# Equivalence between 2 solutions

by CAF123
Tags: equivalence, solutions
 PF Gold P: 2,315 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, mk|v|, 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.
 Homework Sci Advisor HW Helper Thanks P: 9,933 You need to post all your working.
PF Gold
P: 2,315
 Quote by haruspex You need to post all your working.
Sure, I have attached my working.

EDIT: The images turned out quite faint - i'll re upload with the working in pen.
Attached Thumbnails

 PF Gold P: 2,315 Equivalence between 2 solutions Working in pen: Attached Thumbnails
 Homework Sci Advisor HW Helper Thanks P: 9,933 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/k2 has dimension LT and g/(k(g+kv)) has dimension T.
PF Gold
P: 2,315
 Quote by haruspex 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/k2 has dimension LT and g/(k(g+kv)) has dimension T.
I see. But what did I do wrong? All I did was differentiate wrt v, and so the terms not containing v just vanished.
EDIT: I made an error there - i see. Thanks for pointing this out. I'll recheck things.
 PF Gold P: 2,315 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..
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HW Helper
Thanks
P: 9,933
 Quote by CAF123 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..
That's not what I get. Pls post the corrected steps.
 PF Gold P: 2,315 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.
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