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# Solve for v...?

p = 1 / [ √(1 - (v^2/c^2)) ]

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### 3 Answers

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- :-)Lv 41 decade agoFavorite Answer
p = 1 / [ √(1 - (v^2/c^2)) ]

p [ √(1 - (v^2/c^2)) ] = 1

squaring on both sides to get rid of the sq.root

p^2 (1- (v^2/c^2)) = 1

1 - (v^2/c^2) = 1 / p^2

v^2/c^2 = 1 - (1/p^2)

v^2 = c^2 (1 - (1/p^2))

Apply sq.root on both sides to get v

v = √ (c^2 (1 - (1/p^2)))

v = c √ (1 - (1/p^2))

- lenpol7Lv 71 decade ago
p = 1 / [ â(1 - (v^2/c^2)) ]

1/p = [ â(1 - (v^2/c^2)) ]

(1/p)^2 = 1- (v^2/c^2)

(1/p)^2 - 1 = -v^2/c^2

c^2[(1/p)^2 - 1] = -v^2

v^2 = [c^2 - c^2/p^2]

v = +/- sqrt[c^2 - c^2/p^2]

- 1 decade ago
p = 1 / [ â(1 - (vÂ²/cÂ²)) ]

Square this equation,

pÂ² = 1 / (1-vÂ²/cÂ²)

Cross multiply

1-vÂ²/cÂ² = (1/pÂ²)

vÂ²/cÂ² = 1 - (1/pÂ²)

vÂ² = cÂ² [1 - (1/pÂ²)]

take square root

v = c â[1 - (1/pÂ²)]

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