So I encountered this question:
Given that the relationship between distance (m) and velocity (v) of an object is $$v^2 = 1 - m^3$$
Find the acceleration of the object when $m=1$
By taking the derivative of each side with respect to $t$
$2v \frac{dv}{dt} = -3m^2 \frac{dm}{dt} $
and we know that $\frac{dv}{dt}$ = acceleration, $\frac{dm}{dt}$ is equal to $v$, then:
$2v a = -3m^2 v$
and by solving for $a$ without dividing by $v$ since $v$ can be zero, $a = -3/2$
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 $a$
That's how our teacher solved it. My questions are: is it permissible to substitute the value for a variable and keep the other, even if i know its value?
And why did we treat $v$ here as any other number other than zero?