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Basic physics understanding. Could someone explain?

  1. May 24, 2013 #1
    I'm trying to get a grip on some classic physics by watching a Stanford lecture. I've made it through the first one, and now in the second one all the professor ever talks about is 'derivatives' without actually explaining what they are. I guess the Stanford population would be assured with some previous math knowledge.

    So anyway, apparently we can work out velocity from position with v = dx / dt, that's velocity equals position over time. Makes sense. We can further work out acceleration with dv / dt, velocity over time. Now why on earth are there these d's in front of the variables? What exactly is meant by the word 'derivative', and what is a vector? If these questions are too naggy, and I should perhaps be looking at some basic maths before I look at classic physics, could anyone recommend some good starting points (wikipedia is useless for this sort of thing).
  2. jcsd
  3. May 24, 2013 #2


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    Homework Helper

    Master your school math so that you are ready for these more advanced topics when you get to them. This is Calculus and it relies on most of the math you are learning now. If you think you are ready, then I recommend MIT's 18.01SC course, it is free and does a good job, I think.

    I won't say more because there are many threads on these forums that can help.
  4. May 24, 2013 #3


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    You need to get acquainted with elementary calculus (dx/dt is the derivative of distance with respect to time).
    A vector is a quantity which has both magnitude and direction, for example velocity has a speed (how fast) and a direction.
  5. May 24, 2013 #4


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    Hello Ultermarto1,

    The Stanford professor whose class you've watched is using a basic Calculus tool. A derivative of a function like the position [itex]x(t)[/itex] is defined as the following limit:

    [itex]\frac{dx}{dt}=\mathrm{lim}_{t→t_{o}}\frac{x(t) - x(t_{o})}{t - t_{o}}[/itex]

    And this happens to be exactly the definition of the instantaneous velocity, [itex]v(t)[/itex] of the particle which follows the trajectory given by [itex]x(t)[/itex]. So

    [itex]v(t) = \frac{dx}{dt}(t)[/itex]

    Notice that this is different com the mean velocity, given by

    [itex]\bar{v} = \frac{\Delta x}{\Delta t} ≠ \frac{dx}{dt}(t)[/itex]

    In fact, if you substitute the last two equations above on the definition of Derivative I wrote in the begining of this post, you will notice that the instantaneous velocity [itex]v(t)[/itex] is precisely the limit of the mean velocity, [itex]\bar{v}[/itex] when the interval of time under consideration is taken to zero:

    [itex]\frac{dx}{dt}= v(t) = \mathrm{lim}_{\Delta t→ 0}\frac{\Delta x}{\Delta t}[/itex]

    This is basic Calculus! I would tell you to look for a nice book for starters on this subject and learn it before you go deeper into more complicated physics. ;)

  6. May 24, 2013 #5
    I'm afraid I'm in college, and not studying any maths at this point. I'll be sure to make a start on calculus, though, if it'll help.
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