Velocity Calculation with c^2 Substitution

ccb056
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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
 
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You don't.. you get c.

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

- Warren
 
ccb056 said:
... you plug in c^2 for both V1 and V2 ...
This will give a nonsensical result. You must plug in speed, not squared speed.
 
What I meant to say was (3*10^8)^2
 
ccb056 said:
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:
<br /> \frac{2*(3*10^{8})^{2}}{1+(3*10^{8})^{2}}<br />

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.
 
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