Cool classical physics proof I came up with today

[realfuzzhead]

Check out this pdf on this website. No signing up or anything, It took me the better half of today to do this. Anyways, this might be well known for people above my basic physics knowledge, but I came up with this all by myself knowing a few simple equations


http://www.keepandshare.com/doc/3569845/fun-with-physics-pdf-february-18-2012-7-59-am-579k?da=y

 

[willem2]

Equation 1. on page 6 is wrong.

$$V_{avg} = \frac {V_0 + V} {2}$$

is only valid if the acceleration is constant.

The right result is easy to find with calculus by integrating v = (v_0 + at + (1/2)Jt^2).
The distance is equal to the area under this curve.
(of course Archimedes found the area under a parabola in the 3rd century BC without calculus)

 

[pbuk]

That's pretty good! Unfortunately an error creeps in on page 5: I will use your numbering...

p5,1) This expression for V_ave is only true when V is increasing linearly i.e. when its acceleration is constant.

p5,3) Again because acceleration is not constant this is not the right equation. Instead we need $D = D_0 + \int V(t)dt$. From what you have written I am not sure if you have come across integration yet: if not, it is enough to know for now that it is the inverse of differentiation.

Now we already have $V(t) = V_0 + A_0t + \frac 1 2 Jt^2$, integrating this we have $D(t) = D_0 + V_0t + \frac 1 2 A_0t^2 + \frac 1 6 Jt^3$.

Note that we can check this by differentiation:
$$\begin{align}D(t) &= D_0 + V_0t + \frac 1 2 A_0t^2 + \frac 1 6 Jt^3 \\ V(t) &= \frac d{dt} D(t) = V_0 + A_0t + \frac 1 2 Jt^2 \\ A(t) &= \frac d{dt} V(t) = A_0 + Jt \\ \frac d{dt} A(t) &= J\end{align}$$

 

[realfuzzhead]

THANK YOU GUYS! Damnit.. I thought that last part was a little bit to easy. I just started calculus this week so I have not learned integration yet..

 

[pmacias]

That's great to hear! It's always exciting when we are able to come up with our own proofs and discoveries in physics. I will definitely take a look at the pdf and see what you have come up with. Even if it may be well known to others, it's still a valuable learning experience for you to have come up with it on your own. Keep exploring and pushing the boundaries of classical physics!