Finding velocity function from a model of force

AI Thread Summary
The discussion revolves around deriving the velocity function from a force model involving a resistive force of -βv². The initial equation derived is ma = -βv², leading to a separable differential equation. The user attempts to solve it and arrives at a function for velocity but encounters a division by zero issue when plugging in t=0. A participant points out a mistake in the rearrangement of the equation, clarifying that the correct form is v(t) = [β/m * t + 1/v0]⁻¹. The user acknowledges the error and appreciates the correction, indicating a resolution to their confusion.
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Homework Statement



Hi, I've got an assignment which involves finding the function of velocity against time, given the information that the particle is moving with intial velocity of v0 and that it experience a resistive force of -βv2

Homework Equations



From the problem I've derived model of: ma = -βv2

The Attempt at a Solution



re-arranging the model, I have equation of:

dv/dt = -β/m * v2

then, using techniques from seperable equation,

dv/v2 = -β/m * dt

-1/v = -β/m * t + C1

pluging in v(0) = v0

-1/v0 = -β/m * 0 + C1
Or
C1 = -1/v0

Now, plugging in C and rearraging the equation, I obtained

-1/v = -β/m * t - 1/v0

v = m/(β * t) + v0

So as you can see, I'm able to derived the velocity function, but the problem lies to the fact that if I try to plug in t=0 for the velocity, then the function will not work as division by zero is not allowed

Therefore, I'm thinking that something is wrong with my velocity function but up till now I can't still figure it out

Thank you in advance!
 
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Your calculus is fine, you just forgot how to do division:\frac{1}{v(t)} = \frac{\beta}{m} t + \frac{1}{v_0}\Rightarrow v(t) = \left[ \frac{\beta}{m} t +\frac{1}{v_0} \right]^{-1} \neq \frac{m}{\beta t} + v_0
 
Ah yes thank you very much at pointing out my mistake, guess I spend too much time thinking how to actually solve the differential equation that I forgot about that. In any case, thank you for helping me again :)
 
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