Deriving $x=ut+0.5 a t^2$ in different ways I was playing around with the equation in the title when I realized it isn't as simple as I thought. I basically changed the parameters in the derivation given in the book because I was wondering why $t=0$ in particular had to be the starting point. The result was the following, which I am unable to explain.
If we let the times be $t_1$ and $t_2$ instead of $0$ and $t$, then we get $\Delta x = v_{av} \Delta t=(\frac{v_a+v_b}{2})(t_2-t_1)$ where $a,b$ are arbitrary. This leads to the weird result $\Delta x = (\frac{v_0+at_a+v_0+at_b}{2})(t_2-t_1)=(v_0+\frac{1}{2}a(t_a+t_b))(t_2-t_1)=v_0\Delta t+\frac{1}{2} a(t_a+t_b)\Delta t$. *
My question is, is this a correct result? If so, how useful is it? If you replace $a,b$ by $1,2$ it becomes the general case of $\Delta x = v_0 t+\frac{1}{2} at^2$.
* $v_{av}$ means average velocity
 A: $a$ and $b$ can't be arbitrary.  The equation
$$v_{av} = \frac{\Delta x}{\Delta t} = \frac{x_2-x_1}{t_2-t_1}$$
implies (under the assumption that $a=\frac{dv}{dt}$ is constant) that $v_{av} = \frac{v_1 + v_2}{2}$.
If $v_a = v(t_a)$ and $v_b=v(t_b)$ were allowed to be arbitrary, then $v_{av}$ would not be well-defined unless $v$ happened to be constant. \
So the way to derive the correct equation starting from $\Delta x = v_{av}\Delta t$ is
$$\Delta x = v_{av}\Delta t = \frac{v_1 + v_2}{2}\Delta t = \frac{2v_1 + a\Delta t}{2}\Delta t = v_1\Delta t + \frac 12a(\Delta t)^2$$
where $\Delta t = t_2 - t_1$.

To confirm our result, we can simply derive the correct equation from first principles.  Consider a particle undergoing constant acceleration over the time interval $[t_1, t_2]$.  Then, from the definition of acceleration, we have
$$a = \frac{dv}{dt}$$ and thus
$$\int_{t_1}^{t}a dt' = \int_{t_1}^{t} \frac{dv}{dt'}dt' \\
a\int_{t_1}^{t}dt' = \int_{v(t_1)}^{v(t)}dv \\
a(t-t_1) = v(t) - v(t_1)$$
where $t\in [t_1, t_2]$.  Then, integrating again, we get
$$\int_{t_1}^{t} a(t' - t_1)dt' = \int_{t_1}^t [v(t')-v(t_1)]dt' \\
a\left[\frac{t^2-t_1^2}{2}-t_1(t-t_1)\right] = \int_{t_1}^{t} \frac{dx}{dt'}dt' - v(t_1)(t-t_1) \\
a\left[\frac 12(t-t_1)^2\right] = x(t)-x(t_1) - v(t_1)(t-t_1)$$
Then setting $t=t_2$ and solving for $\Delta x = x(t_2)-x(t_1)$, we get
$$\Delta x = v(t_1)\Delta t+\frac 12a(\Delta t)^2$$
where $\Delta t = t_2-t_1$.
