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After years of procrastinating i've decided not to "move ahead" with physics without getting this ridiculously trivial question clear!*I know i had asked a similar question as silly and stupid as this one, however for some reason this equation keeps haunting me cause i see it almost everywhere in kinematics! *

We know that the position of a particle at any time is given by $x = x_{0} + v_{0}t + \frac{1}{2}at^2$.I'am aware that $x_{0}$ and $v_{0}$ are obtained by solving for the constants of integration at $t = 0$. But why should i even care what the velocity or the acceleration is?What purpose do they serve?What has these two quantities got to do with anything?

Ok.to be more "technical" Why should one "involve" velocity and acceleration when the equation is for determining postion at any given point in time?

To be honest i've tried to digest what this equation is really telling me by watching and reading numerous materials both online and offline regarding fundamentals of physics.Nothing seems to be working.I've decided to leave it to you as i seriously lack intuition.Let me reiterate by saying i've no problem with deriving this equation.

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2 Answers

up vote 2 down vote accepted

Suppose you are holding an apple. You obviously know where the apple is, and as long as you don't drop the apple you can predict it's future position. But suppose you now drop the apple and you want to predict where it will be in one second, two seconds or even ten seconds if you're standing on a tall building. As soon as you drop the apple it starts accelerating downwards due to gravity. Therefore to predict the position of the apple 1 second after it leaves your hand you have to know the acceleration.

If you repeat the experiment on the moon you need to know that the acceleration on the moon is different, because of it's lower gravity, and if you feed this lower acceleration into your equation you'll find the apple has moved a smaller distance after 1 second, as indeed you'll know from watching videos of the lunar astronauts.

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Fine.I understand.This only applies when dealing with vertical motion right?Where else would you use accleration due to gravity? –  alok Feb 6 '12 at 19:32
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Note that the equation you give for position only applies under conditions of constant acceleration.

That is you already have a particle under acceleration.

Now, by definition

$$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2} $$

In other words you care about acceleration and velocity because the position of the particle depends on them.

You wouldn't expect the position of a car traveling at 30 kph to have the same time dependence as one traveling at 100 kph.

Nor would you expect a ball in microgravity (i.e. experiencing nearly no acceleration) to exhibit the same path as one in a 1 g field (i.e. experiencing $10\text{ m/s}^2$ acceleration).

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Now my concern is why doesn't it say "postion at any given time under the influence of gravity?"If that's the case what equation would you use to determine the postion of a particle at any given time moving horizontaly? –  alok Feb 6 '12 at 19:34
    
@alok: That depends. Is there air resistance? If not then you can use the same equation with $a \equiv 0$, which reduces to the constant velocity equation $x = x_0 + vt$. You should understand that physicists don't just memorize a lot of equations. They known that $f = ma$ and that acceleration is the second time derivative of position. –  dmckee Feb 6 '12 at 19:37
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And it doesn't say motion under the effect of gravity because it is also motion in a spacecraft under ion drive (or any other constant thrust dive that uses negligible reaction mass) and a car stopping on locked up brakes and ... Basically "constant acceleration" is more general than "gravity near the surface of the Earth" (since gravity in general is not a constant acceleration). –  dmckee Feb 6 '12 at 19:39
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