Instantaneous velocity calculation from position data

[fog37]

Hello Forum,
Given position and related time data, what is the most correct way to calculate the instantaneous speed? For example, given the data (x1, t1) , (x2, t2), (x3, t3) , (x4, t4), (x5, t5), is the instantaneous speed at time t3 given more correctly by

$$v(t_3) = \frac{(x_4-x_2)} {(t_4-t_2)}$$
$$v(t_3) = \frac {(x3-x2) } {(t3-t2)}$$
$$v(t_3) = \frac {(x4-x3)} {(t4-t3)}$$

and why?

Thanks!

 

[Orodruin]

It is impossible to give a general answer to your question.

 

[boneh3ad]

Are you familiar with differentiation and, as a follow-on, finite differences?

 

[fog37]

Yes, differentiation is applicable when the position is given as a function of time. I will look into finite differences. My understanding is that there are different types of finite difference. I guess I will figure out which one is most appropriate for the calculation of the instantaneous speed. Any suggestion in the meantime?

Thanks

 

[boneh3ad]

Differentiation is simply the rate of change of one variable with respect to another. It doesn't have to be position or velocity and it doesn't have to be with respect to time. In this case, you are interested in time rate of change, but derivatives are much more broadly applicable than that.

Anyway, you seem to have some sense of the relationship between derivatives and the quantity you want here. Assuming you understand that, then your next question should be about approximating derivatives when you have data at a discrete set of points in time. That is where finite differences come into play.

 

[Orodruin]

I guess I will figure out which one is most appropriate for the calculation of the instantaneous speed.

My point is that you cannot say that one will be more appropriate than another. It depends on your sampling points as well as on how the velocity varies with time. I can come up with different situations where any of the given differences would be the best approximation.