An object moves with speed $v_{x}=+0.9$ in the Home Frame. In an Other Frame moving at $\beta=0.60$ relative to the Home Frame, the object's $x$-velocity $v_{x}^{\prime}$ is
A) $v_{x}^{\prime}=1.5$
B) $1

Question

An object moves with speed $v_{x}=+0.9$ in the Home Frame. In an Other Frame moving at $\beta=0.60$ relative to the Home Frame, the object's $x$-velocity $v_{x}^{\prime}$ is
A) $v_{x}^{\prime}=1.5$
B) $1<v_{x}^{\prime}<1.5$
C) $0.9<v_{x}^{\prime}<1$
D) $0.3<v_{x}^{\prime}<0.9$
E) $v_{x}^{\prime}<0.3$
F) $v_{x}^{\prime}<0$

Quizplus · Accepted Answer

To find the object's velocity in the Other Frame, we use the formula for velocity transformation in special relativity: $v_{x}^{\prime} = \frac{v_{x} - v}{1 - \frac{v_{x}v}{c^2}}$, where $v$ is the velocity of the Other Frame relative to the Home Frame, $v_{x}$ is the object's velocity in the Home Frame, and $c$ is the speed of light. Given $v_{x} = +0.9c$ and $v = 0.6c$, the transformed velocity $v_{x}^{\prime}$ will be less than $v_{x}$ but greater than 0, as the relative motion reduces the object's speed in the Other Frame but does not reverse its direction. This places $v_{x}^{\prime}$ in the range $0.3 < v_{x}^{\prime} < 0.9$, making option D correct.

An Object Moves with Speed vx=+0.9v_{x}=+0.9vx​=+0.9 In the Home Frame β=0.60\beta=0.60β=0.60 Relative to the Home Frame, the Object's

An Object Moves with Speed $v_{x}=+0.9$ In the Home Frame $\beta=0.60$ Relative to the Home Frame, the Object's