# Kinematics with non constant acceleration II [closed]

I'm getting crazy with this problem and I think that it's pretty simple.

An helicopter's helix is spinning at initial speed $w_0=200\ rpm$, all of a sudden the motor stops and it decreases its velocity with a not constant acceleration of $\alpha=-0,01\cdot w \frac{rad}{s^2}$.

The question is: how many revolutions will it make until it stops?

I know that it can't be solved using regular kinematics formulas because the acceleration is not constant so I tried with the next chain rule:

$$a(x) = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v$$

that I've seen in another question where the acceleration depended on the $x$, and you can solve the problem with $a(x)dx = vdv$ but in my case it depends on the velocity $w$ and I don't know what to do!

I'm trying to get something like $a(w)dw= ...$ but I don't know how to arrive. This is my closest attempt:

$$\alpha=\frac{dw}{dt}\cdot\frac{dw}{dw}\rightarrow \alpha dw=\phi dw$$

and here I'm stucked.

I will appreciate any help! Thank you!

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## closed as off-topic by John Rennie, David Z♦Mar 26 '14 at 17:20

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You have a differential equation that says $$a(x) = -0.01*w = \frac{d w}{d t}$$ What you did with the change of variables is correct, so $w$ cancels on either side. Otherwise you have a first order differential equation to solve.
I am searching for $\phi$ so you are telling me that $\alpha = \phi$ if $w$ is cancelled? I am really confused! – Marc C Mar 26 '14 at 17:25
$dw/dt = dw/dx * dx/dt = w dw/dx$, $w$ cancels, now you are left with an integral until $w = 0$. – webb Mar 26 '14 at 17:34