Relative Velocity: Deriving Velocity from Time & Distance

[Goodver]

if we know the relations between time and distance at the moving frame and stationary frame why can't we derive the velocity in the moving frame in a such way.

Please point out my mistake. See attachment.

 

[Dale]

You are forgetting the relativity of simultaneity. Mathematically, your expression for velocity is incorrect: $v=dx/dt \ne dl_0/dt_0$

 

[Goodver]

You are forgetting the relativity of simultaneity. Mathematically, your expression for velocity is incorrect: $v=dx/dt \ne dl_0/dt_0$

Thanks! Sorry I could not get where exactly the mistake you mentioned.

v = dx / dt
vo = dxo / dto

is not this correct?

Instead of x I am using l

 

[Dale]

L is not x, it is not merely a variable substitution you are doing. It is a completely different quantity.

What is the formula for x? How does that compare to the formula you have for l?

 

[Dale]

Sorry my previous responses were perhaps overly brief, my attention was split.

One of the most important things in science is learning what the various equations represent and what the underlying assumptions are for that equation.

The time dilation equation and the length contraction equation are special cases, and do not always apply. The general formula is the Lorentz transform, and I strongly recommend that you use it instead of the shortcut formulas. Here is the Lorentz transform for one spatial dimension:
$t' = (t-vx/c^2)/\sqrt{1-v^2/c^2}$
$x' = (x-vt)/\sqrt{1-v^2/c^2}$

If you have the quantities $\Delta t=t_b-t_a$, $\Delta x=x_b-x_a$, $\Delta t'=t'_b-t'_a$, and $\Delta x'=x'_b-x'_a$ then it is easy to show that:
$\Delta t' = (\Delta t-v\Delta x/c^2)/\sqrt{1-v^2/c^2}$
$\Delta x' = (\Delta x-v\Delta t)/\sqrt{1-v^2/c^2}$

If $\Delta x = 0$, then you get the time dilation formula that you posted, however, in this problem $\Delta x\ne 0$, so the time dilation formula does not apply. Similarly, the length contraction formula assumes that the object whose length is being measured is at rest in one of the frames, and since the runner is not at rest in either frame the length contraction formula also does not apply.

Instead, we have $v_0=\Delta x/\Delta t$. In the other frame we have:
$$v'_0=\frac{\Delta x'}{\Delta t'} = \frac{(\Delta x-v\Delta t)/\sqrt{1-v^2/c^2}}{(\Delta t-v\Delta x/c^2)/\sqrt{1-v^2/c^2}} = \frac{v_0\Delta t-v\Delta t}{\Delta t-v \, v_0\Delta t/c^2} = \frac{v_0-v}{1-v_0 v/c^2} $$