# Integrate Equation: Find Position with Initial Drag Force & Velocity

• Ashley1nOnly
In summary, the conversation is discussing how to integrate the equation V/(-g-kv) dv in order to find the position of a potato being shot up in the air with air resistance and gravity affecting it. The methods of u-substitution, integration by parts, and partial fractions are discussed but deemed unsuccessful. The conversation then suggests using polynomial long division or rearranging the equation to a form that is integrable.
Ashley1nOnly

## Homework Statement

I'm trying to integrate this equation so that I can find my position using my initial drag force and initial velocity.

V/(-g-kv) dv

## The Attempt at a Solution

U substitution won't work
Integration by parts won't work
Partial fractions won't work.
It doesn't match any trigonometric equations

I'm not sure how to integrate it.
Basically a potato is being shot up in the air with air resistance and gravity affecting it.
This is what I started with
F=ma
F=-mg-kmv
M(dv/dt)= -mg-kmv
dv/dt=-g-kv
dv/(-g-kv)=dt
well v=dx/dt so dt=dx/v

dv/(-g-kv)=dx/v
Separation of variable from D.E
Vdv/(-g-kv)=dx
Now I want to integrate the left side which I don't know how to do.Thanks for any help

Last edited:
Ashley1nOnly said:

## Homework Statement

I'm trying to integrate this equation so that I can find my position using my initial drag force and initial velocity.

V/(-g-kv) dv

## The Attempt at a Solution

U substitution won't work
Integration by parts won't work
Partial fractions won't work.
It doesn't match any trigonometric equations

I'm not sure how to integrate it.
Basically a potato is being shot up in the air with air resistance and gravity affecting it.
This is what I started with
F=ma
F=-mg-kmv
M(dv/dt)= -mg-kmv
dv/dt=-g-kv
dv/(-g-kv)=dt
well v=dx/dt so dt=dx/v

dv/(-g-kv)=dx/v
Separation of variable from D.E
Vdv/(-g-kv)=dx
Now I want to integrate the left side which I don't know how to do.Thanks for any help

Do you know have to compute the indefinite integral
$$\int \frac{dx}{x}\quad ?$$

Ln(x)

Ashley1nOnly said:
dv/(-g-kv)=dt
This is the equation you should integrate.
Are you asked to find the displacement as a function of time?

If the
V dv
-------------
(-g-kv)

If the v wasn't at the stop I could integrate it but I don't know what to do with the V at the top

I'm not given time which is why I had to use the substitution dt=dx/v. Because I am given initial velocity and k

cnh1995 said:
This is the equation you should integrate.
Are you asked to find the displacement as a function of time?
I did that integral first then realized I didn't have time so I had to take a different route

Ashley1nOnly said:
I did that integral first then realized I didn't have time so I had to take a different route
Ok.
So,
v*dv/(g+kv)=-dx
Hint: Isolate the v in the denominator (get rid of the k attached to it).

Ashley1nOnly said:
If the
V dv
-------------
(-g-kv)

If the v wasn't at the stop I could integrate it but I don't know what to do with the V at the top

Take first out the minus sign from the denominator. Do then the substitution ##x=g+kv##.

Buffu
Ashley1nOnly said:
If the
V dv
-------------
(-g-kv)

If the v wasn't at the stop I could integrate it but I don't know what to do with the V at the top

just a suggestion :- try binomial expansion on the denominator.

So I did some rearrangements and got

dv
--------- =- dx
g/v +k

Then integrating both sides I get
Ln((g/v)+k)+c=-x

Assuming c is the initial velocity

Ashley1nOnly said:

## Homework Statement

I'm trying to integrate this equation so that I can find my position using my initial drag force and initial velocity.

V/(-g-kv) dv
First off, this isn't an equation -- an equation always has '=' somewhere in it.
Second, are V and v different quantities?
Ashley1nOnly said:

## The Attempt at a Solution

U substitution won't work
Substitution will work if V and v are different. If they are different, I'm assuming that V is a constant, and v is the variable velocity. If that's the case, substitution will work.
Ashley1nOnly said:
Integration by parts won't work
Partial fractions won't work.
It doesn't match any trigonometric equations

I'm not sure how to integrate it.
Basically a potato is being shot up in the air with air resistance and gravity affecting it.
This is what I started with
F=ma
F=-mg-kmv
M(dv/dt)= -mg-kmv
dv/dt=-g-kv
dv/(-g-kv)=dt
well v=dx/dt so dt=dx/v

dv/(-g-kv)=dx/v
Separation of variable from D.E
Vdv/(-g-kv)=dx
Now it appears that V and v both represent velocity, a variable. Writing both V and v in the same expression is very confusing.
Ashley1nOnly said:
Now I want to integrate the left side which I don't know how to do.
Write the equation as ##\int \frac {v~dv}{g + kv} = -\int dx##. On the left side, use polynomial long division.

Sorry when I type in the first letter it always wants to capitalize it. Well I shall search for understanding on polynomial long division. Thank you

Ashley1nOnly said:
So I did some rearrangements and got

dv
--------- =- dx
g/v +k

Then integrating both sides I get
Ln((g/v)+k)+c=-x

Assuming c is the initial velocity
That's not correct.
Try the methods in #9 and #12. If it's not too much, you can try this too.. Take the k out from the denominator and you'll have the equation in the form y*dy/(a+y). Just a simple rearrangement in the numerator and it will be integrable.

## 1. What is the equation used to find position with initial drag force and velocity?

The equation used is x = (v0/mg) * (1 - e^(-gt/m)) where x is the position, v0 is the initial velocity, m is the mass of the object, g is the acceleration due to gravity, and t is the time elapsed.

## 2. How does initial velocity and drag force affect the position of an object?

The initial velocity determines the starting point of the object's motion, while the drag force acts as a resistance force that slows down the object's movement. Together, they determine the final position of the object at a given time.

## 3. What is the significance of the term "e" in the equation?

The term "e" is a mathematical constant known as Euler's number, approximately equal to 2.71828. It is used in the equation to represent the natural growth rate and decay of a system, in this case the position of an object.

## 4. What happens to the position as time increases in the equation?

As time increases, the position of the object will approach a limiting value known as the terminal position. This is the point at which the drag force and gravity are balanced, resulting in a constant velocity.

## 5. Can this equation be used for any object in motion or are there limitations?

This equation can be used for objects in free fall or objects experiencing a constant drag force. However, it does not take into account other factors such as air resistance or changing drag forces, so it may have limitations in more complex situations.

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