Space cruisers A and B are moving parallel to the positive direction of an x axis. Cruiser A is faster, with a relative speed of $\mathbf{v}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{900}\mathbf{c}$, and has a proper length of $\mathit{L}\mathbf{=}\mathbf{200}\mathit{m}$. According to the pilot of A, at the instant (t = 0) the tails of the cruisers are aligned, the noses are also. According to the pilot of B, how much later are the noses aligned?

The relative speed of Cruiser A is $v=0.900c$

The proper length of Cruiser A is $L=200\text{m}$

The length observed by pilot B and pilot A is found by Length contraction formula. The duration for alignment of noses is found by difference in lengths by both pilots by relative speed.

The length of cruiser for pilot B is given as:

$L_B=\frac{L}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $c$is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$.

Substitute all the values in the above equation.

$$\begin{gathered} L_B=\frac{200\text{m}}{\sqrt{1-\frac{{\left(0.900c\right)}^2}{c^2}}} \\ {L}_B=458.8\text{m} \end{gathered}$$

The length of cruiser for pilot A is given as:

$L_A=L\sqrt{1-\frac{v^2}{c^2}}$

Substitute all the values in the above equation.

$$\begin{gathered} L_A=\left(200\text{m}\right)\left(\sqrt{1-\frac{{\left(0.900c\right)}^2}{c^2}}\right) \\ {L}_A=87.8\text{m} \end{gathered}$$

The duration for alignment of noses of cruiser is given as:

$t=\frac{L_B-L_A}{v}$

Substitute all the values in the above equation.

$$\begin{gathered} t=\frac{458.8\text{m}-87.8\text{m}}{\left(0.900c\right)\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)} \\ t=1.37\times {10}^{-6}\text{s} \end{gathered}$$

Therefore, the noses of cruisers aligned for $1.37\times {10}^{-6}\text{s}$.