MHB How do I plot a distance-time graph using TikZ?

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The discussion focuses on plotting a distance-time graph using TikZ while verifying speed calculations over specific time intervals. The user calculates average speeds for two segments: the first 6 seconds at 5 meters per second and the last 6 seconds at 10 meters per second, confirming that the latter is twice the former. The user seeks confirmation on the accuracy of their distance-time graph, which is affirmed by others in the thread. The TikZ code provided is noted as effective for creating the graph. Overall, the user successfully demonstrates their understanding of distance, time, and speed relationships in the context of graphing.
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Problem

View attachment 6052

Plot the above given information in a Distance time graph.

View attachment 6051

View attachment 6050

Where do I need help

Hoping my graph is correct

Show that the speed in the final 6 seconds is twice the speed in first 6 seconds

But I'm having trouble here , I know $ distance=\frac{distance}{time} $

Many Thanks :)
 

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The average speed $\overline{v}$ over some time interval $\Delta t$ is given by the change in distance $d$ divided by the change in time.

So for the first 6 seconds (assuming the initial position is at 0), we have:

$$\overline{v}_1=\frac{\Delta d}{\Delta t}=\frac{30-0}{6-0}$$

And for the last 6 seconds, we have:

$$\overline{v}_2=\frac{\Delta d}{\Delta t}=\frac{90-30}{12-6}$$

Can you now show that $\overline{v}_2=2\overline{v}_1$?
 
MarkFL said:
The average speed $\overline{v}$ over some time interval $\Delta t$ is given by the change in distance $d$ divided by the change in time.

So for the first 6 seconds (assuming the initial position is at 0), we have:

$$\overline{v}_1=\frac{\Delta d}{\Delta t}=\frac{30-0}{6-0}$$

And for the last 6 seconds, we have:

$$\overline{v}_2=\frac{\Delta d}{\Delta t}=\frac{90-30}{12-6}$$

Can you now show that $\overline{v}_2=2\overline{v}_1$?

Thank you very much :)

First 6 seconds

$$\overline{v}_1=\frac{\Delta d}{\Delta t}=\frac{30-0}{6-0}$$

$$\overline{v}_1=\frac{30-0}{6-0}=\frac{30}{6}$$=5 meters per second

Last 6 seconds

$$\overline{v}_2=\frac{90-30}{12-6}=\frac{60}{6}=$$10 meters per second

Yes Now it is shown that first 6 second speed=2*last 6 second speed

Have I plot the distance time graph correctly? :)

Many Thanks :)
 
Yes, your plots look good to me. (Star)

Using our new TikZ feature:

\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}
\addplot coordinates {(0,0) (2,10) (4,20) (6,30) (8,50) (10,70) (12,90)};
\end{axis}
\end{tikzpicture}
 
MarkFL said:
Yes, your plots look good to me. (Star)

Using our new TikZ feature:

\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}
\addplot coordinates {(0,0) (2,10) (4,20) (6,30) (8,50) (10,70) (12,90)};
\end{axis}
\end{tikzpicture}

Thank you very much MarkFL ! (Happy) (Smile) (Party) , And that's when TIKz comes handy! :)
 
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