Question

Reference frame S’ is to pass reference frame S at speed v along the common direction of the and x axes, as in Fig. 37-9. An observer who rides along with frame S’ is to count off a certain time interval on his wristwatch. The corresponding time interval $\mathbf{∆}\mathit{t}$ is to be measured by an observer in frame S. Figure 37-22 gives $\mathbf{∆}\mathit{t}$versus speed parameter $\mathit{\beta }$for a range of values for . The vertical axis scale is set by ta 14.0 s. What is interval $\mathbf{∆}\mathit{t}$ if v = 0.98c?

Short answer

The time interval for observer is $40s$.

Step-by-step solution

Identification of given data

The speed of reference frame S’ is $v=0.98c$

The duration for the reference frame S’ is $t=8s$

The time dilation is used to find the duration for particle before decay from detector.

Determination of time interval for observer

The time interval for observer is given as:

$∆t=\frac{t}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$

Here, $c$is the speed of light and its value is $3\times {10}^{8}m/s$.

Substitute all the values in equation.

$$\begin{gathered} ∆t=\frac{8s}{\sqrt{1-{\left({\displaystyle \frac{0.98c}{c}}\right)}^2}} \\ ∆t=40s \end{gathered}$$

Therefore, the time interval for observer is 40 s .