I know that v(t)=dx/dt
Then what is v(x) and how?
Is it also dx/dt or something else?
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).
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)?
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?
Yes. And what is the integral (also called "anti-differentiation) of [itex]2t^2[/itex]? That is, what function has [itex]2t^2[/itex] as its derivative?
You mean ##v(x)## not ##x(t)##, right?
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.
If v(t)= dx/dt (change in position wrt time)
then what is v(x) (change in position wrt position?)
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).
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.
No you didn't. You took the equation v(t)=t^2 and differentiated it to get
a(t) = 2t because dv/dt gives acceleration.
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
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.
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?
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.
Thank You. This made things more clear.
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