Free fall with air resistance, must find velocity

In summary, the problem is to find the velocity as a function of time for an object in free fall with an initial velocity of zero, using the formula a = g - kv. The solution involves solving a first-order differential equation using integrating factors and results in the equation v = \frac{g}{k}(1-e^{-kt}). There are alternative methods to solve the problem, such as using integration by parts or solving for a homogeneous or non-homogeneous equation. However, it is important to consider all terms in the equation and not neglect any, as it can lead to incorrect solutions.
  • #1
Merbdon
4
0

Homework Statement



The problem is to find the velocity as a function of time, given the following;
the object is in free fall and its initial velocity is zero. The formula given for the air resistance is: a = g - kv, where g is acceleration d/t gravity, k is a constant, and v is velocity.

Homework Equations


The problem gives the hint to solve using integration by parts, with u = g - kv


The Attempt at a Solution


The solution given in the back of the book is v = [itex]\frac{g}{k}[/itex](1-e-kt)
I have no idea how to arrive at this solution.
 
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  • #2
Merbdon said:

Homework Statement



The problem is to find the velocity as a function of time, given the following;
the object is in free fall and its initial velocity is zero. The formula given for the air resistance is: a = g - kv, where g is acceleration d/t gravity, k is a constant, and v is velocity.

Homework Equations


The problem gives the hint to solve using integration by parts, with u = g - kv


The Attempt at a Solution


The solution given in the back of the book is v = [itex]\frac{g}{k}[/itex](1-e-kt)
I have no idea how to arrive at this solution.

[itex]g - kv[/itex] is the expression for the net downward acceleration of the object, not just the deceleration due to wind resistance.

Set up the differential equation relating velocity and time, to begin with.
 
  • #3
Merbdon said:

Homework Statement



The problem is to find the velocity as a function of time, given the following;
the object is in free fall and its initial velocity is zero. The formula given for the air resistance is: a = g - kv, where g is acceleration d/t gravity, k is a constant, and v is velocity.

Homework Equations


The problem gives the hint to solve using integration by parts, with u = g - kv


The Attempt at a Solution


The solution given in the back of the book is v = [itex]\frac{g}{k}[/itex](1-e-kt)
I have no idea how to arrive at this solution.

You can use differential form of acceleration, ie [itex]\large{a = \frac{dv}{dt}}[/itex]
 
  • #4
You don't have to integrate by parts, BTW. This is an easily separable first order differential equation. Maybe a simple substitution to clarify the form, but that's it.
 
  • #5
Thanks for the replies everyone, but I'm still not seeing it. I am not trying to copy and paste an answer here, I'm doing this on my own time, trying to study before I go back to school in the Fall.
Where does the e come from? Is there some basic calculus that I'm overlooking? It looks to me that solving the problem should go something like this:

[itex]\frac{dv}{dt}[/itex]= g - kv

∫[itex]^{t}_{0}[/itex]dvdt = ∫[itex]^{t}_{0}[/itex](g - kvt)dt

v = gt - kvt → v = [itex]\frac{gt}{1 + kt}[/itex]

Which makes sense, but doesn't match the answer in the book. Is the book incorrect, or am I? What am I missing here?

Thanks again for the help everyone.
-Merb
 
  • #6
When you neglect the gravity term, you should immediately see that the solution of
[itex]\frac{dv}{dt}=-v[/itex]
is
[itex]v=e^{-t}[/itex]
because of the property of exponential functions:
[itex]\frac{d(e^{-t})}{dt}=-e^{-t}[/itex]

Look up how to solve first order ode's using integrating factors and solving homogeneous and non-homogeneous equations. Hope this helps.
 
  • #7
Why would you neglect the gravity term?
 
  • #8
Merbdon said:
Thanks for the replies everyone, but I'm still not seeing it. I am not trying to copy and paste an answer here, I'm doing this on my own time, trying to study before I go back to school in the Fall.
Where does the e come from? Is there some basic calculus that I'm overlooking? It looks to me that solving the problem should go something like this:

[itex]\frac{dv}{dt}[/itex]= g - kv

∫[itex]^{t}_{0}[/itex]dvdt = ∫[itex]^{t}_{0}[/itex](g - kvt)dt

v = gt - kvt → v = [itex]\frac{gt}{1 + kt}[/itex]
This integration is wrong. v is a function of t and you cannot treat it as a constant as you did on the right.

Which makes sense, but doesn't match the answer in the book. Is the book incorrect, or am I? What am I missing here?

Thanks again for the help everyone.
-Merb
From
[tex]\frac{dv}{dt}= g- kv[/tex]
you get
[tex]\int\frac{dv}{g- kv}= \int dt[/tex]
to integrate on the left, let u= g- kv so that du= -k dv, dv= (-1/k)du
[tex]-\frac{1}{k}\int\frac{du}{u}= \int dt[/tex]
so that
[tex]-\frac{1}{k}ln u= ln(u^{-1/k})= t+ c[/tex]
where c is the constant of integration. Taking the exponential of both sides
[tex]u^{-1/k}= e^{t+ c}= Ce^t[/tex]
Take the -k th power of both sides to get
[tex]u= g- kv= Ae^{-kt}[/tex]
where [itex]A= C^{-k}[/itex]

Finally, solve for v:
[tex]v(t)= \frac{Ae^{kt}- g}{k}[/tex]
 
  • #9
Thank you very much Halls, that was a huge help. (It is a definite integral, though, evaluated at an arbitrary time t and assuming v=0 at t=0. I did away with the constant of integration.) I still can't get the answer that is given in the back of the book, however. When I manipulate it to a form similar to the answer above, I get v=[itex]\frac{g}{k}[/itex](1-[itex]\frac{1}{g}[/itex]e-kt) (the exp should not be divided by g). Another strange thing is that both the answer I got using your method and the answer in the book check out when you differentiate with respect to t and set the result equal to a=g-kv, so both appear to be valid, but that can't be right. Any ideas?
 
  • #10
From
[tex]\frac{dv}{dt}= g- kv[/tex]
you get
[tex]\int\frac{dv}{g- kv}= \int dt[/tex]
to integrate on the left, let u= g- kv so that du= -k dv, dv= (-1/k)du
[tex]-\frac{1}{k}\int\frac{du}{u}= \int dt[/tex]


How would you solve differently if you had m(dv/dt)= mg- kv2
 
  • #11
[itex]\int \frac{1}{1-ax^2}dx=\frac{1}{\sqrt{a}}arctanh(\sqrt(a)x)[/itex]

There is probably a nice variable transformation to get an easier intermediate integral in case you don't know the above integral.
 
  • #12
Thanks!
 

1. What is free fall with air resistance?

Free fall with air resistance is a type of motion where an object is falling towards the ground due to the force of gravity, but is also experiencing air resistance or drag from the surrounding air. This results in a slower and more gradual descent compared to an object in a vacuum.

2. How does air resistance affect the velocity of an object in free fall?

Air resistance acts in the opposite direction of an object's motion and increases as the object's velocity increases. This means that the faster an object falls, the more air resistance it experiences, which ultimately slows down its velocity.

3. How can the velocity of an object in free fall with air resistance be calculated?

The velocity of an object in free fall with air resistance can be calculated using the equation v = gt - (kv/m), where v is the velocity, g is the acceleration due to gravity, t is the time, k is the air resistance coefficient, and m is the mass of the object.

4. What factors can affect the air resistance coefficient in free fall?

The air resistance coefficient, or drag coefficient, can be affected by several factors including the shape and size of the object, the density and viscosity of the air, and the velocity of the object. These factors can change the amount of air resistance an object experiences and thus affect its velocity in free fall.

5. How does the mass of an object affect its velocity in free fall with air resistance?

The mass of an object does not directly affect its velocity in free fall with air resistance. However, a heavier object will experience more air resistance and therefore have a lower final velocity compared to a lighter object with the same shape and size.

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