How Can You Solve the Differential Equation a=dv/dt=-Cv^2 for v(t) and x(t)?

AI Thread Summary
To solve the differential equation a(t) = -Cv^2(t), the key approach is to use separation of variables. By rearranging the equation, one can express it as dv/v^2 = -C dt, allowing for integration on both sides. The integration leads to an expression for v(t) in terms of time t, which can then be further manipulated to find x(t) using the relationship between velocity and distance. The discussion highlights the confusion around integrating v^2(t) and emphasizes the importance of correctly applying integration techniques. Ultimately, the solution involves understanding the relationship between acceleration, velocity, and distance through proper integration.
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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.
 
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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 ?
 
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?
 
what??
notice that:
a=\frac{dv}{dt}=-Cv^2
how can you solve this differential equation?

hint: separation of variable!
 
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