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Equivalence between 2 solutions 
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#1
Feb613, 04:24 AM

PF Gold
P: 2,281

1. The problem statement, all variables and given/known data
A particle of mass ##m## is projected from ##x(0) = x_o## in the vertical direction, with an initial velocity, ##\dot{x}(0) = v_o##. It is subject to gravity and linear drag, mkv, against the motion. 1) Show that the body follows ##v(t)## such that:$$v(t) = g/k + (v_o + g/k)e^{kt}$$ 2) Use ##\ddot{x} = v dv/dx ## to find a soln to the motion subject to the initial condition. Show the equivalence between the result attained and the one shown in 1). 2. Relevant equations Separable Diff Eqns, Newton 2nd 3. The attempt at a solution 1) is fine. I don't really know how to show the equivalence, but I have tried two different ways, where one way I get nothing near equivalence and the other I recover the exp term above, but not the g/k in front. Method 1). After solving ##\ddot{x} = v dv/dx, ##I get ##x = x(v)##. Then by the chain rule, ##dx/dt = dx/dv\,dv/dt##. So I differentiated my soln in 2) wrt v and then multiplied this by the derivative of v wrt t in 1). This gives me nothing near equivalence, although I am not sure why. Method 2). Put v = v(t) in 1) into my x(v). This gives x(t). Then simply differentiate wrt t. I get the second term in 1) (exp term) but not the g/k. Both methods seem valid, (are they?) but I don't get the result. I could recheck my algebra again, but both methods yield a (g +kv)^3 and that is not present in 1) and it doesn't cancel. Many thanks. 


#2
Feb613, 04:51 AM

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You need to post all your working.



#3
Feb613, 11:23 AM

PF Gold
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EDIT: The images turned out quite faint  i'll re upload with the working in pen. 


#4
Feb613, 11:39 AM

PF Gold
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Equivalence between 2 solutions
Working in pen:



#5
Feb613, 04:41 PM

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I don't see where the second equation on page 2 comes from. It's dimensionally wrong. k has dimension 1/T, LHS = dx/dv has dimension T, on RHS v/k^{2} has dimension LT and g/(k(g+kv)) has dimension T.



#6
Feb613, 04:46 PM

PF Gold
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EDIT: I made an error there  i see. Thanks for pointing this out. I'll recheck things. 


#7
Feb613, 04:52 PM

PF Gold
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Fixing my error above only removes the v/k term that I had in front of the exp term so I recover the ##(v_o + v_T)exp.. ##term but not the other one..



#8
Feb613, 05:24 PM

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#9
Feb713, 11:59 AM

PF Gold
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So taking the derivative of the first line:
dx/dv = 1/k + g/(k(g+kv)) Multiply this with dv/dt obtained from A) (shown in the image) gives me the term (v_o v_T)exp but not the '+v_T' term. Many thanks. 


#10
Feb713, 03:16 PM

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#11
Feb713, 03:49 PM

PF Gold
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Ok, I will post all my working momentarily. But how would you get that +v_T? In the dx/dv, I have a g/k(g+kv) term , so how when multiplied by dx/dt will I get anything near a simple +v_T?
EDIT: Working attached And that second line should read ##\frac{1}{k} + \frac{g}{k(g+kv)}, ## not ##\frac{1}{k} + \frac{g}{g(g+kv)}, ## 


#12
Feb713, 04:55 PM

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You're missing that you have v on both sides of the equation (as dx/dt on the left). You need to rearrange it so that v is only on the left. Btw, the expression for dx/dv can be simplified a lot.



#13
Feb813, 01:33 AM

PF Gold
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I see, thanks. But what about method 2) (also contained in same link). I don't think I made the same mistake there. Thanks



#14
Feb813, 05:01 AM

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In the very first line after "METHOD2" you have the wrong sign on the g/k term at the end of the line. It must be v_{T}, so that it tends to v_{T} as t tends to infinity, and v_{T} = g/k. If you fix that there'll be some cancellation.



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