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Please help me understand this

  1. Nov 23, 2004 #1
    Uniform motion:

    dv/dt = 0

    ds/dt = v (a constant)

    integral(dx) = v. integral(dt) => x = x(subsript 0) +vt ???

    Uniform accelerated motion: dv/dt = constant

    Equations of motion are:

    dv/dt = a => v = v(subscript 0) + a.t ???

    dx/dt = v => x = x(subscript 0) + v(subscript 0).t + 1/2.a.t^2 ???

    v.dv/dt = a => v^2 = v(subscript 0)^2 + 2.a.(x-x(subcript 0)) ???

    I don't understand the lines with question marks next to them.
  2. jcsd
  3. Nov 23, 2004 #2

    matt grime

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    They've just got an integral equation, from a differential equation, and solved it, albeit with the standard abuse of notation.
  4. Nov 23, 2004 #3
    I still don't understand.

    How do you get from ds/dt = v (a constant) to integral(dx) = v. integral(dt) to x = x(subsript 0) +vt?
  5. Nov 23, 2004 #4
    Can you explain it step by step?
  6. Nov 23, 2004 #5
    Please spoon-feed me.
  7. Nov 23, 2004 #6

    matt grime

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    Have you not done basic differential equations (linear first order)? Sorry, I'm not the best person to spoon feed answers to anyone, least of all on a subject like this. Good luck finding yourself a teacher, but it ain't going to be me.
  8. Nov 23, 2004 #7
    ^ I've done normal differentiation and partial differentiation. Is that what you mean?
  9. Nov 23, 2004 #8
    ds/dt = v,

    => dx = vdt,

    => integral between x and x(subscript 0) of dx =v. integral of dt between x and x(subscript 0)

    => x = x(subscript 0) + vt

    Is this correct?
  10. Nov 23, 2004 #9

    matt grime

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    No, I mean differential equations. Linear first order. Look it up on the web, get an A-level text book out of the library, if you're not from the UK, then I don't know what level you meet it at or what you'd call it - calc 101 perhaps. Latexing up notes for you is more effort than it merits, sorry, but _you_ should put the effort in on this one.
  11. Nov 23, 2004 #10

    matt grime

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    And no that last post of yours isn't correct.
  12. Nov 23, 2004 #11
    does the integral of dx = x?
  13. Nov 23, 2004 #12

    matt grime

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    Seeing as you're making the effort:

    the *indefinite* integral of 1 dx (with respect to x) is x+k where k is a constant.
  14. Nov 23, 2004 #13
    I'm a bit confused. Here's our syllabus:

    Definition, domain, range; odd, even, periodic; polynomials,
    factorisation; rational functions, partial fractions; exponential and
    logarithmic functions; circular and hyperbolic functions; inverse function;
    limits and continuity; composite functions.

    DIFFERENTIATION (six lectures)
    Definition, chain rule, implicit, parametric, product rule; stationary
    values, points of inflection, curve sketching; Maclaurin and Taylor
    expansions, polynomial approximation; l’Hopital’s rule.

    Definition, partial differentiation, total derivative.

    COMPLEX NUMBERS (three lectures)
    Definition, conjugate, representation; roots of complex numbers, de
    Moivre’s theorem.

    How do they expect us to understand this stuff when we haven't even done differential equations?
  15. Nov 23, 2004 #14
    We don't do differential equations until the next symester. I'll be back tomorrow to check this thread (I have to go home now).
  16. Nov 23, 2004 #15

    matt grime

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    differential equations come *after* differentiation.
  17. Nov 23, 2004 #16


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    Okay, I'll do a bit of spoon-feeding.
    1)Suppose the velocity of an object is constant, that is:
    Here, v is a constant, wheras the function x(t) is the position of the object as a function of time.
    2. If we integrate both sides of an equality in the same way, we still have an equality:
    Here, we perform the integration over the interval [tex]0\leq{t}\leq{T}[/tex]
    where T is some arbitrarily chosen instant.
    3. Left-hand side:
    (Why is this true?)
    4. Right-hand side:
    (Why is this true?)
    5. Equating the results from 3. and 4., we gain:
    or, since T is arbitrary, we might as well call it "t":
    The last step could also have been reached by introducing a dummy variable [tex]\tau[/tex] and performed the integration:
  18. Nov 24, 2004 #17
    Thank you my friend. I get it now.
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