Confusion regarding notation of instantaneous velocity wrt something.

[AakashPandita]

I know that v(t)=dx/dt

Then what is v(x) and how?

Is it also dx/dt or something else?

 

[Useful nucleus]

Take the simplest example, motion in one dimension with constant acceleration. Write down the two equations describing v(t), x(t) and try to eliminate the parameter t from the two equations to get v(x).

 

[AakashPandita]

First of all thanks for replying.

I took the equation v(t)=2t^2 and differentiated it to get x(t)=2t.
But again the same confusion cropped up. If v(t)=dx/dt then what is x(t)?

 

[jtbell]

To get v(t) from x(t) you differentiate x(t).

To get x(t) from v(t) you use the operation that "undoes" differentiation. Which operation is that?

 

[AakashPandita]

integration

 

[HallsofIvy]

Yes. And what is the integral (also called "anti-differentiation) of 2t^2? That is, what function has 2t^2 as its derivative?

 

[DrewD]

...then what is x(t)?

You mean $v(x)$ not $x(t)$, right?

 

[Chestermiller]

You remember back in algebra you learned that rate times time equals distance: rt = d

Well in this case, x replaces the distance d, and v replaces the rate r. If the velocity is changing with time, you need to express the relationship in terms of differentials: v dt = dx. But, otherwise, it is basically the same thing you learned in Algebra I.

 

[AakashPandita]

If v(t)= dx/dt (change in position wrt time)
then what is v(x) (change in position wrt position?)
=dx/dx=1?

 

[NascentOxygen]

If v(t)= dx/dt (change in position wrt time)
then what is v(x) (change in position wrt position?)
=dx/dx=1?

You need to think carefully before writing your questions, because a poorly considered question won't encourage the help you may be hoping for.

v(t) is velocity as it varies with time, and v(t)=dx/dt

But more precisely, v(t)=d x(t) / dt

x(t) being displacement as it varies with time

Following the usual convention, v(x) must be velocity as it varies with displacement

Exercise: If x(t)=t^2, and you already showed v(t)= dx(t)/dt = 2t, then determine v(x).

 

[jtbell]

More conceptually, imagine that you are driving a car down the road. At 1:00 PM you look at the instrument panel and observe that the speedometer reads 50 km/hr and the odometer reads 23765.8 km. Then v(t) = 50 km/hr for t = 1:00 PM, and v(x) = 50 km/hr for x = 23765.8 km.

 

[NascentOxygen]

I took the equation v(t)=2t^2 and differentiated it to get x(t)=2t.

No you didn't. You took the equation v(t)=t^2 and differentiated it to get

a(t) = [/size] ^{2t because dv/dt gives acceleration.}[/size]

 

[AakashPandita]

If v(t)=dx/dt
v(x)= d?/d?

 

[DrewD]

Have you read the previous posts? If you are still confused, ask a follow up question. Please don't just repeat the question. When I first saw this thread I was excited because I have seen a lot of students struggle with this, but you must actually make an attempt to learn

 

[Chestermiller]

Please go back to your advanced algebra textbook and look up what the notation f(x) means.

f(x) means "the parameter f expressed as a function of the parameter x"

f(t) means "the parameter f expressed as a function of the parameter t"

So, v(t) represents the velocity v expressed as a function of the time t

v(x) represents the velocity v expressed as a function of the distance x

In both these cases, v is the same parameter.

 

[AakashPandita]

Okay now I understand.I finally worked it out. Thank you very much for talking some sense into me.
v(t) and v(x) are both equal to dx/dt but both the functions define velocity wrt different parameters while dx/dt simply means change in x in very small interval of time.
Am I right?

 

[Chestermiller]

okay now i understand.i finally worked it out. Thank you very much for talking some sense into me.
V(t) and v(x) are both equal to dx/dt but both the functions define velocity wrt different parameters while dx/dt simply means change in x in very small interval of time.
Am i right?

YES. Perfect!

 

[NascentOxygen]

If v(t)=dx/dt
v(x)= d?/d?

This is a good question, and has been addressed. But is there an alternative approach?

Normally, we are given x as a function of time, viz., x(t), and to determine d$x(t)$/dt we differentiate w.r.t. time. But you could take that x(t) and re-arrange it into time as a function of x, t(x) (at least, you can for some functions x(t), otherwise, use a restricted range.)

Now, given t(x), how to determine velocity, dx/dt?

v(x) = 1/(d$t(x)$/dx)

I.e., v(x) = the reciprocal of d$t(x)$/dx

You can see that this has units of m/sec, i.e., velocity.

 

[AakashPandita]

Thank You. This made things more clear.