Explain $\Delta x = v_0t + \tfrac{1}{2}gt^2$ please? [duplicate]

Question

$g = \Delta v/t$, so $\Delta v = gt$. $v = v_0 + \Delta v$, so $v = v_0 + gt$. So if $\Delta x = vt$, then $\Delta x$ should be $v_0t + gt$. Why the $\tfrac{1}{2}gt^2$? I'm really confused, so this question is also probably very confusing. Sorry and thanks in advance!

Answer 1

Indeed your first suggestion is wrong :$ \Delta x = v_o t + gt $

Instead it should be $ \Delta x = v_o t + gt^{2} $(You can recheck it)

Where you are wrong is here:

According to your question v is the final velocity since $(v=v_{0}+gt)$

So $\Delta x\neq vt$ but instead it should be $\Delta x =v_{average}\times t$

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 2

Let's take the first equation of motion which is :

\begin{equation} v=u+at \end{equation} Integrate this equation to get: \begin{equation} \int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt \end{equation} this gives: \begin{equation} x=ut+\frac{1}{2}at^2+x_0 \end{equation} The integration constant can be done away by putting the proper limits on $x$.(Assuming the acceleration is not a function of time which in this case is $g$)

Answer 3

I don't have a conceptual answer as to why it is that way. But mathematically, your first suggestion is wrong: $ \Delta x = v_o t + gt $.

A unit analysis will show you why:

$$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds$$ $$ meters \not= meters + \frac{meters}{seconds}$$

And we can see that the assumption is simply not true by units. But this yields the right units:

$$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds^2$$

We know the SI unit for change in distance is meters, which requires g to be multiplied by squared time, or:

$$ \Delta x = v_o t + gt^2 $$

The square and half will go away when you derive with respect to time, which yields velocity in its correct units. If you derived it once more with respect to time you would get g, also in its right units.

Answer 4

This is a trivial kinematic deduction.

\begin{align}s(t_2) &=\int_{t_1}^{t_2}~v(t)~\mathrm dt +s(t_1)\\ &= \int_{t_1}^{t_2}~\left\{v(t_1)+\int_{t_1}^{t}~a(t')~\mathrm dt'\right\}\mathrm dt+ s(t_1) \;.\end{align}

Integrating this, we would get

$$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a(t_1)\frac{\{t_2-t_1\}^2}{2} + \dot a(t_1)\frac{\{t_2-t_1\}^3}{6} + \ldots $$

(This is actually Taylor's series expansion of $s(t)$ at $t_1\;.$)

From this, you would get the desired result by putting the higher order derivatives($\gt 2$) of $s$ equal to zero to get the desired relation viz:

$$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a\frac{\{t_2-t_1\}^2}2\;.$$