Why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$?

Question

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

Answer 1

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.

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

Answer 2

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

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

Hence;

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

Answer 3

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 $$u=0$$

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