# F(v) related to x(t)

1. Oct 11, 2006

### Tonyt88

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).

2. Oct 11, 2006

### quasar987

It's only a matter of solving the differential equation

$$-m\alpha v^2=m\frac{dv}{dt}$$

and then

$$v=\frac{dx}{dt}$$

3. Oct 11, 2006

### Tonyt88

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?

4. Oct 11, 2006

### quasar987

You forgot the constant of integration. The solution to the differential equation for v is

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

And plugging v(0)=$v_0$ gives C=1/$v_0$. So

$$v(t)=\frac{1}{\alpha t}+\frac{1}{v_0}$$

And now your have to solve the differential equation.

$$\frac{dx}{dt}= \frac{1}{\alpha t}+\frac{1}{v_0}$$

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

Makes sense?

5. Oct 11, 2006

### Tonyt88

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?

6. Oct 11, 2006

### quasar987

It looks fine.