Answer
0.981c
Work Step by Step
Let the positive direction be the direction of particle 2, and find the velocity of particle 2 as measured by particle 1. The answer will be positive.
Let the first particle’s reference frame be S and the lab’s reference frame be S’.
The velocity of the lab relative to the first particle is v = +0.82c. The velocity of the second particle relative to the spacecraft is u’=+0.82c.
Use the relativistic velocity addition formula, equation 26–10, to find the velocity of the second particle as measured by the first particle.
$$u=\frac{v+u’}{1+\frac{vu’}{c^2}}$$
$$u=\frac{0.82c+0.82c}{1+(0.82)(0.82)}=0.981c$$