91Ó°ÊÓ

We have given,

In S frame the first explosion time =$1\mathrm{s}$

In the S frame the second explosion time =$3\mathrm{s}$

Position of first explosion in S' frame =$4\mathrm{m}$

Position of second explosion in S' frame =$-4\mathrm{s}$

We have to find the speed of frame S' with respect to S frames.

Use the Galilean Transformation Equation for frame$S\text{'}$

$$\begin{gathered} {\mathrm{x}}_1^{\text{'}}={\mathrm{x}}_1+{\mathrm{vt}}_1 \\ 4\mathrm{m}={\mathrm{x}}_1+\mathrm{v}×1\mathrm{s} \\ {\mathrm{x}}_1=4-\mathrm{v}.....................................\left(1\right) \end{gathered}$$

Similarly,

$$\begin{gathered} {\mathrm{x}}_2^{\text{'}}={\mathrm{x}}_2+{\mathrm{vt}}_2 \\ -4\mathrm{m}={\mathrm{x}}_2+\mathrm{v}×3\mathrm{s} \\ {\mathrm{x}}_2=-4-3\mathrm{v}.....................................\left(2\right) \end{gathered}$$

Since, they both event happens at one place,

$$\begin{gathered} {\mathrm{x}}_1={\mathrm{x}}_2 \\ 4-\mathrm{v}=-4-3\mathrm{v} \\ \mathrm{v}=-4\mathrm{m}/\mathrm{s} \\ \left|\mathrm{v}\right|=4\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given,

In S frame the first explosion time =$1.0\mathrm{s}$

In the S frame the second explosion time =$3\mathrm{s}$

Position of first explosion in S' frame =$4\mathrm{m}$

Position of second explosion in S' frame =$-4\mathrm{m}$

We have to find the two explosion position S frames.

Using the Galilean Transformation Equation for frame$S\text{'}$

$$\begin{gathered} {\mathrm{x}}_1^{\text{'}}={\mathrm{x}}_1+{\mathrm{vt}}_1 \\ 4\mathrm{m}={\mathrm{x}}_1+\mathrm{v}×1\mathrm{s} \\ {\mathrm{x}}_1=4-\mathrm{v} \end{gathered}$$

putting the value of V,

${\mathrm{x}}_1=4+4=8\mathrm{m}$