Why is this integral not returning to the original equation when derived?

Question

A projectile is fired into fluid at a rate of $60$ (nevermind the units on this one.)

It decelerates such that $a=(-.4)v^3$. This is all fine and dandy. The book provides this solution. $$a=\frac{dv}{dt}$$ $$dt=\frac{dv}{a}$$ $$\int_0^tdt=\int_{60}^v\frac{1}{-.4v^3}dv$$

Skipping a few steps, we have: $$t=\frac{1}{.8}(\frac{1}{v^2}-\frac{1}{60})$$ $$v=[(0.8)t+\frac{1}{60}]^{-1/2}$$

But when I derive this with respect do t, ie $\frac{dv}{dt}$, it comes to $$\frac{-0.4}{(0.8t+1/60)^{3/2}}$$which of course does not =$-.4v^3$.

What's wrong? I can't come back to the equations the work appears to be derived from.

Thanks!

Answer 1

Your derivation actually shows you do have $a = -0.4v^3$.

Remember in your previous work that $$ v = \frac{1}{(0.8t + \frac{1}{60})^\frac{1}{2}} $$