MHB 4.1.26 graph of velocity over acceleration graph

AI Thread Summary
The discussion focuses on plotting a velocity graph over a given piece-wise linear acceleration graph. The user has drawn their own velocity graph in red, seeking to convert it into a TikZ format. They clarify that the acceleration graph is represented in yellow, and the red graph illustrates the velocity derived from integrating the acceleration function. The integration process is detailed, showing how to derive the velocity equation from the acceleration. The conversation emphasizes the need to continue this process for subsequent sections of the graph.
karush
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ok not finding this easy but the red is mine drawn over the given book graph

also want to convert the whole thing to tikx graph
 

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karush said:
ok not finding this easy but the red is mine drawn over the given book graph

also want to convert the whole thing to tikx graph

Is the problem's given acceleration graph the piece-wise linear graph in yellow?

Why the graph in red? What's its purpose?
 
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yes only the red in mine

we are asked to plot velocity(red) over the given graph of acceleration
 
for the given piece-wise linear acceleration graph in yellow, the velocity graph is as shown ...
 

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karush said:
ok not finding this easy but the red is mine drawn over the given book graph

also want to convert the whole thing to tikx graph

We can do for instance:
\begin{tikzpicture}[xscale=.3, >=stealth]
\draw[ystep=0.5,help lines] (0,-2.5) grid (45,2.5);
\draw[->] (-2,0) -- (47,0) node
{(s)};
\draw[->] (0,-2.4) -- (0,2.9) node[above] {$a$ (m/s$^2$)};
\draw
foreach \i in {5,10,...,45} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} }
foreach \i in {-2,2} { (0.3,\i) -- (-0.3,\i) node
{$\i$} }
(0,0) node[below left] {$0$};
\draw[red, ultra thick]
(5,2) parabola (0,0)
(5,2) parabola (10,0)
(15,-2) parabola (10,0)
(15,-2) -- (25,-2)
(25,-2) parabola (30,0)
(35,2) parabola (30,0)
(35,2) -- (40,2)
(40,2) parabola (45,0);
\end{tikzpicture}

I guess we still need to add the velocity graph.
For the section up to 10 seconds, we have the parabola given by:
$$a(t) = 2 - \frac{2}{25}(t-5)^2 = -\frac{2}{25}t^2+\frac 45 t$$
Integrate it, to find:
$$v(t) = \int_0^t a(t)\,dt = \int_0^t \left[-\frac{2}{25}t^2+\frac 45 t\right]dt
= \left[-\frac{2}{3\cdot 25}t^3 + \frac 25 t^2\right]_0^t = -\frac{2}{75}t^3 + \frac 25 t^2$$

Putting it in a graph, we get:
\begin{tikzpicture}[xscale=.3, yscale=.3, >=stealth]
\draw[help lines] (0,-2.5) grid (45,15);
\draw[->] (-2,0) -- (47,0) node
{(s)};
\draw[->] (0,-2.4) -- (0,15.9) node[above] {$v$ (m/s)};
\draw
foreach \i in {5,10,...,45} { (\i,0.3) -- (\i,-0.3) node[below] {$\i$} }
foreach \i in {-2,5,10,15} { (0.3,\i) -- (-0.3,\i) node
{$\i$} }
(0,0) node[below left] {$0$};
\draw[cyan, ultra thick] plot[domain=0:10, variable=\t] (\t, {-(2/75)*\t^3 + (2/5)*\t^2 });
\end{tikzpicture}

Repeat to find the later sections...
And integrate again to find the x graph...​
 
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