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Actually, this is the original question:

A particle moves along the x-axis so that the distance traveled in time $t$ is given by $x=2t + \cos 3t$. Find the distance between the first two positions of rest.

The ans given is 0.37 unit.

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    I found the solution already, thanks everyone.2011-09-22

2 Answers 2

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Introduce a time scaling by means of $t:={\tau\over3}$. Then we have to study the function $x(\tau):={2\over 3}\tau+\cos\tau$ with derivative $\dot x(\tau)={2\over3}-\sin\tau\ .$ Beginning at $\tau=0$ the first time of rest is at $\tau_1=\arcsin{2\over3}$, and the second at $\tau_2=\pi-\arcsin{2\over3}$. In between $\dot x(\tau)$ is negative; therefore the distance $d$ traveled in the time interval $[\tau_1,\tau_2]$ is given by $\eqalign{d =x(\tau_1)-x(\tau_2)&={2\over3}\bigl(2\arcsin{2\over3}-\pi\bigr)+\cos\bigl(\arcsin{2\over3}\bigr)-\cos\bigl(\pi-\arcsin{2\over3}\bigr)\cr &={4\over3}\arcsin{2\over3}-{2\pi\over3}+{2\over3}\sqrt{5}\cr &\doteq 0.369287\ \cr }$

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As pointed out by Srivatsan's comment, the points of rest, where $x'(t)=0$, can be found to be: $ 2-3\sin(3t)=0 \; \; \to \; \; t=\cases{ &$\frac13 \left(2\pi n +\sin^{-1}\left(\frac23\right)\right)$\\ &$\frac13 \left(2\pi n +\pi +\sin^{-1}\left(\frac23\right)\right)$ } $

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    What about $t=(1/3)(\pi-\arcsin(2/3))$? Oh, I think you may just have a plus sign where you need a minus in the second case.2012-07-24