# Find position function of freefall with air resistence ODE

• 1s1
In summary, this conversation provides a summary of how to find the position of an object released from rest. First, you need to solve an equation of motion with the constraint that the velocity is zero at the start. Then, you use the relationship between velocity and position to get x(t).

## 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?

1s1 said:

## 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.

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?

1s1 said:
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...

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 $x(t)$!

Zondrina said:
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.

vanhees71 said:
Given $v$, it's very clear, how to get $x(t)$!
Indeed. Finding ##\Gamma(x(t)+1)## is easy once we have ##v##.

vanhees71 said:
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.