# Derive Velocity Equation

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.

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 $$\int v^2 (t) dt$$ ?

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?

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

hint: separation of variable!