# Equivalence between 2 solutions

• CAF123
In summary: And there's another sign error in the last line on page 3, where you have -gT on the left, but it should be +gT.I see, thanks. I've attached the images with the corrections. I still can't get the +v_T term. I get the ##v_o e^{x/g}## term but not the other.I see, thanks. I've attached the images with the corrections. I still can't get the +v_T term. I get the ##v_o e^{x/g}## term but not the other.I think you are using the wrong value of vT at the start of the 2nd line on page 2,
CAF123
Gold Member

## Homework Statement

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).

## Homework Equations

Separable Diff Eqns, Newton 2nd

## 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.

You need to post all your working.

haruspex said:
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.

#### Attachments

• Method 1.jpg
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• Method 1) part2.jpg
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• Method 2.jpg
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Last edited:
Working in pen:

#### Attachments

• 001.jpg
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• 002.jpg
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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/k2 has dimension LT and g/(k(g+kv)) has dimension T.

haruspex said:
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.

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..

CAF123 said:
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.

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.

CAF123 said:
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.

The procedure you describe works for me, so you will need to post ALL your working.

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)}, ##

#### Attachments

• PEN.jpg
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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.

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

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 vT, so that it tends to vT as t tends to infinity, and vT = -g/k. If you fix that there'll be some cancellation.

## 1. What is meant by "equivalence" between 2 solutions?

Equivalence between 2 solutions refers to the point at which the two solutions have equal concentrations of the same solute. This means that the number of moles of the solute in each solution is the same, resulting in a balanced chemical reaction.

## 2. How is equivalence determined between 2 solutions?

Equivalence is determined by using a process called titration, where a known volume of one solution is added to a known volume of the other solution until an indicator or other measurement method indicates that the two solutions are in equilibrium. The volumes and concentrations of the solutions can then be used to calculate the equivalence point.

## 3. What is the significance of finding the equivalence point between 2 solutions?

The equivalence point is significant because it allows for the calculation of the concentration of an unknown solution, as well as the determination of the stoichiometric relationships between the reactants and products in a chemical reaction. It is also a key step in many analytical and quantitative chemistry experiments.

## 4. Can two solutions with different chemical compositions have an equivalence point?

Yes, two solutions with different chemical compositions can still have an equivalence point. This is because equivalence is determined by the concentration of a specific solute, not the overall composition of the solution. As long as the two solutions have the same concentration of the solute being measured, an equivalence point can be reached.

## 5. What factors can affect the equivalence point between 2 solutions?

The equivalence point can be affected by a variety of factors, including the accuracy of the titration process, the concentration and volume of the solutions being used, and the sensitivity of the indicator or measurement method being used to determine the equivalence point. Temperature and other environmental conditions can also play a role in the accuracy and precision of the equivalence point determination.

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