- For time $t \geq 0$, the velocity of a particle moving along the $x$-axis is given by $\mathrm{v}\left(t\right) = \sin\left(t\right)$, where $t$ is measured in seconds and $\mathrm{v}\left(t\right)$ is in meters per second.
- The initial position of the particle at time $t = 0$ is $x = 8$.
- What is the total distance the particle traveled from time$t = 0$ to $t = 2\pi$?
So, I tried solving this by doing$\int_{0}^{2\pi} \left\vert\sin\left( t\right)\right\vert\mathrm{d}t$, in which I got $2$, but the answer is wrong.
I suspect it's something to do with the $t = 0$ and $x = 8$ but I don't know how to use that in this context.
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$\begingroup$As $Sin (t)$ has symmetry about $\pi/2$ and its range is positive between $(0,\pi)$ and negative between ($\pi,2\pi$), you should integrate between $0$ and $\pi / 2$ and then multiply by 4. The question is about distance and not displacement.
$\endgroup$ 5 $\begingroup$Given $v(t)=\mathrm{sin}\ t$, so $\int_{0}^{x} dx=\int_{0}^{t} \mathrm{sin}(t)dt$$$x(t)=1-\mathrm{cos}\ t$$BUT, here $x(t)$ is the displacement.
Now, to find total distance you've to carefully check if ${x(t)}$ falls down from a value in the above equation, i.e. it definitely rises for t$\in(0,\pi)$ but after that, it starts falling.
Can you take it from here?
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