Velocity of a Particle Consider a particle moving in a straight line from the fixed point $(0,2)$ to another $(\pi,0)$. The only force acting on the particle is gravity. 
How would we parametrically define the motion of the particle with time?

From kinematics, I found that
$\hspace{150pt} y(t)=2-\dfrac{gt^2}{2}$
The slope can be found to be 
$\hspace{140pt} m=\dfrac{0-2}{\pi-0}=-\dfrac{2}{\pi}$
Since the path the particle moves on is a straight line, we have
$\hspace{132pt}y=mx+b=-\dfrac{2}{\pi}x+2$
so $\hspace{150pt}x(t)=\dfrac{\pi gt^2}{4}$
Therefore the parametric equations for the position of the particle are
$\hspace{150pt}x(t)=\dfrac{\pi gt^2}{4}$
$\hspace{150pt} y(t)=2-\dfrac{gt^2}{2}$

Does this seem correct? Any suggestions?
 A: The assumption $y=2-\frac 12g\,t^2$ is wrong.
In fact the energy conservation law says that $$v^2-v_0^2=2g\,(y_0-y)$$If $y=f(x)$, then $v^2=\dot x^2+\dot y^2=\dot x^2+f'(x)^2\,\dot x^2\;$ hence $$\dot x^2=\frac {v_0^2+2g\,[f(x_0)-f(x)]}{1+f'(x)^2}$$If $f(x)=kx+q$ and one puts $x_0=0$ and $v_0=0$, after solving the ode it follows that $$x=-\frac 12g\frac{k}{1+k^2}t^2$$and so$$y=-\frac 12g\frac{k^2}{1+k^2}t^2+q$$considering $\dot x>0$.
Note that $$\lim_{k\rightarrow\infty}\,\frac{k^2}{1+k^2}=1$$ as to the vertical motion.
A: For constrained motion, you cannot take $y=2-\frac 12gt^2$.  The acceleration will be gravity projected onto the path.  As the slope is $-\frac 2{\pi},$ the acceleration will be $g \frac 2{\sqrt{4+\pi^2}}$ so we have $y=2-\frac 12g  \frac 2{\sqrt{4+\pi^2}}t^2$.  Then your calculation of $x$ from $y$ is fine, giving $x=\frac 14{\pi g \frac 2{\sqrt{4+\pi^2}}}t^2$
A: $\newcommand{\+}{^{\dagger}}%
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The Action $S$ is given by $\pars{~\mbox{the motion occurs along the line}\
y = 2\,\pars{1 - x/\pi}~}$
$$
S = \int_{t_{0}}^{t_{1}}\bracks{{1 \over 2}\, m\pars{\dot{x}^{2} + \dot{y}^{2}} - mgy - m\pi\mu\pars{y + {2 \over \pi}\,x - 2}}\,\dd t
$$
The equations of motion are:
$$
m\ddot{x} = - 2m\mu\,,\qquad
m\ddot{y} = -mg - m\pi\mu
\quad\imp\quad
\pi\ddot{x} - 2\ddot{y} = 2g
$$
$$
\pi\,\dot{x}\pars{t} - 2\dot{y}\pars{t}
=
\overbrace{\bracks{\pi\,\dot{x}\pars{0} - 2\dot{y}\pars{0}}}^{\ds{\equiv \beta}} + 2gt\,,\quad
\pi\,x\pars{t} - 2y\pars{t}
=
\overbrace{\bracks{\pi\,x\pars{0} - 2y\pars{0}}}^{\ds{\equiv \alpha}} + \beta t
+
gt^{2}
$$
Then, we have the equations
$$
\left\lbrace%
\begin{array}{rcrcl}
\pi x\pars{t} & - & 2y\pars{t} & = & \alpha + \beta t + gt^{2}
\\
2 x\pars{t} & + & \pi y\pars{t} & = & 2\pi
\end{array}\right.
$$
\begin{align}
x\pars{t} & = {\pars{\alpha + \beta t + gt^{2}}\pi + 4\pi\over \pi^{2} + 4}
=
{\pi \over \pi^{2} + 4}\,\pars{\alpha + 4} + {\pi \over \pi^{2} + 4}\,\beta t
+ {\pi \over \pi^{2} + 4}\,gt^{2}
\\[3mm]
y\pars{t} & = {2\pi^{2} - 2\pars{\alpha + \beta t + gt^{2}}\pi \over \pi^{2} + 4}
=
{2\pi \over \pi^{2} + 4}\,\pars{\pi - \alpha} - {2\pi \over \pi^{2} + 4}\,\beta t
- {2\pi \over \pi^{2} + 4}\,gt^{2}
\end{align}

$$
\color{#0000ff}{\large%
\begin{array}{rcl}
x\pars{t} & = & x\pars{0} + \dot{x}\pars{0}t + {\pi \over \pi^{2} + 4}\,t^{2}
\\[3mm]
y\pars{t} & = & y\pars{0} + \dot{y}\pars{0}t - {2\pi \over \pi^{2} + 4}\,t^{2}
\end{array}}
$$

