Need help with acceleration/time physics...
Skorche Wrote:Once again, while you didn't use any calculus, the formula d = ut + 1/2atÂ² is derived using ideas from calculus.
But it isn't just the area of a trapezoid, it is the area of a right trapezoid which is essentially a rectangle and a triangle. I don't see where calculus comes in here (if this is calculus then I can claim I started learning calculus in 3rd grade ).
It's because you are relating the area under a curve on the velocity graph to the distance that makes it calculus. Calculus was developed partly to solve problems just like this.
Yes, finding the area of a rectangle and a triangle is easy, but would you have ever guessed that doing so relates the velocity to the distance? Unless you were told so, I'd suspect that you wouldn't have.
Yes, finding the area of a rectangle and a triangle is easy, but would you have ever guessed that doing so relates the velocity to the distance? Unless you were told so, I'd suspect that you wouldn't have.
Scott Lembcke  Howling Moon Software
Author of Chipmunk Physics  A fast and simple rigid body physics library in C.
unknown Wrote:yeah but the idea of vbar, average time for an entire journey is not derived from calculus, neithe was anything in my post. (the one that I didnt make a stupid mistake in that is)
The fact that you use an average velocity over an infitesimal period of time to approximate a function is calculus. When you use the "average" velocity formula in a numerical integrator, it becomes trapezoidal integration.
no its not the average over âˆ‚t, its the mean of two numbers, u and v, initial velocity and current velocity (this applies only to a constant acceleration).
If you look at http://en.wikipedia.org/wiki/Equations_o...equation_2
it stating the equation v bar = s/t (thought it calls v bar average velocity), which for a constant acceleration alone = (u+v)/2.
This was all known before calculus techniques and does not make any assumptions about graphs or areas under graphs.
as I showed in a previous post the two equations
v = u + at
s = 1/2 * (u + v) * t
are both derived from basic mathematics and the resulting formula for t in terms of s and a was derived from simple algebraic manipulation.
so you dont need calculus for somthing this simple!
If you look at http://en.wikipedia.org/wiki/Equations_o...equation_2
it stating the equation v bar = s/t (thought it calls v bar average velocity), which for a constant acceleration alone = (u+v)/2.
This was all known before calculus techniques and does not make any assumptions about graphs or areas under graphs.
as I showed in a previous post the two equations
v = u + at
s = 1/2 * (u + v) * t
are both derived from basic mathematics and the resulting formula for t in terms of s and a was derived from simple algebraic manipulation.
so you dont need calculus for somthing this simple!
Sir, e^iÏ€ + 1 = 0, hence God exists; reply!
I see where I was wrong. However, I think Unknown is also right that you are looking at the average velocity which has nothing to do with the graph of the velocity.
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