Question

Reference frame ${\mathbf{S}}^{\mathbf{\text{'}}}$passes reference frame${\mathbf{S}}^{}$ with a certain velocity as in Fig. 37-9. Events 1 and 2 are to have a certain spatial separation$\mathbf{∆}\mathbf{x}\mathbf{\text{'}}$ according to the${\mathbf{S}}^{\mathbf{\text{'}}}$ observer. However, their temporal separation$\mathbf{∆}\mathbf{t}\mathbf{\text{'}}$ according to that observer has not been set yet. Figure 37-30 gives their spatial separation$\mathbf{∆}\mathbf{x}$ according to the${\mathbf{S}}^{}$ observer as a function of $\mathbf{∆}\mathbf{t}\mathbf{\text{'}}$for a range ofrole="math" localid="1663054361614" $\mathbf{∆}\mathbf{t}\mathbf{\text{'}}$ values. The vertical axis scale is set by ${\mathbf{\Delta x}}_{\mathbf{a}}\mathbf{=}\mathbf{10.0}\mathbf{\text{\hspace{0.17em}m}}$.What is$\mathbf{\Delta x}\mathbf{\text{'}}$ ?

Short answer

The value of${\mathrm{\Delta x}}^{\text{'}}$ is $0.78\text{\hspace{0.17em}m}$.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The spatial separation between the event 1 and 2 is ${\mathrm{\Delta x}}^{\text{'}}$.
• The temporal separation between the event 1 and 2 is ${\mathrm{\Delta t}}^{\text{'}}$.
• The value of the vertical axis is ${\mathrm{\Delta x}}_{\mathrm{a}}=10.0\text{\hspace{0.17em}m}$.

Significance of the slope of a graph

The slope of a graph is described as the steepness measure of a graph. The slope mainly stays constant.

Determination of the value of Δx'

According to the line of the graph, it consists of an equation. The equation of the ${\mathrm{\Delta x}}^{\text{'}}$is expressed as:

$\begin{array}{c}\mathrm{\Delta x}={\mathrm{v\Delta t}}^{\text{'}}+{\mathrm{\gamma \Delta x}}^{\text{'}}\\ ={\mathrm{s\Delta t}}^{\text{'}}+\mathrm{a}\end{array}$

Here,$\mathrm{a}$is the y-intercept of the graph and$\mathrm{\gamma }$is the x-intercept.

Equalling both the sides of the above equation, the value of${\mathrm{\Delta x}}^{\text{'}}$is expressed as:

$\begin{array}{c}{\mathrm{\gamma \Delta x}}^{\text{'}}=\mathrm{a}\\ {\mathrm{\Delta x}}^{\text{'}}=\frac{\mathrm{a}}{\mathrm{\gamma }}\end{array}$

From the graph, it can be observed that the value of y-intercept is about $2.0\text{\hspace{0.17em}m}$and $\mathrm{\gamma }$also comes to a value of $2.54$.

Substitute$2.0\text{\hspace{0.17em}m}$ for$\mathrm{a}$ and $2.54$for $\mathrm{\gamma }$in the above equation.

$\begin{array}{c}{\mathrm{\Delta x}}^{\text{'}}=\frac{2.0\text{\hspace{0.17em}m}}{2.54}\\ =0.78\text{\hspace{0.17em}m}\end{array}$

Thus, the value ofrole="math" localid="1663054925380" $∆x\text{'}$ is$0.78\text{\hspace{0.17em}m}$