Calculate Vertical Displacement with Drag Force Equation | Physics Homework

[John Lam]

Homework Statement

Assuming that the drag force magnitude is given by the equation D= bv, where b is the drag parameter and v is the instantaneous velocity magnitude of the object.

(a) Show that the vertical displacement through which a dropped object must fall from to reach X% of its terminal velocity is given by the equation:
Δy= (v^2ty/g)*[(X/100)-ln((100-x)/100))
where vty is the object's terminal velocity.

Homework Equations

v=(-mg/b)*(1-e^(-bt/m))

The Attempt at a Solution

I honestly ran around in circles with this problem trying to integrate the given equation desperately trying to figure out what this problem was asking for exactly. Then I integrated v=(-mg/b)*(1-e^(-bt/m)). No luck there attempting to recreate the given equation for about three hours now.

 

[TSny]

Welcome to PF!

If you had an expression for dy/dv, maybe you could integrate it to get a relation between y and v.

Recall the chain rule: dy/dt = dy/dv ⋅ dv/dt

 

[bigguccisosa]

In your relevant equations you have the equation for velocity as it depends on time. Note that as t approaches infinity limit, v = -\frac{mg}{b}. This is terminal velocity. You want to find how far the distance must be to reach -\frac{x}{100}\frac{mg}{b}. Using this expression, you can solve for time needed to reach that speed. Finding distance is straightforward from there.

 

[John Lam]

But how would I lead it all back to proving this Δy= (v^2ty/g)*[(X/100)-ln((100-x)/100))

 

[bigguccisosa]

You are looking for the distance required to reach a specific velocity, yes? If you know the time required and you have the function of the velocity, how would you then go from velocity to distance?

 

[haruspex]

In your relevant equations you have the equation for velocity as it depends on time. Note that as t approaches infinity limit, v = -\frac{mg}{b}. This is terminal velocity. You want to find how far the distance must be to reach -\frac{x}{100}\frac{mg}{b}. Using this expression, you can solve for time needed to reach that speed. Finding distance is straightforward from there.

Yes, but TSny's method avoids the need to find the time.

 

[John Lam]

Since v=(-x/100)*(mg/b), I used kinematics equation d=vit+1/2 at^2, assuming a=0, I found t to be -100d*b/mgx

 

[haruspex]

Since v=(-x/100)*(mg/b), I used kinematics equation d=vit+1/2 at^2, assuming a=0, I found t to be -100d*b/mgx

That is a SUVAT formula. Those are only valid for constant acceleration.

 

[John Lam]

What equation should be used then to isolate t?

 

[haruspex]

What equation should be used then to isolate t?

There are several ways open to you.
You can start with the differential equation for the acceleration, then use TSny's method to make it a diffeential equation only involving velocity, distance, and a derivative of one with respect to the other (so no time in the equation). Solve that.
Since you are given the solution for velocity as a function of time, you can integrate that. You posted that you tried that but have not posted your working. If you post it we can lead you through it.

 

[Melika]

how would you go from t=(-m/b) ln((100-x)/(100)) to the vertical displacement?

 

[haruspex]

how would you go from t=(-m/b) ln((100-x)/(100)) to the vertical displacement?

Which post is that a response to? (Please use the Reply/Quote buttons)

 

[Melika]

Which post is that a response to? (Please use the Reply/Quote buttons)

The original post i guess. I'm working on the same problem.
The time it takes to reach X% of its terminal velocity is t=(-m/b) ln((100-x)/(100)).
Vy=(mg/b)(e^(-bt/m)-1)
If i plug in t to this equation i get
Vy=(mg/b)(e^(ln((100-x)/(100)) -1)
But I'm not sure what to do next? i know i need to integrate velocity to get displacement but I'm not sure how

 

[haruspex]

i know i need to integrate velocity to get displacement but I'm not sure how

You don't know how to integrate $\frac{mg}b(1-e^{-\frac{bt}m})$ with respect to t?