Distance as function of velocity

[E&M]

Homework Statement

Given that a = -g[1+(v/v_t)²]

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

Homework Equations

v = dx/dt

The Attempt at a Solution

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

 

[cepheid]

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).

 

[E&M]

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

 

[vela]

The Attempt at a Solution

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

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

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

Then what?

 

[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/(v_t)² $$\int$$v²dt

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