Question

Two events in reference frame $S$ occur $10\mu s$ apart at the same point in space. The distance between the two events is $2400m$in reference frame $S\text{'}$.

a. What is the time interval between the events in reference frame S′?

b. What is the velocity of$S\text{'}$relative to $S$?

Short answer

a. The time interval between the events in the reference frame $S\text{'}$is $12.8\mu s$.

b. The velocity of $S\text{'}$ relative to $S$is$0.480c$.

Step-by-step solution

Part (a) Step 1: Given Information

We need to find the time interval between the events in reference frame$S\text{'}$.

Part (a) Step 2: Simplify

Consider :

$$\begin{gathered} ∆x\text{'}=2400m \\ t=10\times {10}^{-6}s \end{gathered}$$

We finding time interval between the events in reference frame $S\text{'}$:

$$\begin{gathered} ∆t^2={\left(c∆t\right)}^2-{\left(∆x\right)}^2={\left(c∆t\text{'}\right)}^2-{\left(∆x\text{'}\right)}^2 \\ {\left(c∆t\text{'}\right)}^2={(c\times 10\times {10}^{-6})}^2+{\left(2400\right)}^2{\left(c∆t\text{'}\right)}^2=\left({\left(3.0\right)}^2\right)\times {10}^6+{\left(2.4\right)}^2\times {10}^6 \\ {\left(c∆t\text{'}\right)}^2=\left(3.0^2+2.4^2\right)\times {10}^6 \\ {\left(c∆t\text{'}\right)}^2=3.841\times {10}^3 \\ ∆t\text{'}=\frac{3.841\times {10}^3}{3\times {10}^8}=12.8\mu s \end{gathered}$$

Part (b) Step 1: Given Information

We need to find the velocity of $S\text{'}$relative to$S$.

Part (b) Step 2: Simplify

The velocity of $S\text{'}$relative to $S$, as:

$t\text{'}=\gamma \left(t-\frac{\nu x}{c^2}\right)\to v=0.480c$