Defining velocity as a function of distance

[Marcell]

Homework Statement

The equation is given:

v = f(t) = (mg/b)(1 - e^((-b/m)t))

Where v is velocity, m is mass, g is 9.81 and b is a constant of proportionality

I need to find the function k(x), where x is distance from said body's initial position.

2. The attempt at a solution

v = f(t) = dx/dt
dx = f(t)dt
x = F(t)

and here I got stuck :/

 

[PeroK]

Homework Statement

The equation is given:

v = f(t) = (mg/b)(1 - e^((-b/m)t))

Where v is velocity, m is mass, g is 9.81 and b is a constant of proportionality

I need to find the function k(x), where x is distance from said body's initial position.

2. The attempt at a solution

v = f(t) = dx/dt
dx = f(t)dt
x = F(t)

and here I got stuck :/

Do you know how to integrate?

 

[Marcell]

Do you know how to integrate?

I like to think so, but you're making me uncertain.

 

[haruspex]

I like to think so.

So integrate:

dx = f(t)dt

 

[SammyS]

Homework Statement

The equation is given:

v = f(t) = (mg/b)(1 - e^((-b/m)t))

Where v is velocity, m is mass, g is 9.81 and b is a constant of proportionality

I need to find the function k(x), where x is distance from said body's initial position.

I assume that the k(x) you are finding is a function giving velocity as a function of x, the distance from said body's initial position.

Is that correct?

 

[Ray Vickson]

Homework Statement

The equation is given:

v = f(t) = (mg/b)(1 - e^((-b/m)t))

Where v is velocity, m is mass, g is 9.81 and b is a constant of proportionality

I need to find the function k(x), where x is distance from said body's initial position.

2. The attempt at a solution

v = f(t) = dx/dt
dx = f(t)dt
x = F(t)

and here I got stuck :/

If you follow the suggestion in #4 you will obtain a formula of the form $x(t) = F(t)$ for some computable function $F(t)$. Then, if the question really is to determine $v$ as a function of $x$, you need to solve $F(t) =x$ to find $t$ in terms of $x$----say $t = T(x).$ Then substitute that formula for $t$ into the $v(t)$ equation, to get $v$ as a function of $x$. That will be your $k(x)$ function.

However, if the problem is exactly as you have stated it, the solution will involve the non-elementary "Lambert W function."

 

[Marcell]

I assume that the k(x) you are finding is a function giving velocity as a function of x, the distance from said body's initial position.

Is that correct?

Yes!

If you follow the suggestion in #4 you will obtain a formula of the form $x(t) = F(t)$ for some computable function $F(t)$. Then, if the question really is to determine $v$ as a function of $x$, you need to solve $F(t) =x$ to find $t$ in terms of $x$----say $t = T(x).$ Then substitute that formula for $t$ into the $v(t)$ equation, to get $v$ as a function of $x$. That will be your $k(x)$ function.

However, if the problem is exactly as you have stated it, the solution will involve the non-elementary "Lambert W function."

Am I supposed to integrate from 0 to t? or just do an infinite integral?

Thanks for all your help~!

 

[PeroK]

Yes!
Am I supposed to integrate from 0 to t? or just do an infinite integral?

Thanks for all your help~!

Displacement is the (signed) area under a velocity-time graph.

In general, most integrals in physics are definite, as they represent physical quantities.

 

[Marcell]

Displacement is the (signed) area under a velocity-time graph.

In general, most integrals in physics are definite, as they represent physical quantities.

Ok I see.

I integrated the equation from 0 to t and got:

$$\dfrac{gm\mathrm{e}^{-\frac{bt}{m}}\left(\left(bt-m\right)\mathrm{e}^\frac{bt}{m}+m\right)}{b^2}$$

So now I 'just' need to solve for t and substitute it back into the original equation to get k(x)?

 

[PeroK]

Ok I see.

I integrated the equation from 0 to t and got:

$$\dfrac{gm\mathrm{e}^{-\frac{bt}{m}}\left(\left(bt-m\right)\mathrm{e}^\frac{bt}{m}+m\right)}{b^2}$$

So now I 'just' need to solve for t and substitute it back into the original equation to get k(x)?

That doesn't look right. What is $k(x)$?

 

[Marcell]

That doesn't look right. What is $k(x)$?

k(x) is velocity in terms of distance from the initial position.

I know that in terms of time velocity is v(t) = (mg/b)(1 - e^((-b/m)t)).

 

[PeroK]

k(x) is velocity in terms of distance from the initial position.

I know that in terms of time velocity is v(t) = (mg/b)(1 - e^((-b/m)t)).

Okay, once you've fixed your integral, look at what Ray said in post #6. Is this a question you made up yourself?

 

[Marcell]

Okay, once you've fixed your integral, look at what Ray said in post #6. Is this a question you made up yourself?

More or less yeah.

What is wrong my integral? I used https://www.integral-calculator.com/ to get it...

Also, if its not too much to ask, could I please get some more guidance on the whole Lambert W function?

 

[PeroK]

More or less yeah.

What is wrong my integral? I used https://www.integral-calculator.com/ to get it...

Also, if its not too much to ask, could I please get some more guidance on the whole Lambert W function?

https://en.wikipedia.org/wiki/Lambert_W_function

 

[SammyS]

There is another approach. Following it will give x as a function of v, but it has the same issue as described by Ray in post (#6).

It goes like this:

Using the chain rule express the acceleration $\ a\$ as:

$\displaystyle \frac{dv}{dt} = \frac{dv}{dx} \, \frac{dx}{dt}$

$\displaystyle = v\, \frac{dv}{dx} \ \quad$ because $\ \displaystyle \frac{dx}{dt}= v$​

So we have: $\displaystyle \ a = v \, \frac{dv}{dx} \, .$

With acceleration as a function of $\ t\,,$ this would not be of much use. However, as it turns out, differentiating the velocity ( given as a function of $\ t \$ by the function $\ f(t)\$ ) gives a result for acceleration, which together with the expression for velocity can be manipulated to give acceleration in terms of velocity together with some constants. To be specific, one obtains the following.

$\displaystyle a = g - \frac{b}{m} v \quad \quad$ ( a familiar looking result )​

So far my few attempts obtaining an expression for position $\ x\$ in terms of velocity $\ v\$ have given me results which appear to me have inconsistencies. However, the form of the results tend to confirm that the velocity function in the form of $\ k(x) \$ does involve the Lambert W function.

 

[SammyS]

Ok I see.

I integrated the equation from 0 to t and got:

$$\dfrac{gm\mathrm{e}^{-\frac{bt}{m}}\left(\left(bt-m\right)\mathrm{e}^\frac{bt}{m}+m\right)}{b^2}$$

So now I 'just' need to solve for t and substitute it back into the original equation to get k(x)?

That does look like it's correct. Let's rewrite it a bit.

$\displaystyle x= \left(\frac{gm}{b^2}\right) \mathrm{e}^{-(b/m)t}\left(\left(bt-m\right)\mathrm{e}^{(b/m)t}+m\right) \$​

Those exponentials cancel for the most part. Multiply through by the common factors and simplify. I suppose we should call this $\ F(t)\$, where $\ F'(t) = f(t)\$.

$\displaystyle F(t)= \left(\frac{mg}{b}\right)t-\left(\frac{m^2g}{b^2}\right)+\left(\frac{m^2g}{b^2}\right)\mathrm{e}^{-(b/m)t} \$​

.
You will not be able to solve this for $\ t\$ using elementary functions. However, you can sovle the given velocity expression (Post #1) for $\ t \$ and plug that into this expression to get $\ x\$ in terms of $\ v\$. That's also not invertable using elementary functions.