Integrating the force equation when velocity drops to zero

[Speags]

i have a force acting on and object and I've gotten the question down to when i have to intergrate but I'm stuck on the intergral
$$F=-(f+cv^2)$$
$$m \frac{dv}{dx} \frac {dx}{dt} = -(f+cv^2)$$
f and c are constants
$$mvdv=-(f+cv^2)dx$$
$$\frac{mvdv}{f+cv^2}=-dx$$
now how do i intergrat them from $$v_{o} to 0 and 0 to x$$

 

[maverick280857]

This is the kind where the numerator has to be expressed as a constant times the derivative of the denominator. Find the constant and you're through. Simply put, differentiate the denominator and observe the similarity with the numerator.

 

[wais]

To integrate this equation, we can use the method of partial fractions. First, we can rewrite the equation as:

\frac{mvdv}{f+cv^2}=-dx

Next, we can use the substitution u = f+cv^2, du = 2cv dv to rewrite the integral as:

\frac{m}{2c}\int\frac{du}{u}=-\int dx

Solving for u, we get:

u = f+cv^2 = f+\frac{m}{2c}\frac{du}{dv}

Substituting this back into the integral, we get:

\frac{m}{2c}\int\frac{du}{u}=-\int dx

Solving for v, we get:

v = \sqrt{\frac{2c}{m}}\sqrt{u-f}

Now, we can integrate this equation from v_{o} to 0 and 0 to x to find the final solution for v(x):

v(x) = \sqrt{\frac{2c}{m}}\left(\sqrt{f}-\sqrt{f+cx^2}\right)

This is the final solution for the velocity of the object as a function of its position. To find the position of the object, we can simply integrate the velocity function with respect to x:

x(t) = \frac{1}{c}\left(\sqrt{f}x-\frac{m}{4c}x^2+\frac{m}{4c}x_{o}^2\right)

where x_{o} is the initial position of the object.