Free fall with air resistance, must find velocity

[Merbdon]

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 = \frac{g}{k}(1-e^-kt)
I have no idea how to arrive at this solution.

 

[Curious3141]

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 = \frac{g}{k}(1-e^-kt)
I have no idea how to arrive at this solution.

g - kv 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.

 

[cupid.callin]

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 = \frac{g}{k}(1-e^-kt)
I have no idea how to arrive at this solution.

You can use differential form of acceleration, ie \large{a = \frac{dv}{dt}}

 

[Curious3141]

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.

 

[Merbdon]

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:

\frac{dv}{dt}= g - kv

∫^{t}_{0}dvdt = ∫^{t}_{0}(g - kvt)dt

v = gt - kvt → v = \frac{gt}{1 + kt}

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

 

[bigfooted]

When you neglect the gravity term, you should immediately see that the solution of
\frac{dv}{dt}=-v
is
v=e^{-t}
because of the property of exponential functions:
\frac{d(e^{-t})}{dt}=-e^{-t}

Look up how to solve first order ode's using integrating factors and solving homogeneous and non-homogeneous equations. Hope this helps.

 

[Merbdon]

Why would you neglect the gravity term?

 

[HallsofIvy]

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:

\frac{dv}{dt}= g - kv

∫^{t}_{0}dvdt = ∫^{t}_{0}(g - kvt)dt

v = gt - kvt → v = \frac{gt}{1 + kt}

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

Finally, solve for v:
v(t)= \frac{Ae^{kt}- g}{k}

 

[Merbdon]

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=\frac{g}{k}(1-\frac{1}{g}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?

 

[sunnyceej]

From
\frac{dv}{dt}= g- kv
you get
\int\frac{dv}{g- kv}= \int dt
to integrate on the left, let u= g- kv so that du= -k dv, dv= (-1/k)du
-\frac{1}{k}\int\frac{du}{u}= \int dt

How would you solve differently if you had m(dv/dt)= mg- kv²

 

[bigfooted]

\int \frac{1}{1-ax^2}dx=\frac{1}{\sqrt{a}}arctanh(\sqrt(a)x)

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

 

[sunnyceej]

Thanks!