dynamical list problem

[Thomas Klemm]

(01-04-2019 02:34 PM)Albert Chan Wrote:  It might help calculations if list is travelled (time distance):

{1 2 1 3 2} => {(0 0) (1 1) (2 3) (3 4) (4 7) (5 9)}

For speed of any distance, just interpolate for the time.

We can extend this table a bit to make sure that the difference between the travelled distance is 1m:

\(\begin{matrix}
\textbf{time } t & \textbf{distance } s(t)\\
0 & 0 \\
1 & 1 \\
1\frac{1}{2} & 2\\
2 & 3 \\
3 & 4 \\
3\frac{1}{3} & 5 \\
3\frac{2}{3} & 6 \\
4 & 7 \\
4\frac{1}{2} & 8 \\
5 & 9
\end{matrix}\)

And then calculate the difference of the times that are 5 meters apart:

\(\begin{matrix}
3\frac{1}{3} &- & 0 &=& 3\frac{1}{3} \\
3\frac{2}{3} &- & 1 &=& 2\frac{2}{3} \\
4 &- & 1\frac{1}{2} &=& 2\frac{1}{2} \\
4\frac{1}{2} &- & 2 &=& 2\frac{1}{2} \\
5 &- & 3 &=& 2
\end{matrix}\)

Either calculate the partial sum of the times (as suggested by Albert) and then calculate the difference or then add up 5 time entries (as I suggested).
You will end up with the same result.

Cheers
Thomas