Intereting Posts

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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.

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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]

$$

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