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Physics and g = 2s t^-2 - A-Man - 11-18-2012

So, I was studying for my physics exam (2d kinematics) and that formula in the title is driving me crazeh. Where did that 2 come from? We all know gravity is acceleration, and acceleration is only "s t^-2" (without the 2). Why are we multiplying by 2 with gravity? is there anything special about the gravity acceleration? And why is it that when we wanna get the distance we use "gt^2/2" ?? again, where did this extra 2 in the denominator come from? (i do remember that this 2 didn't exist for gravity in the past lessons) So yeah, please help!!!

RE: Physics and g = 2s t^-2 - zort - 11-18-2012

Acceleration is 2*s/t^2, not s/t^2. You're right, there is nothing special about acceleration that is caused by gravity. The 2 comes from the area of the right triangle with side lengths t and at, but you've probably seen that derivation. If not, you may content yourself with the observation that s=a*t^2/2=(a*t/2)*t, which is the average speed in the interval times the total time.

RE: Physics and g = 2s t^-2 - A-Man - 11-18-2012

(11-18-2012, 05:25 PM)zort Wrote:  Acceleration is 2*s/t^2, not s/t^2. You're right, there is nothing special about acceleration that is caused by gravity. The 2 comes from the area of the right triangle with side lengths t and at, but you've probably seen that derivation. If not, you may content yourself with the observation that s=a*t^2/2=(a*t/2)*t, which is the average speed in the interval times the total time.
Oh! I see now. Damn it man. Thanks a lot ^^. And darn you useless physics teacher... Anyways, ~Solved

RE: Physics and g = 2s t^-2 - Silverthorn - 11-18-2012

edit: eh, looks like I was too slow again :p

You can also solve it via a very simple differential equation :p

a is the derivative of the velocity v, therefore: v = Integral(a dt) = a*t + C[sub]1[sub]
We'll just omit C[sub]1[sub] because we assume that the object doesn't have any initial movement.
The velocity v is the derivative of the position s, therefore: s = Integral(v dt) = 1/2*a*t² + C[sub]2[sub]
Let's say we start from the origin, so that this C[sub]2[sub] becomes zero as well.
Plug in Earth's acceleration g for a and we'll get: s = 1/2*g*t²
When you now bring g to the left side and s to the right, you get: g = 2s*t-2


Quote:We all know gravity is acceleration, and acceleration is only "s t^-2"
That is correct for units. Acceleration is distance per times squared. You won't get any information about potential factors that might pop up randomly (compare to Coulomb's law where a sudden 4*π falls from the sky; same for Schrödinger's equation with ħ/(2m)). Most of the time, it works, but you have to know where to pay attention to special cases ;)