# Tried to find the acceleration by doing ma=mg-mkv^2

1. Aug 17, 2012

### XtremePhysX

1. The problem statement, all variables and given/known data

http://i.imgur.com/kK7xE.jpg

2. Relevant equations

In the picture.

3. The attempt at a solution

I tried to find the acceleration by doing ma=mg-mkv^2.

Last edited by a moderator: Aug 17, 2012
2. Aug 17, 2012

### voko

Re: Mechanics

I suggest that you change the direction of x, to be consistent with the questions.

Now, with the equation of the acceleration and the forces, can you answer (i)?

3. Aug 17, 2012

### XtremePhysX

Re: Mechanics

Can you explain to me the concept of terminal velocity please ?

4. Aug 17, 2012

### dydxforsn

Re: Mechanics

That's the right Newton's 2nd law for the ascension (who knew how to spell this word?..) From that you can parts (ii) and (iii) (hint for part (ii): What happens at the terminal velocity that would allow you to solve for it from that equation?)

For part (iv) you are going to need to do something to the left-hand side of the equation:$$\frac{dv}{dt} = -\frac{g}{{V_t}^{2}} ({V_t}^2 + v^2)$$ Note: For the love of god change upper-case 'V' to '$V_t$' to avoid confusion with lower-case 'v'...
What are you going to have to do to the left-hand side? Well you want just v's and x's, t's are bad because you want some final form x(v). A rule from standard calculus will allow you to write this side in terms of just 'v' and 'x'. Hint: It's a rule so divisive that Joe Biden was in the news for even mentioning it recently.

Edit #1: I notice your mentioning of terminal velocity and perhaps some misunderstanding regarding it. Terminal velocity is the velocity at which a particle is going so fast (due to a velocity independent accelerating force) that some resistive force (that's a function of 'v') will get so large as to perfectly oppose it. In other words, it's a velocity at which there is zero acceleration on the particle. Notice that they make you calculate it on the descent first in part (ii) before part (iii). You need to use the same value you get in part (ii) for part (iii), it's not physically going to change during ascent/descent, but it makes more sense to calculate it for the descent equation because in that equation it's actually something reachable....

Edit #2: Further hint for the left-hand side stuff I was mentioning. You are going to want some form on the left-hand side that contains $\frac{dx}{dt}$ which you are simply going to call 'v'. That's how you get rid of the 't' after you apply the infamous rule I hinted at.

Edit #3: Woops, I put that the left-hand side stuff was for part (iii), I meant to say that it's for part (iv)..

Last edited: Aug 17, 2012
5. Aug 17, 2012

### dydxforsn

Re: Mechanics

First notice the edits to my last post

For part (v) you're going to want to forget about the solution x(v) you found for part (iv) and go back to the equation: $$\frac{dv}{dt} = -\frac{g}{{V_0}^2} ({V_0}^2 + v^2)$$ and actually solve for v(t). Here you want to use '$\frac{{V_o}}{2}$' and 0 as your limits of integration for your velocity integral, you can then solve for your upper limit of your time integral which will simply be the time in question.

Part (vi) is really similar to part (iii).

And finally for part (vii) you are going to want to go back to your solution x(v) and find x(0.1$V_0$). Well... that would be for ascension, I think it wants it for the descent in which case you will have to derive part (iv) for the descent and solve for x(0.1$V_o$) from that equation.. Ugh.

6. Aug 17, 2012

### voko

Re: Mechanics

Observe that the air resistance (drag) grows as the square of the velocity. Which means that at a certain velocity the drag will be equal to the pull of gravity, the net acceleration will be zero, and the velocity will stop growing. This is the terminal velocity.

Last edited: Aug 17, 2012
7. Aug 18, 2012

### XtremePhysX

Re: Mechanics

So how do I do part (i) ?

8. Aug 18, 2012

### voko

Re: Mechanics

"Which means that at a certain velocity the drag will be equal to the pull of gravity, the net acceleration will be zero, and the velocity will stop growing." What does that mean with regard to the equation of motion?

9. Aug 18, 2012

### XtremePhysX

Re: Mechanics

$$mkV_{t}^{2}=mg$$ therefore $$kV_{t}^{2}=g$$ ?

10. Aug 18, 2012

### voko

Re: Mechanics

Correct.

11. Aug 18, 2012

### XtremePhysX

Re: Mechanics

I think once I get part the second part, which is the acceleration, the part after that is just changing the acceleration to vdv/dx and integrating.
But how do I get (iii) ? the acceleration?

12. Aug 18, 2012

### voko

Re: Mechanics

Open the parentheses in (iii), what do you get?

13. Aug 18, 2012

### XtremePhysX

Re: Mechanics

$$v\frac{dv}{dx}=-g-g\frac{v^{2}}{V_{t}^{2}}$$

14. Aug 18, 2012

### voko

Re: Mechanics

How did you get the term on the left? The rightmost term, apply the identity you derived for (ii).