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I'm preparing for my exam, but I have difficulties in perceiving why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$?

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    $\begingroup$ Integrate acceleration twice wrt time. $\endgroup$ – jinawee Dec 9 '13 at 11:51
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It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration.

enter image description here

Some say, a good plot is worth a million words! :)

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    $\begingroup$ The positive reaction to this answer is way better than what I was expecting! Thanks to everyone. $\endgroup$ – Ali Dec 9 '13 at 17:54
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    $\begingroup$ @Ali Part of it I think is that clarity in technical writing is a jewel to behold. $\endgroup$ – Selene Routley Dec 10 '13 at 13:36
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We can also do it using calculus,the displacement of the particle is given by

$$v=u+at$$

$$\rightarrow v\,dt = u\,dt + at\,dt$$

$$\rightarrow \int_0^t v\,dt = \int_0^t u\,dt + \int_0^t at\,dt$$

$$\rightarrow s = ut + \frac{1}{2}at^2$$

If $$u=0$$

$$\rightarrow s=\frac{1}{2}at^2$$

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$\Delta x =v_{average}\times t$

In uniform acceleration $v_{average}$ becomes $\dfrac{v-v_{0}}{2}$

Hence;

$\Delta x =\dfrac{v-v_{0}}{2}t=\dfrac{1}{2}\dfrac{v-v_{0}}{t}t^{2}=\dfrac{1}{2}at^{2}$

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