91Ó°ÊÓ

Suppose a particle is moving with speed u^' in x^' in an inertial frame S^'. If the frame S^' is moving with a velocity v with respect to another frame S, then the velocity of the particle with respect to frame S is given by,

$\mathbf{u}\mathbf{=}\frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{v}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}}$.....................(1)

Here, the speed of the light is c.

Rearrange the equation (1) for u^'.

$\begin{array}{rcl}\mathrm{u}\left(1+\frac{\mathrm{u}\text{'}\mathrm{v}}{{\mathrm{c}}^2}\right)& =& {\mathrm{u}}^{\text{'}}+\mathrm{v}\\ \frac{{\mathrm{uu}}^{\text{'}}\mathrm{v}}{{\mathrm{c}}^2}-{\mathrm{u}}^{\text{'}}& =& \mathrm{v}-\mathrm{u}\\ \mathrm{u}\text{'}\left(\frac{\mathrm{uv}}{{\mathrm{c}}^2}-1\right)& =& \mathrm{v}-\mathrm{u}\\ {\mathrm{u}}^{\text{'}}& =& \frac{\mathrm{v}-\mathrm{u}}{\left({\displaystyle \frac{\mathrm{uv}}{{\mathrm{c}}^2}}-1\right)}\end{array}$

Substitute 0.82c for u, and -0.82c for v in equation (2).

$$\begin{gathered} {\mathrm{u}}^{\text{'}}=\frac{-0.82\mathrm{c}-0.82\mathrm{c}}{{\displaystyle \frac{\left(-0.82\mathrm{c}\right)\left(0.82\mathrm{c}\right)}{{\mathrm{c}}^2}}-1} \\ {\mathrm{u}}^{\text{'}}=0.98\mathrm{c} \end{gathered}$$

The expression to calculate the time is given by,

data-custom-editor="chemistry" $∆{\mathrm{t}}^{\text{'}}=\frac{∆\mathrm{x}}{{\mathrm{u}}^{\text{'}}}$

Substitute 350m fordata-custom-editor="chemistry" $∆\mathrm{x}$ and 0.98c for u^' in equation (3).

$\begin{array}{rcl}∆{\mathrm{t}}^{\text{'}}& =& \frac{350\mathrm{m}}{0.98\mathrm{c}}\\ & =& \frac{350\mathrm{m}}{0.98\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}\\ & =& 1.2×{10}^{-6}\mathrm{s}\\ & =& 1.2\mathrm{μ²õ}\end{array}$

Therefore, the time taken by the object to pass the ship is data-custom-editor="chemistry" $1.2\mathrm{μ²õ}$.