginning and at the end of s to be a0 and a1. Hence the acceleration at the distance s1 is (for s1<s or s1=s)

    a = a0 + (a1-a0)/s * s1

    Now I would like to know the velocity of the object at s1 and the time it takes to get there. Note that the acceleration only changes linearly with the distance but not with the time. And how can you derive the functions a(t), v(t) and s(t) with the information given? What am I missing?

    If you need more information on the problem, feel free to ask.
    Any help would be much appreciated. Thanks.
    Last edited: Sep 30, 2004
  2. jcsd
  3. Sep 30, 2004 #2


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    Write [tex]a=\frac{d^2 s_1}{dt^2} [/tex] and setup the differential equation for [itex] s_1(t)[/itex]. If the coefficient of s1 is negative, it looks like you will get something like a harmonic oscillator whose equilibrium point is undergoing constant acceleration.
  4. Sep 30, 2004 #3


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    This definitely will NOT be SMH! The exponent of the exponential in the solution of the DE should be real, not imaginary as required for SMH. Therefore the position should increase exponentially with time.
  5. Oct 1, 2004 #4
    First of all, thanks for replying.

    My problem lies in setting up the DE. Normally, t is the only independent variable. However, in this situation, s is also independent in a(s) and therefore s(t) will contain the function a(s). I'm stuck at this point.

    Could you please elaborate on setting up the differential equation?
  6. Oct 1, 2004 #5


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    s is certainly NOT independent of acceleration! I don't think that's what you meant to say. What you are saying is that a is dependent on s. Okay, that's precisely what differential equations are designed for:

    By definition, [itex]\frac{d^2s_1}{dt^2}= a[/itex]. Here, you are saying that
    "a = a0 + (a1-a0)/s * s1" . ( I was a bit puzzled as to what "s" and "s1" were. I would consider it more usual to use "s1" to mean a fixed distance and "s" to mean the variable but you have those reversed here. Okay, as long as it is clear that s is a constant and s1 is the variable.) You have [itex]\frac{d^2s_1}{dt^2}= a_0+ \frac{a_1-a_0}{s} s_1[/itex]. That's a reasonably straightforward linear, non-homogeneous, differential equation.

    Rewrite it as [itex]\frac{d^2s_1}{dt^2}- \frac{a_1-a_0}{s}s_1= a_0[/itex].

    The "characteristic equation" for that is [itex]r^2- \frac{a_1-a_0}{s}= 0[/itex] which has solutions [itex]r= \sqrt{\frac{a_1-a_0}{s}}[/itex] so the general solution to the corresponding homogenous equation (= 0 rather than = a0) is
    [itex]s_1(t)= C_1e^{\sqrt{\frac{a_1-a_0}{s}}t}+ C_2e^{-\sqrt{\frac{a_1-a_0}{s}}t}[/itex].

    Since a0 is a constant, we can try s(t)= A, a constant as the "particular solution" of the entire equation. Since the second derivative of the constant A is 0, the equation becomes [itex]- \frac{a_1-a_0}{s}A= a_0[/itex] so [itex]A= \frac{a_0s}{a_1-a_0}[/itex].

    Adding that to the general solution to the homogeneous equation,
    [itex]s_1(t)= C_1e^{\sqrt{\frac{a_1-a_0}{s}}t}+ C_2e^{-\sqrt{\frac{a_1-a_0}{s}}t}+ \frac{a_0s}{a_1-a_0}[/itex].

    You can now calcuate the velocity and acceleration at any time t by differentiating that.
    Last edited: Oct 1, 2004
  7. Oct 1, 2004 #6
    That was exactly what I was looking for, thank you. I think that I wasn't able to solve this because I'm fairly new to calculus.
  8. Oct 1, 2004 #7


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    Is it given or obvious that [tex]a_1-a_0 > 0[/tex]?
    Or is this just being assumed by everyone?
  9. Oct 2, 2004 #8
    It is not given that a1-a0=0. Where does he state that?
  10. Oct 2, 2004 #9


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    I asked about a1-a0 > 0.
    If a1-a0 >0, then the coefficient of s1 is positive and, in HallsofIvy's calculation, r^2=(a1-a0)/s is positive (since a distance s is positive). So, r is real and one obtains the exponential solutions shown.
    If, however, a1-a0 < 0, then the coefficient of s1 is negative... so r is imaginary, leading to sinusoidal solutions. This is the possiblity I alluded to in my first response.
  11. Oct 3, 2004 #10
    I think s1 must be able to remain positive even when a1-a0<0. That would mean that the acceleration decreases with the distance and nothing else. I don't see why a1-a0<0 leads to s1<0, could you explain that?

    In fact, as I said in the initial post, the problem I'm trying to solve is for 0<s1<s only, because acceleration only occurs in the section s, and it increases or decreases during the period it moves through s, depending on the coefficients a0 and a1, which specify the acceleration at the beginning and at the end of s.

    However, I've run into another problem. Whenever I plot s1(t) HallsOfIvy gave me, I get the wrong graph. s1(0) should be 0, but it isn't. Could you please explain what C1 and C2 stand for and what values they have?

    Thank you.
  12. Oct 3, 2004 #11


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    C1 and C2 are constants that must be determined by your initial conditions. To complete this problem you need to specify 2 on them. An obvious one is s1(0)=0. This condition will force
    [tex] C1 + C2 +\frac{a_0s}{a_1-a_0}=0 [/tex]

    Your problem statement does not specify, or even give a feel for what the second condition might be. I can think of 2 obvious ways.

    1. Specify an initial velocity

    [tex] \dot{s1}(0) = V_0 [/tex]

    2. Specify the total time (T) required to travel from 0 to S

    [tex] s1(T) = s [/tex]

    Either will work, the 2nd leads to some ugly expressions for your final solution. Go with the first.
    Last edited: Oct 3, 2004
  13. Oct 9, 2004 #12
    Ok. Thank you for the hint. Everything has worked out concerning the DE.

    One more question has come up:

    When I set the initial values so that a0>a1, which means that acceleration decreases with the distance travelled, I get complex values for s[t] and its derivatives. What does the imaginary part of the number represent? What is the distance (or velocity or acceleration) actually? The absolute value of the complex number or only the real part?
  14. Oct 9, 2004 #13


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    I believe that there is some miscommunication here.
    You first wrote
    and, later, I replied
    In my statement, "the coefficient of s1" means (a1-a0)/s, that is, "that factor which multiplies s1".... not s1 itself.
  15. Oct 10, 2004 #14
    robphy, of course you're right, I misunderstood you.

    Back to my question:
    What does the imaginary part of the solution represent?
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