MHB How does tikx help to create graphs for a car's motion?

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Physics
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
from tikx package...

\begin{tikzpicture}
%\draw (0,5) -- (6,5);
\draw [thick] (0,1) -- (18/2,1);
\node at (0,.8){0};
\node at (18/2,.8){180};
%\draw (1,5)--(1,4)--(2,4);
%\node at (1.5,4.1) {v};
\draw[step=.45 cm,gray,very thin,dashed]
%(-6,0)
grid (18/2,8);
\end{tikzpicture}

At $t=0$ a car is stopped at a traffic light
When the light turns green, the car starts to speed up,
and gains speed at a constant rate until it reaches a speed of $2 m/s \, 8s$ after the light turns green.
The car continues at a constant speed for $60m$
Then the driver sees a read light up ahead at the next intersection, and starts slowing down at a constant rate.
The car stops at the red light, $180 m$ form where it was at $t=0$a) Draw $x_t, v_t,$ and $a_t$ graphs for the motion of the car
b) In a motion diagram show the position, velocity and acceleration of the car.ok I am tying to do this using tikx but got stuck at the beginingI currently have the x-axis at distance but maybe it shud be time
we might need actually 2 graphs

I think the velocity is really $2m/s$
$v = v_0 + at$
then
$\dfrac{v-v_0}{t}=a$
so
$\dfrac{2m/s}{s}=\dfrac{2m}{s^2}=a$
 
Last edited:
Mathematics news on Phys.org
If you have the x-axis as distance, what is the y-axis? I suggest that the x-axis be t, time, and the x-axis be the distance traveled in that time. At constant speed the graph is a straight line, at constant acceleration or deceleration the graph is a parabola.
 
Last edited by a moderator:
why would a constant acceleration be a parabola the slope doesn't change

I think one graph a-t would be the one I should do

here is sample from another problem from google images

View attachment 9249
 

Attachments

  • a-t.jpg
    a-t.jpg
    4.8 KB · Views: 114
Last edited:
The speed of $2\text{ m/s}$ seems rather low for a car — it's a walking speed.

Either way, we could for instance draw the graphs like this:
\begin{tikzpicture}[xscale=0.2,yscale=.04]
%preamble \usepackage{amsmath}
\draw[xstep=2,ystep=10,lightgray,very thin] (0,-20) grid (150,260);
\draw (8,0) -- (8,250) (38,-18) -- (38,100) (150,-18) -- (150,180);
\draw[thick,->] (0,0) node[below] {0 s} -- (8,0) node[below] {8 s} -- (38,0) node[below] {38 s} -- (150,0) node[below] {150 s} -- (155,0) node
{$t$};
\draw[thick,->] (0,0) -- (0,260);
\draw[ultra thick,red] (0,250) -- node[below] {$a_t$} (8,250) node
{$0.25\text{ m/s}^2$} -- (8,0) -- (38,0) -- (38,-18) node
{$-0.018\text{ m/s}^2$} -- (150,-18);
\draw[ultra thick,cyan] (0,0) -- node[above left] {$v_t$} (8,100) node
{2 m/s} -- (38,100) node[above] {2 m/s} -- (150,0);
\draw[ultra thick,blue] (0,0) parabola (8,8) node[above left] {8 m} -- node[above left] {$x_t$} (38,68) node[below right] {68 m} parabola bend (150,180) (150,180) node[above] {180 m};
\end{tikzpicture}​
 
yeah I think it should be 20 not 2
the copy was really hard to read...
that would make the horizontal lite blue line to 8+3=11s
wow thanks for the graph

ok I did this fot a-t and 20 m/s

\begin{tikzpicture}[xscale=0.2,yscale=.04]
%preamble \usepackage{amsmath}
\draw[xstep=2,ystep=10,lightgray,very thin] (0,0) grid (40,160);
%\draw (8,0) -- (8,250) (38,-18) -- (38,100) (150,-18) -- (150,180);
\draw[thick,->]
(0,0) node[below] {0 s}
-- (16,0) node[below] {8 s} -- (11*2,0) node[below] {11 s} -- (15*2,0) node[below] {15 s} -- (40,0) node
{$t$};
\draw[thick,->] (0,0)-- (0,100) node
{20 m/s} -- (0,160);
%\draw[ultra thick,red] (0,250) -- node[below] {$a_t$} (8,250) node
{$0.25\text{ m/s}^2$} -- (8,0) -- (38,0) -- (38,-18) node
{$-0.018\text{ m/s}^2$} -- (150,-18);
\draw[ultra thick,cyan] (0,0) -- node[above left] {$v_t$} (8*2,100) -- (11*2,100) -- (30,0);

% \draw[ultra thick,blue] (0,0) parabola (8,8) node[above left] {8 m} -- node[above left] {$x_t$} (38,68) node[below right] {68 m} parabola bend (150,180) (150,180) node[above] {180 m};
\end{tikzpicture}

clueless about the speed parabola​
 
Last edited:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top