Why Does the Displacement Formula Include a 1/2 Factor in Physics?

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SUMMARY

The displacement formula x = vt + ½t² is derived from the principles of constant acceleration in physics. The ½-factor accounts for the average speed during the time interval, as the velocity is not constant but changes linearly from the initial velocity (v_i) to the final velocity (v_f). When starting from rest, the average speed is half of the final speed, which is why the formula includes the ½. Understanding this derivation clarifies why x = at² alone does not accurately represent displacement under constant acceleration.

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I have just started out with Physics, so please don't expect to much :)

I've tried by myself to figure out where x = vt + ½t² comes from; I know (because I've read) how to deduce it from a v versus t graph, and that it only works with a constant acceleration.

My question is as follows:

If -for any instant in time- :
v = x / t
a = v / t
are true; Then, if i do this:

x = vt and v = at
Then subsitute the second in the first:
x = at²
I obviously miss the ½-factor from the area under the graph..
What does this formula (x=at²) tell me in this case? I don't really see why exactly this wouldn't also give me the displacement..

Could anyone be so kind to clear this up for me :) I'd hate to constantly have this question in the back of my mind.. just knowing that 'this is how it is'..

Thanks in advance!
Evert
 
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shinni said:
If -for any instant in time- :
v = x / t
Since the speed is not constant, [itex]\Delta x / \Delta t[/itex] gives the average speed, not the final speed at time t.

For constant acceleration, the average speed is given by:
[tex]v_{ave} = (v_i + v_f)/2[/tex]

So, if you start from rest (t=0, v_i=0, x=0), then [itex]v_f = 2 v_{ave} = 2x/t[/itex].

That should explain your missing ½-factor.
 

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