Day: May 9, 2024

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202405091824 Pastime Exercise 011

About the graph below, tell some stories as probable as probable can be.

Roughwork.

Assuming linear (/rectilinear) motion in a single direction.

Assuming uniform acceleration, there are two cases: i. zero acceleration and ii. non-zero (constant) acceleration; in the former velocity being (a) constant and the latter (b) non-constant.

Assuming a flat spacetime metric of which the spatial part does not expands with the temporal part.

Imagine a man walking along a straight line from point (x_1,y_1) to point (x_2,y_2). From

\displaystyle{\textrm{Speed }(v)=\frac{\textrm{Distance }(s)}{\textrm{Time }(t)}}

as the path is fixed, i.e., \Delta s=\textrm{Const.}, we see speed v and time t

\begin{aligned} v\uparrow\enspace \Leftrightarrow \enspace& t\downarrow \\ v\downarrow\enspace\Leftrightarrow\enspace& t\uparrow \\ \end{aligned}

in an inverse relationship. As the man keeps his own fair pace and makes himself a good timekeeper, we can treat speed v as an independent variable, and time t a dependent variable.

\textrm{\scriptsize{CASE} \textbf{\texttt{(a)}}}: velocity being constant

Parameterize the Cartesian equation y=mx+c by the parameter t' (*as distinguished from the natural/unit-speed/arc-length parameter time t) so as to write a set of parametric equations:

\begin{aligned} & \begin{cases} s_x(t')=t' \\ s_y(t')=mt'+c \\ \end{cases} \\ & \\ & \begin{cases} v_x(t') =s_x'(t') = 1 \\ v_y(t') = s_y'(t') = m \\ \end{cases} \\ \end{aligned}

\begin{aligned} v^2(t') & = v_x^2(t')+v_y^2(t') \\ v(t') & = \displaystyle{\sqrt{\bigg(\frac{\mathrm{d}s_x(t')}{\mathrm{d}t'}\bigg)^2 +\bigg(\frac{\mathrm{d}s_y(t')}{\mathrm{d}t'}\bigg)^2}} \\ |\mathbf{v}(t')| & = \sqrt{1+m^2} \\ t & = \int_{0}^{t'}|\mathbf{v}(t)|\,\mathrm{d}t \\ & = (\sqrt{1+m^2})t'\\ \end{aligned}

\textrm{\scriptsize{CASE} \textbf{\texttt{(b)}}}: velocity being non-constant

The man begins with initial speed v_1 at start point (x_1,y_1) and ends with final speed v_2 at finish point (x_2,y_2). We have

\displaystyle{\textrm{Acceleration }(a)=\frac{\textrm{Change in speed }(v-u)}{\textrm{Time }(t)}}

Write by SUVAT equations of motion:

\begin{aligned} \mathbf{u} & = (u_x,u_y) \\ \mathbf{v} & = (v_x,v_y) \\ \mathbf{a} & = (a_x,a_y) \\ & = \bigg( \frac{v_x-u_x}{t}, \frac{v_y-u_y}{t}\bigg) \\ \mathbf{s} & = (s_x,s_y) \\ & = \bigg( u_xt+\frac{1}{2}a_xt^2, u_yt+\frac{1}{2}a_yt^2\bigg) \\ & = \bigg(\frac{1}{2}(u_x+v_x)t, \frac{1}{2}(u_y+v_y)t\bigg) \\ & = (x_2-x_1,y_2-y_1) \\ \end{aligned}

Is this time t also a natural parameter?

For a given parametric curve, the natural parametrization is unique up to a shift of parameter.

Wikipedia on Differentiable curve

(to be continued)