Inertial system $\overline{\mathbf{S}}$moves in the $x$direction at speed $\stackrel{}{\frac{\mathbf{3}}{\mathbf{5}}\mathbf{c}}$relative to system$\mathit{S}$. (The$\overline{\mathbf{x}}$axis slides long the$\mathit{x}$axis, and the origins coincide at $\mathit{t}\mathbf{=}\overline{\mathbf{t}}\mathbf{=}\mathbf{0}$, as usual.)

(a) On graph paper set up a Cartesian coordinate system with axesrole="math" localid="1658292305346" $\mathit{c}\mathit{t}$ and $\mathit{x}$. Carefully draw in lines representing$\overline{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}$and$\mathbf{3}$. Also draw in the lines corresponding to $\mathit{c}\overline{\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}$, and$\mathbf{3}$. Label your lines clearly.

(b) In$\overline{\mathbf{S}}$, a free particle is observed to travel from the point $\stackrel{}{\overline{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}}$at time$\mathit{c}\overline{\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{2}$to the point $\overline{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{,}$ at$\mathit{c}\overline{\mathbf{t}}\mathbf{=}\mathbf{+}\mathbf{3}$. Indicate this displacement on your graph. From the slope of this line, determine the particle's speed in $\mathit{S}$.

(c) Use the velocity addition rule to determine the velocity in $\mathit{S}$algebraically,and check that your answer is consistent with the graphical solution in (b).

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

The lines corresponding to $\overline{x}=-3,-2,-1,0,1,2,3$

The lines corresponding to $c\overline{t}=-3,-2,-1,0,1,2,3$

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

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

Here, $\mathbf{u}$ is the velocity framerole="math" localid="1658293037181" $\overline{\mathbf{S}}$ relative to $\mathbf{S}$, $\overline{\mathbf{v}}$is the velocity of particle relative to frame$\overline{\mathbf{S}}$ 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 $\overline{S}$, a free particle is observed to travel from the point $\overline{x}=-2$at time $c\overline{t}=-2$to the point $\overline{x}=2$at $c\overline{t}=3$.

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

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

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

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

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

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