# Equations of motion accounting for drag

Egaston

## Homework Statement

I need something like the equations of motion, but accounting for drag as given by the http://en.wikipedia.org/wiki/Drag_equation" [Broken]. Particularly for:

## Homework Equations

$$& v && = v_0+at \qquad$$
$$& s && = s_0 + v_0t + \tfrac12 at^2 \qquad$$

## The Attempt at a Solution

I have racked my brains for hours on this but can't make any progress. Google doesn't seem to be giving me much either (at least with my search terms). I am not formally educated in physics at all so there might be some obvious solution that I am missing.

I've tried to work out velocity by calculating drag and acceleration as a series of updates and seeing if it approaches any sort of useful value, but no matter what I try the equations always end up approaching either 0 or v0+at when I increase the frequency of the updates.

This isn't really a homework assignment, but this seems to be the only part of the forums suitable for these sorts of questions. Assistance would be really appreciated.

Last edited by a moderator:

Egaston;2156294[tex said:
& v && = v_0+at \qquad[/tex]
$$& s && = s_0 + v_0t + \tfrac12 at^2 \qquad$$

These equations are usually used if the acceleration is constant. By Newton's second law F=ma, if the acceleration is constant, the force must be constant.

In the drag equation, the force depends on the speed, hence the force will change in time as the speed changes in time, hence the acceleration will not be constant in time.

Egaston
Ah, of course. It makes sense when you say it like that. Since that is the case, what should I be looking at instead? Are there generalized forms of these equations for when acceleration is dependent on velocity or time?

Newton's second law: F=ma=mx'' (x'' is the second derivative of position, which is by definition the acceleration)

Then you have to specify what the forces are in the situation you are interested in, eg:
Gravity near the Earth's surface: F=mg
Simple drag: F=-kv=-kx' (x' is the first derivative of position, which is by definition the velocity)

Putting these together you get:
mg-kx'=mx'',
which is an equation containing derivatives of position x (ie. a differential equation in x).

Once you specify the initial position and velocity, you can solve the equation for position x as a function of time.