91Ó°ÊÓ

The frame$\stackrel{¯}{S}$moves in$x$ direction at the speed of$\frac{3}{5}c$ relative to frame$S$ .

The lines corresponding to $\stackrel{¯}{x}=−3,−2,−1,0,1,2,3$

The lines corresponding to $c\stackrel{¯}{t}=−3,−2,−1,0,1,2,3$

The expression to calculate the velocity of the relative to frame$S$ is given as follows.

role="math" localid="1658293002770" $\mathbf{V}\mathbf{=}\frac{\stackrel{\mathbf{¯}}{\mathbf{v}}\mathbf{+}\mathbf{u}}{\mathbf{1}\mathbf{+}\frac{\stackrel{\mathbf{¯}}{\mathbf{v}}\mathbf{u}}{{\mathbf{c}}^{\mathbf{2}}}}$ …… (1)

Here, $\mathbf{u}$ is the velocity framerole="math" localid="1658293037181" $\stackrel{\mathbf{¯}}{\mathbf{S}}$ relative to $\mathbf{S}$, $\stackrel{\mathbf{¯}}{\mathbf{v}}$is the velocity of particle relative to frame$\stackrel{\mathbf{¯}}{\mathbf{S}}$ and $\mathbf{C}$ is the velocity of the light speed.

(a)

The graph in cartesian coordinate system with axes $ct$and $x$is shown below.

(b)

In $\stackrel{¯}{S}$, a free particle is observed to travel from the point $\stackrel{¯}{x}=−2$at time $c\stackrel{¯}{t}=−2$to the point $\stackrel{¯}{x}=2$at $c\stackrel{¯}{t}=3$.

The slope of the line is given as,

$\begin{array}{c}\frac{c}{v}=\frac{9.2}{8.7}\\ \frac{v}{c}=\frac{8.7}{9.2}\\ v=0.95c\end{array}$

Hence the speed of the particle is $0.95c$.

(c)

The frame $\stackrel{¯}{S}$moves in$x$direction at the speed of$u=\frac{3}{5}c$relative to frame $S$.

The velocity of the particle relative to$\stackrel{¯}{S}$frame, $\stackrel{¯}{v}=\frac{4}{5}c$.

Calculate the velocity of the relative to frame $S$,

Substitute $\frac{3}{5}c$ for $u$and $\frac{4}{5}c$ for $\stackrel{¯}{v}$into equation (1).

$\begin{array}{c}V=\frac{\frac{4}{5}c+\frac{3}{5}c}{1+\frac{1}{c^2}\left(\frac{4}{5}c\right)\left(\frac{3}{5}c\right)}\\ V=\frac{\frac{7}{5}c}{1+\frac{1}{c^2}×\frac{12}{25}{c}^2}\\ V=\frac{\frac{7}{5}c}{1+\frac{12}{25}}\\ V=0.95c\end{array}$

Hence, by the velocity additional rule, velocity of particle is same as velocity particle in frame$S$ by graphical solution.