Answer
See answers.
Work Step by Step
a. An Earth observer measures 21.6 ly as the (rest) length. The time is the distance divided by the speed.
$$t_E=\frac{21.6\;ly}{0.950c}=22.74y\approx 22.7y$$
b. The time measured by the spacecraft is the proper time. Use equation 26–1a.
$$\Delta t_0=\Delta t\sqrt{1-v^2/c^2}=(22.74y) \sqrt{1-(0.950)^2}=7.10y$$
c. The spacecraft observers measure the contracted distance. Use equation 26–3a.
$$\mathcal{l} = \mathcal{l}_{o} \sqrt{1-\frac{ v^{2} } { c^{2} }} = 21.6ly\sqrt{1-0.950^2}=6.74ly$$
d. Spacecraft observers calculate their speed as distance divided by time.
$$v=\frac{6.7446ly}{7.10y}=0.950c$$
This is the expected answer.