91Ó°ÊÓ

The velocity of ship A is $v_A=0.900c$.

The velocity of ship B is $v_B=0.800c$.

The velocity of ship C is $v_C=0.800c$.

A speed having a magnitude that is a significant fraction of the speed of light. (particles) with relativistic speed:

Suppose an object is moving with the velocity $u^{\text{'}}$with respect to $S^{\text{'}}$frame, which is moving with the velocity $v$with respect to frame S then the velocity of the object with respect to $u$frame is,

$u=\frac{u\text{'}+v}{1+\frac{u\text{'}v}{c^2}}$

Or

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

As the ship A and ship C approaches ship B at the same speed, we can say that,

$v{\text{'}}_A=-v{\text{'}}_C$ ..… (1)

The velocity of ship A relative to ship B is,

$v{\text{'}}_A^{}=\frac{v_A-v_B}{1-\frac{v_A{v}_B}{c^2}}=\frac{c\left({\mathrm{β}}_A-{\mathrm{β}}_B\right)}{1-{\mathrm{β}}_A{\mathrm{β}}_B}$

Where, $\mathrm{β}=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$.

Similarly, the velocity of ship C relative to ship B is

$v{\text{'}}_C^{}=\frac{v_C-v_B}{1-\frac{v_C{v}_B}{c^2}}=\frac{c\left({\mathrm{β}}_C-{\mathrm{β}}_B\right)}{1-{\mathrm{β}}_C{\mathrm{β}}_B}$

Inserting these values in equation (1),

$$\begin{gathered} \frac{c\left({\mathrm{β}}_A-{\mathrm{β}}_B\right)}{1-{\mathrm{β}}_A{\mathrm{β}}_B}=\frac{-c\left({\mathrm{β}}_C-{\mathrm{β}}_B\right)}{1-{\mathrm{β}}_C{\mathrm{β}}_B} \\ \left({\mathrm{β}}_A-{\mathrm{β}}_B\right)\left(1-{\mathrm{β}}_C{\mathrm{β}}_B\right)=\left({\mathrm{β}}_B-{\mathrm{β}}_C\right)\left(1-{\mathrm{β}}_A{\mathrm{β}}_B\right) \\ {\mathrm{β}}_A-{\mathrm{β}}_A{\mathrm{β}}_B{\mathrm{β}}_C-{\mathrm{β}}_B+{\mathrm{β}}_C{\mathrm{β}}_B^2={\mathrm{β}}_B-{\mathrm{β}}_A{\mathrm{β}}_B^2-{\mathrm{β}}_C+{\mathrm{β}}_A{\mathrm{β}}_B{\mathrm{β}}_C \end{gathered}$$

$$\begin{gathered} \left({\mathrm{β}}_A+{\mathrm{β}}_C\right){\mathrm{β}}_B^2-2{\mathrm{β}}_A{\mathrm{β}}_B{\mathrm{β}}_C-2{\mathrm{β}}_B+\left({\mathrm{β}}_A+{\mathrm{β}}_C\right)=0 \\ \left({\mathrm{β}}_A+{\mathrm{β}}_C\right){\mathrm{β}}_B^2-2\left({\mathrm{β}}_A{\mathrm{β}}_C+1\right){\mathrm{β}}_B+\left({\mathrm{β}}_A+{\mathrm{β}}_C\right)=0 \end{gathered}$$

Inserting the values of$0.90$for${\mathrm{β}}_A$and$0.80c$for ${\mathrm{β}}_c$in above equation,

$$\begin{gathered} \left(0.9+0.8\right){\mathrm{β}}_B^2-2\left(\left(0.8×0.9\right)+1\right){\mathrm{β}}_B+\left(0.9+0.8\right)=0 \\ 1.7{\mathrm{β}}_B^2-3.44{\mathrm{β}}_B+1.7=0 \\ {\mathrm{β}}_B^2-2.024{\mathrm{β}}_B+1=0 \end{gathered}$$

Therefore, the expression you get is in the quadratic form, using the quadratic formula rule to get the solution of the equation.

${\mathrm{β}}_B=\frac{-\left[-2.024\right]Â\pm \sqrt{{\left[-2.024\right]}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}$

$$\begin{gathered} {\mathrm{β}}_B=\frac{2.024Â\pm \sqrt{4.095-4}}{2} \\ =\frac{2.024-0.308}{2} \\ =0.858 \end{gathered}$$

Hence, the speed of ship B such that ship A and ship C approaches ship B at the same speed relative to ship B is $v{\text{'}}_B=0.858c$.

The speed of ship A relative to ship B can be given as,

$$\begin{gathered} {v\text{'}}_A=\frac{v_A-{v\text{'}}_B}{1-\frac{v_A{v\text{'}}_B}{c^2}} \\ =\frac{0.90c-0.858c}{1-\frac{\left(0.90c\right)\left(0.858c\right)}{c^2}} \\ =0.184c \end{gathered}$$

Hence, the relative speed is $0.184c$.