How to find terminal velocity from experimental data?

[greco117]

Homework Statement

So I have a set of data, velocity and time, of a falling object. So the question is: how can I get the value for the terminal speed?

Homework Equations

Since we're talking about terminal velocity an ecuation that descrives a Velocity vs time graph should be of the form:
$V =Vt\left(1-e^\left(\frac{-bt}{m}\right)\right)$

where $Vt$ is ther terminal velocity
$m$ is the mass
$b$ is a constant which depends of the caracteristics of the body
and $t$ is the time

The Attempt at a Solution

So I need to fit my data with an ecuation of this style,my first thought was that I had to do a logistic regression but that didn't look right (the ecuation is similar but is not the same),
Then I thought doing a little of algebra
$Vt-V=Vt\left(e^\left(\frac{-bt}{m}\right)\right)$
And then All I needed to do was an exponential regression but again that doesn't look like the best way because I'm already giving $Vt$ a value

 

[ehild]

If you have got measurement data at time t₁ and 2t₁ you can determine Vt by solving the system of equations
Vt-V₁=V_te^-Bt₁
Vt-V₂=V_te^-2Bt₁.
(B=b/m)

Square the first equation and divide by the second. The exponentials cancels, and you get an equation for V_t in terms of V₁ and V₂, easy to solve.

Find as many pairs as you can, and take the average of V_t-s.

ehild

 

[truesearch]

How many data points do you have?
Do you have enough to plot a graph of v against t?

 

[greco117]

If you have got measurement data at time t₁ and 2t₁ you can determine Vt by solving the system of equations
Vt-V₁=V_te^-Bt₁
Vt-V₂=V_te^-2Bt₁.
(B=b/m)

Square the first equation and divide by the second. The exponentials cancels, and you get an equation for V_t in terms of V₁ and V₂, easy to solve.

Find as many pairs as you can, and take the average of V_t-s.

ehild

So you're saying that I should do the following?
(Vt-V1)^2=Vt^2*e^-2bt
and then divided by
(VT-V2)=Vt*e^-2bt
and then I should get:
(Vt-V1)^2/(Vt-V2)=Vt
but that ecuation is not so easy to solve unless I've misunderstood the procedure

 

[greco117]

How many data points do you have?
Do you have enough to plot a graph of v against t?

yes, I have 56 data points, the problem lies in the graph because the points don't seem to have a tendency to a certain value, I thought about taking the maximum as VT but I don't like that solution.
Any suggestion?

 

[ehild]

So you're saying that I should do the following?
(Vt-V1)^2=Vt^2*e^-2bt
and then divided by
(VT-V2)=Vt*e^-2bt
and then I should get:
(Vt-V1)^2/(Vt-V2)=Vt
but that ecuation is not so easy to solve unless I've misunderstood the procedure

That equation is equivalent to (V_t-V₁)^2=V_t(V_t-V₂).
Expand both sides of the equation. : V_t^2-2V_tV₁+V₁^2=V_t^2-V_tV₂.

V_t^2 cancels. -2V_tV₁+V₁^2=-V_tV₂. Isolate V_t.
V_t=V₁^2/(2V₁-V₂)


ehild

 

[greco117]

thank you!
seems obvious now, that would do it

 

[ehild]

I guess you have got some data-processing program, Excel perhaps? As you have enough data points, it is possible to get the approximate derivative of the V(t) function. V' = V_tBe^-Bt. The plot ln(V') vs time has to be straight line, with slope -V_tB and crossing the vertical axis at V_tB.

ehild