91Ó°ÊÓ

The length of the spaceships is $L_0=1\text{ l²â}$

The speed of spaceships is $u=0.800c$

The speed of messenger is $v^{\text{'}}=0.950c$

The speed of armada of messenger relative to ground station S is found by using the formula for relativistic relative velocity. The length of spaceships for ground observer is found by length contraction formula.

(a)

The speed of the messenger in rest frame is given as:

$\mathrm{v}=\frac{\mathrm{u}−{\mathrm{v}}^{\text{'}}}{1−\frac{{\mathrm{uv}}^{\text{'}}}{{\mathrm{c}}^2}}$

Substitute all the values in the above equation.

$\begin{array}{l}v=\frac{0.800c−0.950c}{1−\frac{\left(0.800c\right)\left(0.950c\right)}{c^2}}\\ v=−0.625c\end{array}$

The length of armada in rest frame of messenger is given as:

$L=L_0\sqrt{1−{\left(\frac{v}{c}\right)}^2}$

Substitute all the values in the above equation.

$\begin{array}{l}L=\left(1\text{ l²â}\right)\sqrt{1−{\left(\frac{−0.625c}{c}\right)}^2}\\ L=0.781\text{ l²â}\end{array}$

The trip time in the rest frame of messenger is given as:

$t=\frac{L}{\left|v\right|}$

Substitute all the values in above equation.

$\begin{array}{l}t=\frac{0.781\text{ l²â}}{|−0.625c|}\\ t=\frac{\left(0.781\text{ l²â}\right)\left(\frac{1c\text{ y}}{1\text{ l²â}}\right)}{0.625c}\\ t=1.25\text{ y}\end{array}$

Therefore, the trip time in the rest frame of messenger is $1.25\text{ y}$.

(b)

The trip time in the rest frame of armada is given as:

$t_a=\frac{L_0}{\left|v\right|}$

Substitute all the values in above equation.

$\begin{array}{l}{t}_a=\frac{1\text{ l²â}}{|−0.625c|}\\ t_a=\frac{\left(1\text{ l²â}\right)\left(\frac{1c\text{ y}}{1\text{ l²â}}\right)}{0.625c}\\ t_a=1.60\text{ y}\end{array}$

Therefore, the trip time in the rest frame of armada is $1.60\text{ y}$.

(c)

The length of armada for observer in ground frame S is given as:

$L_1=L_0\sqrt{1−{\left(\frac{u}{c}\right)}^2}$

Substitute all the values in the above equation.

$\begin{array}{l}{L}_1=\left(1\text{ l²â}\right)\sqrt{1−{\left(\frac{0.800c}{c}\right)}^2}\\ L_1=0.6\text{ l²â}\end{array}$

The trip time for observer in ground frame S is given as:

$t_g=\frac{L_1}{v^{\text{'}}−u}$

Substitute all the values in above equation.

$\begin{array}{l}{t}_g=\frac{0.6\text{ l²â}}{0.950c−0.800c}\\ t_g=\frac{\left(0.6\text{ l²â}\right)\left(\frac{1c\text{ y}}{1\text{ l²â}}\right)}{0.150c}\\ t_g=4\text{ y}\end{array}$

Therefore, the trip time for observer in ground frame S is$4\text{ y}$ .