Find position function of freefall with air resistence ODE

[1s1]

Homework Statement

An object of mass $5$kg is released from rest $1000$m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant $b=50$N-sec/m, determine the equation of motion (for the position $x(t)$) of the object.

Homework Equations

$$m\frac{dv}{dt} = mg-kv$$

The Attempt at a Solution

The given differential equation describes the velocity with respect to time. How do I turn it into a differential equation that describes the position with respect to time?

 

[STEMucator]

Homework Statement

An object of mass $5$kg is released from rest $1000$m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant $b=50$N-sec/m, determine the equation of motion (for the position $x(t)$) of the object.

Homework Equations

$$m\frac{dv}{dt} = mg-kv$$

The Attempt at a Solution

The given differential equation describes the velocity with respect to time. How do I turn it into a differential equation that describes the position with respect to time?

Your relevant equation is a bit off, shouldn't it read $m\frac{dv}{dt} = (-1)(mg+kv)$?

If that's the case then solving it with the constraint that $v(0) = 0$ should simplify things a bit.

 

[1s1]

You may be right about the relevant equation being negative on the RHS. My book has it written as I did, but my prof mentioned that physicists sometimes make sign changes to the constant.

But I am still confused about how to make the function into a function of position? I assume you use the relationship $d=vt$? Or maybe there is an entirely different formula derived for the position rather than the velocity?

 

[STEMucator]

You may be right about the relevant equation being negative on the RHS. My book has it written as I did, but my prof mentioned that physicists sometimes make sign changes to the constant.

But I am still confused about how to make the function into a function of position? I assume you use the relationship $d=vt$? Or maybe there is an entirely different formula derived for the position rather than the velocity?

Notice this was part of your question:

An object of mass 5kg is released from rest

This means at $t=0$, the velocity of the object is zero. That's why I mentioned solving the equation with the constraint $v(0) = 0$. Right now you have an acceleration equation since it involves derivatives of velocity, but if you integrate acceleration, you get velocity;if you integrate vel...

 

[vanhees71]

the sign is given by the direction of the $x$ axis. If you define it to point "down", i.e., in direction of the constant gravitation force, the positive sign $+mg$ is correct. The sign in $-k v$ (assuming $k>0$) is the same for both directions of the $x$ axis, because the friction force must always be such that it tends to decrease $v$.

Further note, that you have a nice inhomogeneous linear differential equation of first order with constant coefficients, which is pretty easy to solve by separation of variables. The gives you $v(t)$.

Then you have to solve another very simple equation
$$v=\frac{\mathrm{d} x}{\mathrm{d} t}.$$
Given $v$, it's very clear, how to get $x(t)$!

 

[Mandelbroth]

Your relevant equation is a bit off, shouldn't it read $m\frac{dv}{dt} = (-1)(mg+kv)$?

Why? Air RESISTANCE implies that it will go against the direction of acceleration. This is a physics problem, primarily.

We're supposed to solve the equation and make a position function, right? I will denote $\frac{dv}{dt}$ as $v'$, for clarity and ease of notation (id est, laziness. It's a friday. Give me a break :tongue:).

I'm assuming we have $k=b$, right? So, we have $$mv'=mg-bv \\ v' +\frac{b}{m}v=g.$$ In this form, we can introduce an integrating factor. Can you take it from there?

Edit: vanhees71 got there first by noting the inhomogeneous equation.

Given $v$, it's very clear, how to get $x(t)$!

Indeed. Finding $\Gamma(x(t)+1)$ is easy once we have $v$.

 

[1s1]

This is very helpful, thanks!

 

[STEMucator]

the sign is given by the direction of the $x$ axis. If you define it to point "down", i.e., in direction of the constant gravitation force, the positive sign $+mg$ is correct. The sign in $-k v$ (assuming $k>0$) is the same for both directions of the $x$ axis, because the friction force must always be such that it tends to decrease $v$.

Further note, that you have a nice inhomogeneous linear differential equation of first order with constant coefficients, which is pretty easy to solve by separation of variables. The gives you $v(t)$.

Then you have to solve another very simple equation
$$v=\frac{\mathrm{d} x}{\mathrm{d} t}.$$
Given $v$, it's very clear, how to get $x(t)$!

I suppose it depends on how you decide to define up and down.

For me that's dependent on the day of the week :P.