Given that the relationship between distance (m) and velocity (v) of an object is \(\displaystyle v^2 = 1 - m^3\)

Find the acceleration of the object when \(\displaystyle m=1\)

By taking the derivative of each side with respect to \(\displaystyle t\)

\(\displaystyle 2v \frac{dv}{dt} = -3m^2 \frac{dm}{dt} \)

and we know that\(\displaystyle \frac{dv}{dt}\) = acceleration, \(\displaystyle \frac{dm}{dt} = v\), then:

\(\displaystyle 2v a = -3m^2 v\)

and by solving for \(\displaystyle a\) without dividing by \(\displaystyle v\) since \(\displaystyle v \) can be zero(assuming that we don't know yet, \(\displaystyle a = -3/2\)

But hold on... we've just substituted m with 1, this means that we have to substitute v with zero, but if we do this we would turn out with 0 = 0, without solving for a

If we don't substitute v with 1, we wouldn't be able to solve for acceleration if we cancel v's with dividing, and it doesn't really make sense to keep v in there while we know that v = 0 when m = 1. My teacher clarified that by going back to the definition of derivative, that it's after all nothing but a limit and a limit is an approximation of some f(x) when x approaches some value, so a limit which equals 0 would be in real something like 0.0001 or 0.000000001 in a way we would be able to cancel v with v. But that doesn't really make sense to me. If that's true, when to use the approximate definition of limits and when to use the accurate? And what can we do for this equation?