Motion Analysis (1 Viewer)

leehuan · Aug 5, 2016

$\text{The acceleration }a\, \textit{ms}^{-2}\text{ of a particle moving in a straight line is given by }a=3\left(1-x^2\right)\text{, where }x\text{ metres is the displacement of the particle to the right of the origin.}\\ \text{Initially, the particle is at the origin moving at a velocity of }4\textit{ ms}^{-1}$

$\text{Proven: }v^2=16+6x-2x^3$

$\text{Q: Will the particle ever return to the origin? Justify.}$

Yeah, please help. I was never good at the analysis part once it got a bit chaotic.

InteGrand · Aug 6, 2016

leehuan said:
$\text{The acceleration }a\, \textit{ms}^{-2}\text{ of a particle moving in a straight line is given by }a=3\left(1-x^2\right)\text{, where }x\text{ metres is the displacement of the particle to the right of the origin.}\\ \text{Initially, the particle is at the origin moving at a velocity of }4\textit{ ms}^{-1}$

$\text{Proven: }v^2=16+6x-2x^3$

$\text{Q: Will the particle ever return to the origin? Justify.}$

Yeah, please help. I was never good at the analysis part once it got a bit chaotic.

$\noindent Yes, it returns to the origin. Let $f(x)\equiv 16+6x-2x^3$ (the function for $v^2$). Since $f(2) = -2\times \left(-3\times 2\right) = 12>0$ and $f(3) = -2 \times \left(27 -9-8\right) < 0$, $f$ (being a continuous function) attains the value $0$ for some $x^{*}\in \left(2,3\right)$. At this place, we must have the acceleration be negative, since $|x|>1$ here. Now, this essentially means the particle moves forwards at the start until it reaches this $x^{*}$. Since the velocity is $0$ here and the acceleration negative, the velocity must become negative after the time at the particle reaches the point $x^{*}$. In other words, the particle will begin to move backwards after this point in time (i.e. $v=-\sqrt{f(x)}$ after this point in time).$

$\noindent Now as the particle starts moving back, its \emph{speed} will pick up, because the value of $|f(x)|$ is increasing as $x$ decreases from $x^{*}$ initially (as $f$ is a polynomial function). It can be shown that $f(x)$ has only one non-negative zero (e.g. by Descarte's Rule of Signs or calculus or whatever method you like). So as the particle keeps moving back, it keeps picking up speed (until reaching the max. value of the speed on the $x$-axis interval $\left(0,x^{*}\right)$, whose position we can find using calculus) and the speed won't go back to $0$ while $x\geq 0$. Since it keeps picking up speed until getting to the place $x_2$ where the speed is at a local maximum, the particle indeed reaches this place $x_2$.$

$\noindent After reaching here, its speed begins to decrease, but it is still moving backwards (i.e. left) and its speed on the $x$-axis interval $\left[0,x_2\right]$ is bounded above by $f(0) = 16$ (again can be shown via calculus; refer to graph below to see all these results visually). So the particle keeps on moving left and gets to the origin.$

$\noindent Here's a sketch of the $v^2$ vs. $x$ graph:$

http://www.wolframalpha.com/input/?i=sketch+y^2+=+16+6x-2x^3 .

Paradoxica · Aug 6, 2016

InteGrand said:
$\noindent Yes, it returns to the origin. Let $f(x)\equiv 16+6x-2x^3$ (the function for $v^2$). Since $f(2) = -2\times \left(-3\times 2\right) = 12>0$ and $f(3) = -2 \times \left(27 -9-8\right) < 0$, $f$ (being a continuous function) attains the value $0$ for some $x^{*}\in \left(2,3\right)$. At this place, we must have the acceleration be negative, since $|x|>1$ here. Now, this essentially means the particle moves forwards at the start until it reaches this $x^{*}$. Since the velocity is $0$ here and the acceleration negative, the velocity must become negative after the time at the particle reaches the point $x^{*}$. In other words, the particle will begin to move backwards after this point in time (i.e. $v=-\sqrt{f(x)}$ after this point in time).$

$\noindent Now as the particle starts moving back, its \emph{speed} will pick up, because the value of $|f(x)|$ is increasing as $x$ decreases from $x^{*}$ initially (as $f$ is a polynomial function). It can be shown that $f(x)$ has only one non-negative zero (e.g. by Descarte's Rule of Signs or calculus or whatever method you like). So as the particle keeps moving back, it keeps picking up speed (until reaching the max. value of the speed on the $x$-axis interval $\left(0,x^{*}\right)$, whose position we can find using calculus) and the speed won't go back to $0$ while $x\geq 0$. Since it keeps picking up speed until getting to the place $x_2$ where the speed is at a local maximum, the particle indeed reaches this place $x_2$.$

$\noindent After reaching here, its speed begins to decrease, but it is still moving backwards (i.e. left) and its speed on the $x$-axis interval $\left[0,x_2\right]$ is bounded above by $f(0) = 16$ (again can be shown via calculus; refer to graph below to see all these results visually). So the particle keeps on moving left and gets to the origin.$

$\noindent Here's a sketch of the $v^2$ vs. $x$ graph:$

http://www.wolframalpha.com/input/?i=sketch+y^2+=+16+6x-2x^3 .

This is what happens when you focus too much on the singularities of the function. //facedesks

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Motion Analysis (1 Viewer)

leehuan

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InteGrand

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Paradoxica

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