Answer
See explanation.
Work Step by Step
Let the positive direction be the direction in which the observer S’ travels. The velocity of the rocket relative to Earth is u = 0.600c. The velocity of the particle relative to the rocket is v’=0.400c. Use the relativistic velocity addition formula, equation 37.23.
a. $$v_x=\frac{v_x’+u}{1+\frac{uv_x’}{c^2}}$$
$$v_x=\frac{0.400c+0.600c}{1+(0.600) (0.400) }=0.806c$$
b. $$v_x=\frac{v_x’+u}{1+\frac{uv_x’}{c^2}}$$
$$v_x=\frac{0.900c+0.600c}{1+(0.600) (0.900) }=0.974c$$
c. $$v_x=\frac{v_x’+u}{1+\frac{uv_x’}{c^2}}$$
$$v_x=\frac{0.990c+0.600c}{1+(0.600) (0.990) }=0.997c$$
The speed is always less than c.
