## College Physics (4th Edition)

The velocity of particle B as seen by particle A is $-0.99~c$
Let $v_{EA}$ be the velocity of the Earth relative to particle A. Then $v_{EA} = -0.90~c$ Note that $v_{BE} = -0.90~c$ We can find $v_{BA}$: $v_{BA} = \frac{V_{BE}~+~v_{EA}}{1+\frac{(v_{BE})~(v_{EA})}{c^2}}$ $v_{BA} = \frac{(-0.90c)~+~(-0.90c)}{1+\frac{(-0.90c)~(-0.90c)}{c^2}}$ $v_{BA} = \frac{-1.80c}{1.81}$ $v_{BA} = -0.99~c$ The velocity of particle B as seen by particle A is $-0.99~c$.