How can speed at a given position be estimated from a bunch of average speed measures at different positions?

Question

I have loose measures of average speed in different positions (speed is in the x axis given particularities of my own problem)

How can I estimate a fitting curve for 'instantaneous speed at each position' vs. 'position' ?

I guess the first step is to fit an average speed curve. What then? My plot shows that the average speed around 1000m changes from about 1.01m/s to 1.05m/s, which means that instantaneous speed at that region must have been considerably higher to compensate for the 0-800m region moving slowly.

Answer 1

In general the average speed for movement in 1D is calculated by $$v_{\text {avg}}=\frac{x_2-x_1}{t_2-t_1}$$ where the object in question is at position $x_1$ at time $t_1$ and at position $x_2$ at time $t_2$. If we are only given the average speed and one of either $x_1$ or $x_2$ then we can't do anything to determine the other variables. This is probably where your issue is.

However, you have stated in the comments that the time intervals and displacements for the average speeds are calculated relative to the same position and time. Therefore, we know what $x_1$ and $t_1$ are. Your data points then tell us $(v_{\text{avg}},x_2$, and all we don't know is $t_2$. We can use the definition of average speed to determine $t_2$: $$t_2=\frac{x_2-x_1}{v_{\text{avg}}}+t_1$$

Therefore, you can use your data to get $(x_2,t_2)$ points, which you can then use usual methods to determine the instantaneous velocity as a function of $t_2$.

Answer 2

We normally write average speed as integral of instantaneus speed with respect to the time divided by total time interval. However, it's better to write it with respect to position in discrete form here i.e: $$v_{avg}=\frac{1}{\sum_{i}\Delta x_i}\sum_{i}v(x_i)\Delta x_i=\frac{1}{X}\sum_{i}v(x_i)\Delta x_i~~(1)$$ $\Delta x_i$ Denotes the position interval where instantaneus velocity $v(x_i)$ remains the same. If you had a continuous data, I'd have used integral and $d x$ instead of $\Delta x_i$. Equation $(1)$ gives: $$Xv_{avg}=\sum_{i}v(x_i)\Delta x_i ~ ~(2)$$ We have many $v_{avg}$ at different positions. $X$ is known from graph too (it's just the total distance from orgin to the arbitrary point). is it possible to find $v(x_i)$ then? Probably! It will depends on linearity of $v(x_i)$. Let's do it for 2 point: $$X_2v_{2avg}-X_1v_{1avg}=\sum_{j}v(x_j)\Delta x_j-\sum_{i}v(x_i)\Delta x_i ~ ~(3)$$ Left side is known. We can also expand the right side such that we get: $$X_2v_{2avg}-X_1v_{1avg}=\sum_{i}v(x_i)\Delta x_i-\sum_{i}v(x_i)\Delta x_i~+~...~+v(x_{n-1})\Delta x_{n-1}+v(x_n)\Delta x_n ~ ~(4)$$ Let's say that $\Delta x_i$ is the same for all instantaneus velocities (linearity) and we can assume something for its value. The fewer data you have, the bigger $\Delta x_i$ should be. In the case of 2 points, we can always assume $\Delta x_i$ such that equation $4$ reduces to $$\frac{X_2v_{2avg}-X_1v_{1avg}}{\Delta x}=v(x_n) ~ ~(5)$$ This will give us an instanteus velocity at a given position. This instanteus velocity is the same for all points in $\Delta x$. Thus if you are looking for a "real" instanteus velocity, you have to assume $\Delta x_i$ as small as possible.In your case where we have a lot of data, one should use equation 4, without reduction. After all, you can write it between any two arbitrary points, So for every unknown parameter on the right side, there is an independent equation hopefully. (i.e you can assume a small value for $\Delta x_i$ such that many terms for instantaneus velocity at different positions emerge in the right side, you can find them all with other equations)

Note that the equation $(1)$ is like $$v_{avg}=\frac{1}{T}\int_{0}^{T}v(t)dt~~~or~~~~v_{avg}=\frac{1}{T}\sum_{i}v(t_i)\Delta t_i$$ And of course $$v_{avg}=\frac{1}{T}\int_{0}^{T}v(t)dt=\frac{X}{T}$$