Solving x(t) using F(v) for Particle of Mass m

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A particle of mass m is subject to a force F(v) = -m(alpha)v^2. The initial position is zero, and the initial speed is v nought. Find x(t).
 
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It's only a matter of solving the differential equation

[tex]-m\alpha v^2=m\frac{dv}{dt}[/tex]

and then

[tex]v=\frac{dx}{dt}[/tex]
 
Sorry perhaps I'm not catching on, so I have:

-1/v = -(alpha)t

Do I merely say that 1/(alpha t) = dx/dt or perhaps I'm missing something.
Or is my first step incorrect?
 
You forgot the constant of integration. The solution to the differential equation for v is

-(alpha)t = -1/v +C

And plugging v(0)=[itex]v_0[/itex] gives C=1/[itex]v_0[/itex]. So

[tex]v(t)=\frac{1}{\alpha t}+\frac{1}{v_0}[/tex]

And now your have to solve the differential equation.

[tex]\frac{dx}{dt}= \frac{1}{\alpha t}+\frac{1}{v_0}[/tex]

with initial condition x(0)=0 to find x(t).

Makes sense?
 
Okay, so I got:

x(t) = (ln(t)/alpha) + (t/v nought) + C
and C = 0 at x(0)

So I have x(t) = (ln(t)/alpha) + (t/v nought)

Is that correct or did I once again miss something?