Basic physics understanding. Could someone explain?

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Discussion Overview

The discussion revolves around the understanding of basic physics concepts, particularly derivatives, velocity, and acceleration, as introduced in a Stanford lecture. Participants explore the relationship between calculus and physics, and the necessity of mathematical foundations for grasping these concepts.

Discussion Character

  • Conceptual clarification
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about the term 'derivative' and its application in calculating velocity and acceleration, questioning the meaning of the notation used.
  • Another participant emphasizes the importance of mastering school-level math, particularly calculus, to understand advanced physics topics.
  • A different participant explains that a derivative represents the rate of change of a quantity, specifically relating to the concept of instantaneous velocity as the limit of mean velocity.
  • One participant suggests looking for introductory resources on calculus to build a solid foundation before tackling more complex physics concepts.
  • Another participant reiterates the recommendation of MIT's 18.01SC course as a helpful resource for learning calculus.

Areas of Agreement / Disagreement

Participants generally agree on the necessity of understanding calculus for physics, but there is no consensus on the best approach to learning these concepts or the specific resources to use.

Contextual Notes

Some participants mention the assumption that a certain level of mathematical knowledge is required for understanding the physics discussed, which may not be universally applicable. There are also unresolved questions regarding the clarity of definitions and the foundational knowledge needed.

Ultermarto1
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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).
 
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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.
 
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.
 
Hello Ultermarto1,

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

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

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

v(t) = \frac{dx}{dt}(t)

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

\bar{v} = \frac{\Delta x}{\Delta t} ≠ \frac{dx}{dt}(t)

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

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

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. ;)


Zag
 
verty said:
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.
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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