The General Definition of Instantaneous Speed: A Debate on V=dx/dt vs V=x/t

[deda]

Hurkyl probably remebers this post..

The main question of this debate is:
What definition for instaneous speed is more general:
1)$$V=\frac{x}{t}$$
2)$$V=\frac{dx}{dt}$$?

the notification is standard.

 

[selfAdjoint]

Second definition is more general. x/t is only accurate if the x-displacement is strictly linear in time: x(t) = vt + c. The second definition will work for x(t) any differentiable function of time, which is a much bigger class of course, making the definition more general.

 

[master_coda]

How could the first equation possibly be more general? In what situation does the first equation work and not the second?

 

[Njorl]

The first equation is not correct. The second one is. The first equation will produce the proper units, and it may, under lucky circumstances, produce the proprer result, but it is not generally correct. Even in the linear case of x(t)=vt+c it is only correct if c=0.

Njorl

 

[himanshu121]

Originally posted by deda
Hurkyl probably remebers this post..

The main question of this debate is:
What definition for instaneous speed is more general:
1)$$V=\frac{x}{t}$$
2)$$V=\frac{dx}{dt}$$?

the notification is standard.

If notification are standard Then the 2 is also standard look at poll results

 

[deda]

Originally posted by Njorl
The first equation is not correct. The second one is. The first equation will produce the proper units, and it may, under lucky circumstances, produce the proprer result, but it is not generally correct. Even in the linear case of x(t)=vt+c it is only correct if c=0.

Njorl

where from is your conclusion "x=f(t)" when both from the 1st and the 2nd option we have x=f(V,t)?

regarding x(t)=vt+c i can say only this:
dx=Vdt <=>
dx=Vdt+tdV and dV=0 <=>
$$\int_{x_0}^{x}dx=\int_{V_0t_0}^{Vt}d(Vt)$$ <=>
$$x-x_0=Vt-V_0t_0$$ <=>
$$x=Vt$$ and $$x_0=V_0t_0$$
in general for every corresponding x,V and t it's
x=f(V,t)=Vt[/color]
while dx=Vdt is special case of x=Vt when V=const.