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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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  • $\begingroup$ Thanks @pre-kidney for the derivation! To generalize the result slightly, the conditions for velocity and distance can be prescribed at different times (instead of at the same time $t_1$), and we can still determine $x(t)$, right? $\endgroup$ – TSJ Jan 19 '15 at 8:47
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    $\begingroup$ Yes, that is correct. $\endgroup$ – pre-kidney Jan 19 '15 at 8:49
  • $\begingroup$ I just realized this problem is a simple second-order differential equation, $\frac{d^2 x}{dt^2}=a(t)$ with two conditions given, and this method can be used to solve any such equation. $\endgroup$ – TSJ Jan 19 '15 at 8:49
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    $\begingroup$ Yes, and the same method also solves $\frac{d^nx}{dt^n}=a(t)$. $\endgroup$ – pre-kidney Jan 19 '15 at 8:50

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