Answer:
The answer is "0.82 c".
Explanation:
Given:
Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]
Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]
The spacecraft velocity 2 measured by the Earth observation
[tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]
[tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]
