# Derive Velocity Equation

1. Jan 31, 2007

### dagr8est

Given a(t) = -Cv^2(t) where C is a constant, express v(t) and x(t) as explicit functions of time t. Assume v(0) = v0

So I tried integrating both sides of the equation to get:
v(t)+C' = -C integral v^2(t) dt

but then I how am I supposed to integrate v^2(t) dt...

Also, I thought I could do it another way:

velocity
a(t) = -Cv^2(t)
v(t)/t = -Cv^2(t)
-1/(Ct) = v(t)

distance
-1/(Ct) = v(t)
-1/(Ct) = x(t)/t
-1/C = x(t)

Well the distance can't be constant so I guess that doesn't work either. I don't even know why this 2nd method doesn't work... It looks fine to me. Any help would be appreciated.

2. Jan 31, 2007

### Staff: Mentor

You are correct in using integration to go from a(t) to an expression for v(t).

Are you just asking how to do the indefinite integral $$\int v^2 (t) dt$$ ?

3. Jan 31, 2007

### dagr8est

Yeah, I don't get how to integrate that. Wouldn't it be something like v^3(t)/(3a(t)) but where does the a(t) come from? It's not a constant so you just can't add it in right?

4. Jan 31, 2007

### tim_lou

what??
notice that:
$$a=\frac{dv}{dt}=-Cv^2$$
how can you solve this differential equation?

hint: separation of variable!