Integration constant: What am I missing? Deriving some physics formulas with my son, I managed to confuse myself. From:
$$a_0 = \frac{dv}{dt} \implies a_0\, dt = dv \implies \int_{v_0}^{v} dv = \int_{t_0}^{t} a_0\, dt$$
we have:
$$v=v_0 + a_0\Delta t \tag{1}$$
If $t_0 = 0$ we have:
$$v=v_0 + a_0t \tag{2}$$
From (2):
$$v = \frac{dx}{dt} \implies v\, dt = dx \implies \int_{x_0}^{x}dx=\int_{t_0}^{t}v_0+a_0t \, dt$$
Thus,
$$x= x_0+v_0\Delta t+\frac{1}{2}a_0\Delta t^2 \tag{3}$$
Question
What algebraic manipulation would allow me derive (3) from (1), i.e.,
$$\int_{x_0}^{x}dx=\int_{t_0}^{t}v_0+a_0\Delta t \, dt \implies x= x_0+v_0\Delta t+\frac{1}{2}a_0\Delta t^2$$
 A: We have the governing differential equations:
\begin{align}
\dot x = v \\
\ddot x = a
\end{align}
with initial values \begin{align}x(0) &= x_0 \\ \dot x(0) &= v_0 \\ \ddot x &= a_0.\end{align}
This is a nice system of differential equations: to solve it, we only need to integrate.
$$x(t) = x_0 + \int_{s=0}^{s=t} v(s)\, ds = x_0+\int_{s=0}^{t}\left( v_0 +\int_{\tau=0}^s a(\tau)\,d\tau \right)\,ds $$
Since acceleration is constant, this reduces to 
\begin{align} 
x(t) &= x_0+\int_{s=0}^{t}\left( v_0 +\int_{\tau=0}^s a_0\,d\tau \right)\,ds  \\
&= x_0 + \int_{s=0}^{t} v_0 + a_0(s-0) \,ds \\
&= x_0 + (t-0)v_0 + a_0\left[\frac{1}{2}s^2\right]_{s=0}^{t} \\ 
&= x_0 + v_0t + \frac 12a_0t^2.
\end{align}
A: You need initial values to get the $v_0$ and $x_0$. This is an IVP problem, with $v(t_0)=v_0$ and $x(t_0)=x_0$. For example, to get $\int_{t_0}^t a_0\mathrm{d}t=v_0+a_0\Delta t$, you go from $\int_{t_0}^t a_0\mathrm{d}t=a_0\Delta t+C$. Then, solve for $C$ by substituting $t=t_0$ and $v(t_0)=v_0$.
