Hyperdrive accident

The base vectors for coordinate system of the pilot are $\vec a_{x'} = a' \left(\begin{array}{c} \cos \alpha \\ \sin \alpha \end{array}\right), \quad \vec a_{t'} = a' \left(\begin{array}{c} \sin \alpha \\ \cos \alpha \end{array}\right)$ The coordinates $(x, ct)$ and $(x', ct')$ , that describe the same point $X$ in spacetime, are connected by $\begin{aligned} \vec X &= x \vec a_x + ct \vec a_t = x' \vec a_{x'} + ct' \vec a_{t'} \\ \Rightarrow \quad \left( \begin{array}{c} x \\ ct \end{array} \right) &= \frac{1}{\sqrt{\cos 2 \alpha}} \left( \begin{array}{c} \cos \alpha \, x' + \sin \alpha \, ct' \\ \sin \alpha \, x' + \cos \alpha \, ct' \end{array} \right) \\ &= \frac{1}{\sqrt{\cos 2 \alpha}} \left( \begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right) \left( \begin{array}{c} x' \\ ct' \end{array} \right) \end{aligned}$ Inversion of a $2\times 2$ matrix can be done as follows: $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)^{-1} = \frac{1}{ad - bc} \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right)$ Therefore, $\begin{aligned} \left( \begin{array}{c} x' \\ ct' \end{array} \right) &= \frac{1}{\sqrt{\cos 2 \alpha}} \left( \begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array} \right) \left( \begin{array}{c} x \\ ct \end{array} \right) \\ &= \gamma \left( \begin{array}{cc} 1 & -\beta \\ -\beta & 1 \end{array} \right) \left( \begin{array}{c} x \\ ct \end{array} \right) \end{aligned}$ with $\beta = \dfrac{v_1}{c} = \dfrac{1}{2}$ and $\gamma = \dfrac{1}{\sqrt{1 - \beta^2}} = \dfrac{2}{\sqrt{3}}$ . Therefore, the points $B$ and $C$ have the following coordinates in the moving reference frame $\begin{aligned} (x_B', ct_B') &= \left( \frac{8}{\sqrt{3}}, \frac{-4}{\sqrt{3}} \right) \\ (x_C', ct_C') &= \left( 0, \frac{6}{\sqrt{3}} \right) \end{aligned}$ We calculate the velocity of the incoming shaceship by $v_2' = \frac{x_C' - x_B'}{t_C' - t_B'} = \frac{0 - 8}{6 - (-4)} c = \frac{8}{10} c = 0.8 \, c$

Thus, the incoming spacecraft moves from the pilot's viewpoint with 80% of the light velocity. A radar pulse propagating at the speed of light is reflected at the point $D$ at the distance $x '= 4 \, \text{Ls}$ from the ghost driver and reaches after the time $\Delta t' = 4 \, \text{s}$ our pilot at point $E$ . The ship controlled by a robot needs $4 \, \text{Ls} / | v_2 | = 5 \, \text{s}$ for the same route, so that the human pilot only has a warning period of one second.