91Ó°ÊÓ

The given data can be listed below as:

• The velocity of the frame$\mathrm{S}$ is$\mathrm{v}$ .
• The velocity of the frame ${\mathrm{S}}^{\text{'}}$is ${\mathrm{u}}^{\text{'}}$.
• The scale in the vertical axis is ${{\mathrm{u}}^{\text{'}}}_{\mathrm{a}}=−0.800\mathrm{c}$.

The reference frame is described as the position where the geometric points are being identified as physically and mathematically. This frame is set by some set of the reference points.

(a)

According to the graph, when$\mathrm{v}=0$, then $\mathrm{u}=−0.20\mathrm{c}$.

The equation of the velocity of the frame${\mathrm{S}}^{\text{'}}$is expressed as:

${\mathrm{u}}^{\text{'}}=\frac{\mathrm{u}−\mathrm{v}}{1−\mathrm{uv}/{\mathrm{c}}^2}$ …(¾±)

Here, $\mathrm{u}\text{'}$is the velocity of the frame ${\mathrm{S}}^{\text{'}}$, $\mathrm{u}$is the initial and $\mathrm{v}$is the final velocity of the frame. $\mathrm{c}$is the speed of the light.

Substitute $−0.20\mathrm{c}$for role="math" localid="1663049487197" $\mathrm{u}$, $3×{10}^8\text{″¾/²õ}$for $\mathrm{c}$and$0.80\mathrm{c}$ for $\mathrm{v}$in the above equation.

$\begin{array}{c}{\mathrm{u}}^{\text{'}}=\frac{(3×{10}^8\text{″¾/²õ})(−0.20−0.80)}{1+0.20×0.80}\\ =\frac{−3×{10}^8\text{″¾/²õ}}{1+0.16}\\ =\frac{−3×{10}^8\text{″¾/²õ}}{1.16}\\ =−2.5×{10}^8\text{″¾/²õ}\end{array}$

Thus, the value of${\mathrm{u}}^{\text{'}}$ is $−2.5×{10}^8\text{″¾/²õ}$.

(b)

The equation (i) has been recalled below:

${\mathrm{u}}^{\text{'}}=\frac{\mathrm{u}−\mathrm{v}}{1−\mathrm{uv}/{\mathrm{c}}^2}$

Substitute $−0.20\mathrm{c}$for $\mathrm{u}$, $3×{10}^8\text{″¾/²õ}$for $\mathrm{c}$and $3×{10}^8\text{″¾/²õ}$for $\mathrm{v}$in the above equation.

$\begin{array}{c}{\mathrm{u}}^{\text{'}}=\frac{(3×{10}^8\text{″¾/²õ})(−0.20−1)}{1+0.20}\\ =\frac{−3.6×{10}^8\text{″¾/²õ}}{1.20}\\ =\frac{−3.6×{10}^8\text{″¾/²õ}}{1.20}\\ =−3×{10}^8\text{″¾/²õ}\end{array}$

Thus, the value of ${\mathrm{u}}^{\text{'}}$is $−3×{10}^8\text{″¾/²õ}$.