# Is it worth watching Yale Fundamentals of Physics vids?

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1. Jul 20, 2012

### myink

http://oyc.yale.edu/physics/phys-200
http://oyc.yale.edu/physics/phys-201

I'm starting college in September and have never taken Physics before, so I am preparing by reading different books and watching some videos, and was wondering if this course is worth watching because the curriculum for this seems weird. I think in Fundamentals of Physics I, it's Classical Mechanics and Thermodynamics, and in Fundamentals of Physics II, it's E&M and Modern Physics. Is this at all worth watching for introduction to Physics?

2. Jul 20, 2012

### Creel

It's definitely not a bad idea, MIT also has open courseware that is good for a watch. You did say that you had no physics experience under your belt but you didn't mention any math experience. If you are currently in Calc 1, or perhaps under Calc 1, don't get too intimidated by the professors using integration and differentiation to arrive at formulas, because in Phys 1 at most schools you can stick to Algebra/Trig and succeed.

I also highly recommend checking out www.khanacademy.com and see the first few videos on Physics to get a good primer on what to expect. Khan is a lot more basic and should give you a really good foundation before you even get in the door.

3. Jul 20, 2012

### myink

My goodness, I just watched the first video and I understood virtually nothing apart from the formulae and which numbers to plug in for their respective variables.

For example, deriving the first formula they used...taking the integral of acceleration with respects to time, he arrived at $$x(t) = t^2 + c$$, but then multiplied the $$t^2$$ by $$\frac{a}{2}$$ to make the acceleration appear when you took the derivative, I guess. And then he just added $$bt$$ to the entire thing, and I have no clue why. He explained that it was so we can take the second derivative of it, but you can take the second derivative of $$x(t) = \frac{a}{2}\cdot t^2 + c$$ without the $$bt$$ too...maybe there needs to be a constant both times you take the derivative? Not sure why. Then again, I'm not sure why you need acceleration to appear with the velocity when you take the derivative of the position, because if you take the derivative of the velocity, don't you get the acceleration as that result anyway?

Anyway, 75% of the material I had no clue what was going on, with the exception of plugging in numbers to solve for some stuff. I don't think this course is good for introduction to Physics.

4. Jul 20, 2012

### Nano-Passion

I didn't watch the video but what kind of mathematical rule lets you multiply just one of the variables in the equation and not the other? This looks like hand-waving overload.

5. Jul 20, 2012

### myink

I think it was supposed to be part of taking the integral of the second derivative (obtaining the equation for position with respects to time from the equation of acceleration which is just a constant in this case)

6. Jul 20, 2012

### Nano-Passion

Second derivative of what?

Your confusing me, and I think you are misinterpreting the video. If you can link me to the specific video I'll be able to help you.

7. Jul 20, 2012

### myink

The position with respects to time

http://oyc.yale.edu/physics/phys-200/lecture-1

8. Jul 20, 2012

### Orlandeau

He was defining position in terms of acceleration and velocity.

he multiplied by a/2 in accordance with a change of variables. he wanted the second derivative of his position function to equal a(t) = a for constant acceleration. I skipped around the video, hope this clears your problems up.

9. Jul 20, 2012

### Nano-Passion

Taking the derivative of position is meaningless.

This is what he was saying: You have acceleration, and you want to see how position relates to it.

So you ask yourself, what variable when you take the derivative of twice, gives you acceleration. Well, you can take the derivative of at^2/2 twice to get back a. So

x = at^2/2

This isn't the only answer, because there could have been any constant in the above equation to give you acceleration. So in general,

x = at^2/2 + c .. you could have also had x = at^2/2 + bt + c

Check by integration.

I don't like his method though, let us try this method

a = F/m
Take the integral w/ respect to time
v = F/m * t + v_o [ v_o is the constant of integration here ]
Take the integral w.r.t dt again
x = F/m *t^2/2 + v_o(t) + x_o where x_o is the constant of integration here

Now plug in a for F/m

$$x = at^2/2 + v_o t + x_o$$

10. Jul 20, 2012

### myink

Oh boy. Well, now I'm kind of scared I won't make it through college...lol...

11. Jul 20, 2012

### Nano-Passion

Lol why, because you didn't understand it from the first time or because you still need some clarification? Either way it doesn't mean anything.

12. Jul 20, 2012

### myink

I am concerned because when school starts, I will have to learn all this stuff on my own and won't be able to keep up. We have class three times per week, so each topic we learn, I will have to know the topic inside and out within two days or I will be behind the class when we move on. I don't think I could learn how to derive the formulae in this first video course and understand the underlying concepts of them within two days before moving on, much less more difficult topics (the next course in this particular video series seems to be something about vectors in three-dimensional spaces). That, and I will have four other classes to tend to, which will probably require the same amount of time and effort and I have to dedicate to each subject equally, which means less time for each. So be a slightly slow learner, I'm not sure how I'll hold up with more advanced classes in my upper years.

13. Jul 20, 2012

### Nano-Passion

So take less classes. Things take time. A couple of years ago I absolutely knew NOTHING about physics and mathematics. Now I'm studying some subjects by myself. You just have to train your brain to be able to cope with the new line of thinking. The secret to this is to struggle -- spend large amounts of time struggling and you will get better.

I can learn things much faster now in comparison to have a while back -- it is just a byproduct of hard work. Your brain wasn't created to sit down and think abstractly -- it takes TIME.

14. Jul 20, 2012

### Norfonz

Don't sweat it. Learn at your own pace. You may not find introductory physics particularly interesting. I know I didn't get a lot out of it. Just practice problems and do your best. Try to understand the underlying principles so you can solve whichever problems are thrown at you.

15. Jul 20, 2012

### myink

Yes, but all I did during high school up until last month when I graduated was study all day. I never had any friends so all I did was study pretty much every day, including the weekends. It doesn't seem like much fruition has come of it now that I'm starting college, which seems so different from high school since everything in high school was pretty easy and I learned everything the moment I saw it in class.

16. Jul 20, 2012

Those videos really are unbelievable - he teaches you the philosophy of how to think about the material starting from the basics each time to re-derive everything on the spot. Without a doubt the way to fail at this is not to have the maths.

I recommend you watch these:
http://www.pa.msu.edu/courses/phy233b/VideoLectures.html [Broken]
www.pa.msu.edu/courses/PHY234B/VideoLectures/ [Broken]
as well, they are a little more concrete at times with more problems.

As for Shankar turning x(t) = t² + c into x(t) = ½at² + bt + c:
The way to think about it is:
dv/dt = a
v = at + b
dx/dt = at + b
∫(dx/dt)dt = ∫(at + b)
∫dx = ∫(at + b)
x = ½at² + bt + c
He is just saying that when you integrate twice you're going to get an x(t) = t² + c term, but then because we integrated a we have to include ½a. Furthermore since we integrated, we get a constant at each stage of integrating - but after the first integration a t is added to the first constant (to the b). Even if you only knew about differentiation & that acceleration is second derivative of position & thought we were just working backwards here then it just makes sense that hypothetically the bt + c term could be included since it disappears by the second derivative & we end up with constant acceleration. I hope you understand the importance of constant acceleration here & how different the situation would be if there was variable acceleration. Also, a = F/m has not been defined yet.

Just use this thread to throw out your questions as you watch them, & post all the stupid scary thoughts you have during them. I watched these videos without knowing the above & freaked out, then learned the above derivation from calculus & came back to them & got through. Then failed during the later videos Constant failure is a virtue

Last edited by a moderator: May 6, 2017
17. Jul 20, 2012

### Gravitational

Are you planning on taking physics in college?

18. Jul 20, 2012

### Nano-Passion

You've only been studying for a month and your getting discouraged?

One thing you will learn is that things take a lot of time. And that people who make it to the top are the ones who don't expect considerable progress so fast, they are the ones that aren't easily swayed.

Like I told you, it is a skill and it isn't easily acquired. It takes time, more time. You have to keep training your brain from different angles to bring your analytical skills up to par.

19. Jul 21, 2012

### myink

Yes, I declared my major as Physics when I applied to the college I got accepted into
No, I actually took a break from June until now to watch every single movie I hadn't seen in my high school years since I never had time to watch any because of my busy study schedule.

Can't it be detrimental to learn material too slowly in comparison with everyone else if everyone is learning the content rather quickly?

20. Jul 21, 2012

### Nano-Passion

When did you get very serious about school? When did you get serious about math and physics?