Question

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

a. What is the time interval between the events in reference frame $S\text{'}$′?

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

Short answer

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

b. $480c$is the velocity of $S\text{'}$relative to$S$.

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:

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

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

$$\begin{gathered} ∆t^2={\left(c∆t\right)}^2-{\left(∆x\right)}^2={\left(c∆t\text{'}\right)}^2-{\left(∆x\text{'}\right)}^2 \\ {\left(c∆t\text{'}\right)}^2={\left(c\times 10\times {10}^{-6}\right)}^2+{\left(2400\right)}^2 \\ {\left(c∆t\text{'}\right)}^2=\left({\left(3,0\right)}^2\right)\times {10}^6+{\left(2.4\right)}^2\times {10}^6 \\ {\left(c∆t\text{'}\right)}^2=\left(3.0^2+2.4^2\right)\times {10}^6 \\ {\left(c∆t\text{'}\right)}^2=3.841\times {10}^3 \\ ∆t\text{'}=\frac{3.841\times {10}^3}{3\times {10}^8} \\ ∆t\text{'}=12.8\mu s \end{gathered}$$

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 $S$

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