# Drag forces

1. Nov 6, 2004

### UrbanXrisis

Last edited by a moderator: May 1, 2017
2. Nov 6, 2004

### Pyrrhus

Urban, what was the problem with the solution in the other thread?

3. Nov 6, 2004

### UrbanXrisis

I couldn't really follow what was going on. I didn't know there was a solution. In physics, I was given 2 formulas:

dv/dt=-g+bv/m <- this being the differentiable equatioin
and also:
v=mg/b(1-e(-bt/m)) <- which showed the velocity at any time

My homework asks to "solve" for the equations but I'm in high school so it only expects me to jump from

dv/dt=-g+bv/m
to
v=mg/b(1-e(-bt/m))

without actaully solving it
From that, they ask questions like...what if the drag force = -bv^2 and they change it. Then they tell me to find the differential equation for the object's motion. I THINK that all I have to do is sub in the numbers but I'm not quite sure. Cyclovenom, you did a great job but I'm only on the basics

4. Nov 6, 2004

### Pyrrhus

Urban if you want to check your solution then derivate it with respect to time, and it should give the right side of the equation.

5. Nov 6, 2004

### UrbanXrisis

what does that mean? "give the right side of the equation"? Do I have the wrong equations?

6. Nov 6, 2004

### Pyrrhus

For example:

For this differential equation

$$\frac{dy}{dt} = ay - b$$

its solution is: $(y_{o} = 0)$

$$y = \frac{b}{a} - \frac{b}{a}e^{at}$$

so if we substitute in the differential equation we should get both sides the same.

$$\frac{d(\frac{b}{a} - \frac{b}{a}e^{at})}{dt} = a(\frac{b}{a} - \frac{b}{a}e^{at}) - b$$

we get:

$$be^{at} = be^{at}$$

so that's the solution for our differential equation.