Anna and Bob are In identical spaceships, each 100 m long. The diagram shows Bob's, view as Anna's ship passes at 0.8c. Just as the backs of the ships pass one another, both clocks.there read O. At the instant shown, Bob Jr., on board Bob's ship, is aligned with the very front of Anna's ship. He peers through a window in Anna's ship and looks at the clock. (a) In relation to his own ship, where is Bob Jr? (b) What does the clock he sees read?

The given data can be written as:

• The length of each spaceship is ${\mathrm{x}}^{\text{'}}=100\mathrm{m}$.
• The speed of Anna’s spaceship is $\mathrm{v}=0.8\mathrm{c}$.

The relative motion between Anna and Bob is used to calculate the position of Bob and reading of the clock by Bob Jr. The formula of time dilation and length contraction is used for the calculation.

$\mathrm{x}={\mathrm{x}}^{\text{'}}\sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}$

Substitute all the values in the above equation,

$$\begin{gathered} \mathrm{x}=\left(100\mathrm{m}\right)\left(\sqrt{1-\frac{{\left(0.8\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}\right) \\ \mathrm{x}=60\mathrm{m} \end{gathered}$$

Therefore, Bob Jr is data-custom-editor="chemistry" $60\mathrm{m}$behind his own spaceship.

$t^{\text{'}}=\left(\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}\right)\left(-\frac{\mathrm{vx}}{{\mathrm{c}}^2}+\mathrm{t}\right)$

Here, v is the speed of spaceship, c is the speed of light, x is the length of each spaceship.

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{t}}^{\text{'}}=\left(\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.8\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}}\right)\left(-\frac{\left(0.8\mathrm{c}\right)\left(60\mathrm{m}\right)}{{\mathrm{c}}^2}+0\right) \\ {\mathrm{t}}^{\text{'}}=\left(1.67\right)\left(-\frac{\left(0.8\mathrm{c}\right)\left(60\mathrm{m}\right)}{\mathrm{c}\left({\displaystyle \frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\mathrm{c}}}\right)}\right) \\ {\mathrm{t}}^{\text{'}}=267\times {10}^{-9}\mathrm{s} \end{gathered}$$

Therefore, the reading of clock for Bob Jr is $267\times {10}^{-9}\mathrm{s}$.