## College Physics (4th Edition)

The relative speed of the two spaceships as measured by a passenger on either spaceship is $0.946~c$
Let $v_{MA}$ be the velocity of the moon relative to spaceship A. Then $v_{MA} = 0.80~c$ Then $v_{BM} = 0.60~c$ We can find $v_{BA}$: $v_{BA} = \frac{V_{BM}~+~v_{MA}}{1+\frac{(v_{BM})~(v_{MA})}{c^2}}$ $v_{BA} = \frac{(0.60c)~+~(0.80c)}{1+\frac{(0.60c)~(0.80c)}{c^2}}$ $v_{BA} = \frac{1.40c}{1.48}$ $v_{BA} = 0.946~c$ The velocity of spaceship B relative to spaceship A is $0.946~c$ Therefore, the relative speed of the two spaceships as measured by a passenger on either spaceship is $0.946~c$