Velocity Calculation with c^2 Substitution

[ccb056]

If this equation is true:
V=(V1+V2)/(1+(V1*V2)/c^2)

Then why is it when you plug in c^2 for both V1 and V2 you get the total velocity as 2m/s

 

[chroot]

You don't.. you get c.

$$V = \frac{2 c}{1 + c^2/c^2} = \frac{2 c}{2} = c$$

- Warren

 

[turin]

... you plug in c^2 for both V1 and V2 ...

This will give a nonsensical result. You must plug in speed, not squared speed.

 

[ccb056]

What I meant to say was (3*10^8)^2

 

[arildno]

What I meant to say was (3*10^8)^2

It's still meaningless, besides wrong arithmetic:
Putting your value into slots for V1 and V2 yields:
$$\frac{2*(3*10^{8})^{2}}{1+(3*10^{8})^{2}}$$

To clarify, and make a more "accurate" argument:
Let the "velocities" be some big, ugly number on the form: V=kc, k>>1
Then you have by plugging in:
$$\frac{2V}{1+(\frac{V}{c})^{2}}\approx\frac{2c}{k}$$
In your case, k=c; hence the approximate value of 2.