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I have complications to do the following problem:

A particle of mass $m$ moves with constant speed $v$ along the curve $y^{2}=4a(a-x)$. Find its velocity and acceleration vectors.

My first idea was to parameterize the curve given, however did not know how to introduce speed $v$. Therefore I derived with respect to time, the equation of the curve, obtaining:

$$2y\frac{\partial y}{\partial t}=-4a\frac{\partial x}{\partial t}$$

Also, I know that

$$\left( \frac{\partial y}{\partial t}\right)^2 +\left( \frac{\partial x}{\partial t}\right)^2=v^2$$

Thus have two equations relating the $x$ and $y$ components of the velocity, but I have not been able to resolve. Is my method OK? Is there another way? Is it easier to do so using the parametric equations, but then as I enter the speed $v$?

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Let $v_x=dx/dt$ and $v_y=dy/dt$.
We got: $2yv_y=-4av_x$
Rewriting $v_y=-\dfrac{2av_x}{y} \tag{1}$
Also we got : $v_x^2+v_y^2=v^2 \tag{2}$
Subsitute value of $v_y$ in eqn 2.
$v_x^2+{(-\dfrac{2av_x}{y})}^2=v^2$
Solving gives $v_x=\pm \dfrac{vy}{4a^2+1},$
Substitute this value of $v_x$ in eqn 2 gives: $v_y=\mp\dfrac{2av}{\sqrt{4a^2+1}}$
We know $v_x$ and $v_y$. velocity is as we know $\vec v=v_x {\hat i}+v_y\hat j$ and can be found now. It should be clear that $\vec v$ depends upon the $y$ co-ordinate.

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You're on the right track. A couple of notes:

  1. Those are actually total derivatives. You can think of $x(t)$ and $y(t)$ as functions of $t$ alone
  2. you have two equations for two functions. You probably want to isolate them into two equations, each for one function
  3. Think about how you would solve this by elimination
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  • $\begingroup$ I totally agree, however, the problem in trying to solve is that I can not remove the term $ y $ that appears in the first equation derived. $\endgroup$ – Santi Carmesí Feb 25 '14 at 14:36
  • $\begingroup$ @user41316: you won't be able to. The best you'll do is an ODE that gives you $\dot y\,f(y) = {\rm const.}$, which you will then have to integrate on both sides with respect to time. A $u$ substitution of $u = y(t)$ will do the rest. $\endgroup$ – Jerry Schirmer Feb 25 '14 at 14:39
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Whenever a particle is constrained to move along a path, its velocity and acceleration vectors are decomposed along the tangent vector $\vec{e}$ and the normal vector $\vec{n}$ as

$$\boxed{ \begin{aligned} \vec{v} & = v \, \vec{e} \\ \vec{a} & = \dot{v} \, \vec{e} + \frac{v^2}{\rho}\, \vec{n} \end{aligned} }$$

where $v$ is the speed and $\rho$ is the instant radius of curvature. In addition, for a path described by the position vector $\vec{r}(t) = ( x(t), y(t) )$ the path properties are

$$ \begin{aligned} \vec{e}(t) & = \left( \frac{x'}{\sqrt{x'^2+y'^2}}, \frac{y'}{\sqrt{x'^2+y'^2}}\right) \\ \vec{n}(t) & = \left( -\frac{y'}{\sqrt{x'^2+y'^2}}, \frac{x'}{\sqrt{x'^2+y'^2}}\right) \\ \frac{1}{\rho(t)} &= \frac{ y' x'' - y'' x'}{\left( x'^2+y'^2 \right)^\frac{3}{2}} \end{aligned} $$

where $x'=\frac{{\rm d}x(t)}{{\rm d} t}, x''=\frac{{\rm d}^2 x(t)}{{\rm d} t^2}$ and $y'=\frac{{\rm d}y(t)}{{\rm d} t}, y''=\frac{{\rm d}^2 y(t)}{{\rm d} t^2}$

So to solve your problem I propose to parametrize the path $y^2(x)=4 a (a-x)$ as

$$ \vec{r}(t) = \left( a t, 2 a \sqrt{1-t} \right) $$

HINT: $\frac{1}{\rho(t)} = \frac{1}{2 a \left(2 -t\right)^\frac{3}{2}}$, $\dot{v}=0$

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