How to Solve a Challenging Physics Problem with Proportional Acceleration?

[SheldonG]

Homework Statement

A body has acquired a downward velocity $v_0$ and thereafter the acceleration is proportional to the distance traveled. Measure time and distance from the instant and place where the object has acquired the velocity $v_0$ and find the formula or the distance traveled as a function of time.

Homework Equations

Suggestion in text to use

$$\frac{dv}{dt} = v\, \frac{dv}{dy}$$

The Attempt at a Solution

I have found this problem very challenging. I'd appreciate any suggestions. I know it's long and complicated, and I am asking a lot, but if any of you have time, it really would be appreciated.

Using the suggestion in the text, I started with $a = ky$ which means $\frac{dv}{dt} = ky$ or

$$v\frac{dv}{dy} = ky$$

My idea here was to get an expression for v, and then integrate that (as dy/dt) to get y in terms of t. But I have never really seen something like this before in the text. So perhaps I took a wrong turn here. What I tried to do was this: $$v\, dv = ky\, dy$$ and integrate this to get $$v^2/2 = ky^2/2 + C$$.

I know the book says you aren't supposed to view dv/dy (or any derivative) as a fraction, but I couldn't think of anything else to do.

On the off chance this was ok. I found $C = v_0^2/2$ and so end up with $v^2 = ky^2 + v_0^2$.

So (assuming this is right), I have $$\frac{dy}{dt} = \sqrt{ky^2 + v_0^2}$$ I use the theorem on reciprocals to write it as $$\frac{dt}{dy} = \frac{1}{\sqrt{ky^2 + v_0^2}}$$

To integrate this, I set $y = v_0\tan\theta/\sqrt{k}$ and so $dy/d\theta = v_o\sec^2\theta/\sqrt{k}$ and I end up with the integral

$$t = \frac{1}{\sqrt{k}}\int \frac{1}{\cos\theta}\, d\theta = \frac{\log(\cos\theta)}{\sqrt{k}} = \frac{1}{\sqrt{k}}\log\left(\frac{v_0}{\sqrt{v_0^2 + y^2k}}\right) + C$$

So

$$De^{\sqrt{k}t} = \frac{v_0}{\sqrt{v_0^2+y^2k}}$$

and

$D = 1/v_0$.

Here is where I stopped, since it is obvious that this is not going to lead to the correct answer as given in the book, which is

$$y = \frac{v_0}{2\sqrt{k}}(e^{\sqrt{k}t} - e^{-\sqrt{k}t})$$

Thanks again, if anyone has the fortitude to slog through all this.

Sheldon

 

[SheldonG]

Now I see a mistake in my integration of 1/cos. Let me think on this some more, and I will post again. Sorry to bother everyone.

Sheldon

 

[SheldonG]

Yes, that was the problem. Well, I managed on my own for once. Sorry to bother anyone who read this.

Thanks
Sheldon