# Find Distance Function from Acceleration Function

The (non-constant) acceleration as a function of time, $a(t)$, is defined and known over $[t_0, t_2]$. It is also known that $a(t)$ is integrable. Also, $a(t)=\frac{dv(t)}{dt}$ and $v(t)=\frac{dx(t)}{dt}$, where $v(t)$ is the velocity function and $x(t)$ is the distance function. $t_1$ is a known time within $[t_0, t_2]$. Given $v(t_1)$ and $x(t_1)$, is it possible to find $x(t)$ over the entire interval? If so, how can this be done? Rigor would be appreciated.

#### Solutions Collecting From Web of "Find Distance Function from Acceleration Function"

By the Fundamental Theorem of Calculus,
$$v(t)-v(t_1)=\int_{t_1}^{t}a(s)\ ds.$$
Similarly,
\begin{align*} x(t)-x(t_1)&=\int_{t_1}^t v(s)\ ds\\ &=\int_{t_1}^t\left(v(t_1)+\int_{t_1}^{s}a(u)\ du\right)\ ds\\ &= (t-t_1)v(t_1)+\int_{t_1}^t\int_{t_1}^{s}a(u)\ du\ ds. \end{align*}
Rearranging yields the solution
$$x(t)=x(t_1)+(t-t_1)v(t_1)+\int_{t_1}^t\int_{t_1}^{s}a(u)\ du\ ds,\qquad t\in [t_0,t_2]$$