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In my $Theoretical \ Mechanics$ book I recently read, it gives the following description about "the path" in a certain time interval:

$$\vec{r}(t)=\vec{r}(t_0)+\vec{v}(t_0)(t-t_0)+\int_{t_0}^tdt^{\prime}\int_{t_0}^{t^{\prime}}dt^{\prime \prime}a(t^{\prime \prime})$$

where the quantities $r(t_0)$ and $v(t_0)$ are the positions and speed at time $t_0$, the initial conditions.

While $\vec{r}(t_0)+\vec{v}(t_0)(t-t_0)$ is fairly obvious, I am having in difficulty understanding why it puts so many derivatives and integrations in $\int_{t_0}^tdt^{\prime}\int_{t_0}^{t^{\prime}}dt^{\prime \prime}a(t^{\prime \prime})$. Can anyone help me to explain what does it mean? (I know it stands for the integration for the acceleration component, but why it represents it like this?)

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    $\begingroup$ so many derivatives and integrations The primes do not indicate derivatives. $t’$ and $t’’$ are just other time variables. There are only two integrations happening here. $\endgroup$ – G. Smith Aug 3 at 16:46
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As G. Smith suggests in the comments, the primes don't indicate derivatives, they're just used to distinguish the new integration variables $t',t''$ with $t$. To see where this comes from, look at the definition of acceleration:

$$\frac{d^2}{dt^2} \mathbf r(t) \equiv \mathbf a(t)$$

Now integrate both sides with respect to time from $t_0$ to $t$:

$$\int_{t_0}^t \frac{d^2}{dt_1^2} \mathbf r(t_1) dt_1 = \int_{t_0}^t \mathbf a(t_1) dt_1$$

where I've used the integration variable $t_1$ instead of $t'$ to avoid confusion with derivatives. Using the fundamental theorem of calculus, this gives:

$$\mathbf v(t) - \mathbf v(t_0) = \int_{t_0}^t \mathbf a(t_1) dt_1$$

with $\mathbf v \equiv d \mathbf r/dt$ being by definition the velocity. Integrating both sides again similarly results in:

$$\int_{t_0}^t \mathbf v(t_1) dt_1 - \int_{t_0}^t\mathbf v(t_0) dt_1 = \int _{t_0}^t \int_{t_0}^{t_1} \mathbf a(t_2) dt_2 dt_1$$

$$\int_{t_0}^t \frac{d\mathbf r(t_1)}{dt_1} dt_1 - \mathbf v(t_0) \int_{t_0}^t dt_1 = \int _{t_0}^t \int_{t_0}^{t_1} \mathbf a(t_2) dt_2 dt_1$$

$$\mathbf r(t) - \mathbf r(t_0) - \mathbf v(t_0)(t-t_0) = \int _{t_0}^t \int_{t_0}^{t_1} \mathbf a(t_2) dt_2 dt_1$$

Rearranging this gives the final result:

$$\mathbf r(t) = \mathbf r(t_0) + \mathbf v(t_0)(t-t_0) + \int _{t_0}^t \int_{t_0}^{t_1} \mathbf a(t_2) dt_2 dt_1$$

As Chet Miller points out in the comments, the double integral can be further simplified by using integration by parts:

$$\int _{t_0}^t \int_{t_0}^{t_1} \mathbf a(t_2) dt_2 dt_1 = \int _{t_0}^t \left(\int_{t_0}^{t_1} \mathbf a(t_2) dt_2\right) dt_1 = t_1 \left(\int_{t_0}^{t_1} \mathbf a(t_2) dt_2\right) \Bigg\vert_{t_1=t_0}^t - \int_{t_0}^t t_1 \ \mathbf a(t_1) dt_1$$

$$= \int_{t_0}^t (t-t_1) \ \mathbf a(t_1) dt_1$$

This gives the simplified result:

$$\mathbf r(t) = \mathbf r(t_0) + \mathbf v(t_0)(t-t_0) + \int_{t_0}^t (t-t_1) \ \mathbf a(t_1) dt_1$$

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    $\begingroup$ Shouldn't the final answer involve $\mathbf{a}(t_2)$, not $$\mathbf{a}(t_1}? Otherwise, it is not consistent with the original post. Also, it seems that this can be reduced to a single integral by integrating by parts. $\endgroup$ – Chet Miller Aug 4 at 0:36
  • $\begingroup$ @ChetMiller That's right! Thanks for correcting me. True, the final double integral can be simplified. I just didn't bother since that wasn't part of OP's question. $\endgroup$ – Sahand Tabatabaei Aug 4 at 0:40

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