Derive y as function of time in relation to terminal velocity

[Ryker]

Homework Statement

Integrate

$$v_{y} = v_{t} (1 - e^{-(\frac{k}{m})t}))$$

to get $$y$$ as a function of time.

Homework Equations

$$y = v_{t}(t - \frac{m}{k} (1 - e^{-(\frac{k}{m})t})))$$ is the solution.

$$v_{t}=\frac{mg}{k}$$

The Attempt at a Solution

Well, when I integrate I get the following:

$$y = v_{t}(t + \frac{m}{k} e^{-(\frac{k}{m})t})))$$

I don't get where that middle part comes from, which when taken out of brackets equals

$$-\frac {v_{t}m}{k}$$.

Does anyone have any idea what I'm missing or where I may have gone wrong with my integration technique?

 

[kuruman]

I don't get where that middle part comes from, which when taken out of brackets equals

$$-\frac {v_{t}m}{k}$$.

No, it does not. What do you get when you distribute v_T correctly inside the parentheses?

 

[Ryker]

I think that middle part does actually match what I wrote. I didn't mean for the whole thing to be taken out of brackets, just that thing that my solution is missing when compared to the proper one. That's what I meant by "middle part".

 

[kuruman]

Look at the equation for y(t). When t is zero, you get correctly y = 0. When t becomes very large, you get (approximately) y = v_Tt, i.e. the object is moving at constant terminal velocity. So the middle part is needed to give you the correct result for the in-between motion.

 

[Ryker]

So this is a constant "C" that you need to add yourself after integrating? And can you explain in more detail why you need to add specifically that expression? I thought since starting velocity is zero everything (everything, but not more than that) is covered when you integrate that original velocity equation. Why isn't this the case then?

 

[kuruman]

There is no constant of integration C. Show me the integral that you used to get the position y(t) from the velocity v(t), and I will explain why.

 

[Ryker]

I used this one:

$$\int v_{t} (1 - e^{-(\frac{k}{m})t})) = v_{t}t - v_{t} \int e^{-(\frac{k}{m})t} = v_{t}(t + \frac{m}{k} e^{-(\frac{k}{m})t})))$$

 

[kuruman]

First off this is only the right side of the equation that you should be using. Show and integrate both sides. Secondly, what are your limits of integration?

 

[Ryker]

Yeah, sorry, I left out the left side, because it's just

$$\int^t_0 v_{y} dt = y$$.

And the limits would be $$\int^t_0$$.

 

[Ryker]

Facepalm.jpg would be in order here. I just forgot to apply the limits and derived an indefinite instead of a definite integral, didn't I? I'm screaming inside like a madman right now.

 

[kuruman]

You also need limits for the y-dot integral which should be y₀ to y.

 

[Ryker]

Thanks a lot for all of your help. One last question, though. When you mention you must have limits from y₀ to y, do you actually mean from 0 to t, which when integrated return values of y - y₀?

 

[kuruman]

No. On the left side the dy integral goes from y₀ to y and on the right side the dt integral goes from zero to t. The lower limit on y matches the lower limit on t and the upper limits are also matched likewise in pairs. That's why I said "Show and integrate both sides." You start with

dy = v dt

and you integrate both sides with lower and upper limits of integration on both sides.

 

[Ryker]

Hmm, but on the left-hand side you start out with v_y, which you have to integrate in relation to time. So setting that up you get

$$\int v_{y} dt$$

on the left. How would you then implement the limit for displacement (y and y₀) if the variable is time?

 

[kuruman]

Look

$$dy=vdt$$

As time increases, add all increments dy on the left which means adding all increments v dt on the right.

$$\int^{y}_{y_0}dy=\int^{t}_{0} vdt$$

 

[Ryker]

Yeah, I understand that, but does that have any practical implications or is it just for notation purposes? I mean, in our case y₀ = 0, so I'm guessing you would only need to take that into account if it wasn't. But how would you do that then? Subtract y₀ from the result you would get on the right hand side?

 

[kuruman]

Yeah, I understand that, but does that have any practical implications or is it just for notation purposes? I mean, in our case y₀ = 0, so I'm guessing you would only need to take that into account if it wasn't. But how would you do that then? Subtract y₀ from the result you would get on the right hand side?

What do you get when you evaluate

$$\int^{y}_{y_0}dy \:?$$

That's how you do it.

 

[Ryker]

I get y - y₀, right?

 

[kuruman]

Right and that's the left side of the equation. If y₀ is zero in the particular situation you are dealing with, then you put nothing on the left side; if it is not zero, then you put the negative of whatever the problem tells you it is at time t=0. It all depends on the initial conditions of your specific problem. What more is there to say?

 

[Ryker]

Yeah, I guess I put that y₀ = 0 as granted, as it was stated so in the book in the paragraphs preceding the equation, so that's why I didn't really address it at first. But once again, thanks again for your help.