Let's say a particle's velocity is modeled by $v(t)=\sin(t)$. Therefore, assuming the particle starts at position $0$, it's position $x$ can be modeled by $x(t)=\cos(t)$. The total distance traveled by the particle in $t$ seconds can then be modeled by $\int_0^t{|\sin(t')|}\mathrm{d}t'$. In this instance, what is the arc length of the position in physical terms? To be specific, what is the meaning of $a(g)=\int_0^g{\sqrt{1+\sin^2t}}\space \mathrm{d}t$?
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$\begingroup$ This link seems to deal with a similar question. $\endgroup$– MarkFeb 27, 2016 at 5:28
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$\begingroup$ Is it meaningful in this context? The units of the terms in the square root of the arc-length integral don't match, do they? $\endgroup$– NealFeb 27, 2016 at 6:06
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$\begingroup$ In an actual physical problem, you would probably have something like $v(t) = sin(a t)$ with $a t$ being dimensionless, so that $a$ would have units of inverse time. That would take care of the subsequent dimensional issues you notice. $\endgroup$– user_of_mathFeb 27, 2016 at 6:54
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Note that $x(t)=1-\cos t$ instead, so $x(0)=0$ and $v(t)=\sin t$.
Distance travelled $\displaystyle \, =\int_{0}^{t} |v(t)| dt =\int_{0}^{t} |\sin t| \, dt =2\left \lfloor \frac{t}{\pi} \right \rfloor +1- \cos \left( t-\left \lfloor \frac{t}{\pi} \right \rfloor \pi \right)$
Whereas $\displaystyle \int_{0}^{x} \sqrt{1+\sin^{2} x}\, dx = E(x,i)$ is the arc length of the curve $y=\cos x$
answered Feb 27, 2016 at 7:42