A spacecraft traveling at 0.75 c ejects a missile at a speed of 0.25 c (relative to the spacecraft) in the direction of travel. The speed of the missile relative to a stationary observer is

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Quizplus · Accepted Answer

To find the speed of the missile relative to a stationary observer, we use the formula for addition of velocities in special relativity: $v = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}}$, where $v_1 = 0.75c$ is the speed of the spacecraft, $v_2 = 0.25c$ is the speed of the missile relative to the spacecraft, and $c$ is the speed of light. Plugging in the values, we get $v = \frac{0.75c + 0.25c}{1 + \frac{0.75c \cdot 0.25c}{c^2}} = \frac{1c}{1 + \frac{0.1875c^2}{c^2}} = \frac{1c}{1.1875} = 0.84c$.

A Spacecraft Traveling at 0