Distance as function of velocity

1. Feb 19, 2010

E&M

1. The problem statement, all variables and given/known data

Given that a = -g[1+(v/vt)2]

where,
v = velocity of an object
vt = terminal velocity
find x(v)

2. Relevant equations

v = dx/dt

3. The attempt at a solution

i started from a = dv/dt = dv/dx . dx/dt. but could not go to x(v)

Last edited: Feb 19, 2010
2. Feb 19, 2010

cepheid

Staff Emeritus
If 'g' is meant to be acceleration due to gravity, then I should point out that the right hand side of your equation for v doesn't have dimensions of velocity.

Anyway, to solve for x(v), notice that if you have v(t), then you should be able to figure out its inverse function t(v). That way, once you have solved for x(t) by integrating, you can substitute t(v) into it in order to get x(t(v)) = x(v).

3. Feb 19, 2010

E&M

oh yeah my bad. It's v_dot, or acceleration.

4. Feb 19, 2010

vela

Staff Emeritus
Need a bit more detail of what you did. So you had

$$v \frac{dv}{dx} = -g[1+(v/v_t)^2]$$

Then what?

5. Feb 19, 2010

E&M

at this point, i think i can ignore v. dv/dx and simply write a = dv/dt. I integrated after that which gave me an expression for v.

v = -gt - g/(vt)2 $$\int$$v2dt

b/c terminal velocity is a constant, i pulled it out of the integral.