Question

According to observer O, a blue flash occurs at xb = 10.4 m when tb= 0.124 μs, and a red flash occurs at xr= 23.6 m when tr= 0.138 μs. According to observer O’, who is in motion relative to O at velocity u, the two flashes appear to be simultaneous. Find the velocity u.

Answer 1

Answer:

The velocity is  $u=0.3m/s$

Explanation:

From the question we can deduce that both flashes occurred simultaneous to the observer

    So the first derivative of the time difference   $\Delta t' =0$

Generally the Lorentz transformation i mathematically denoted as

               $\Delta t' = \frac{ \Delta t -\frac{v \Delta x}{c^2} }{\sqrt{1-\frac{v^2}{c^2} } }$

=>           $0 = \frac{ \Delta t -\frac{v \Delta x}{c^2} }{\sqrt{1-\frac{v^2}{c^2} } }$

=>           $\Delta t -\frac{u \Delta x }{c^2} = 0$

Now making u the subject

             $u = \frac{c^2 \Delta t }{\Delta x}$

                 $= \frac{t_b - t_r}{x_b - x_r} c^2$

substituting $0.24 \mu s = 0.24*10^{-6}s$ for $t_b$ , $0.138 \mu s = 0.138*10^{-6}s$ for $t_r$ , $10.4 m$ for $x_b$ , $23.6 m$ for $x_r$ and  $3.0*10^8 m/s$ for c

       

              $u = \frac{0.124*10^ {-6} -0.138 *10^{-6}}{10 - 23.6}$

                $u=0.3m/s$

Answer 2

Answer:

u = 95.46 * 10⁶ m/s

Explanation:

For the first observer O,

Time of the blue flash, $t_{b} = 0.124 \mu s$

Time of the red flash, $t_{r} = 0.138 \mu s$

Distance covered before the blue flash, $x_{b} = 10.4 m$

Distance covered before the red flash,  $x_{r} = 23.6 m$

Difference in time observed by the first observer O,  Δt = $t_{r} - t_{b}$

Δt = 0.138 - 0.124 = 0.014  μs

Difference in the distance observed by the observer O, Δx = $x_{r} - x_{t}$

Δx = 23.6 m - 10.4 m

Δx = 13.2 m

Since for the observer O', the two flashes are simultaneous, there will be no time difference between the two flashes. i.e. Δt' = 0

Using the lorent'z transformation equation:

$\triangle t' = \frac{\triangle t - \frac{u \triangle x}{c^{2} } }{\sqrt{1 - \frac{u^{2} }{c^{2} } } }$...................(1)

But Δt' = 0,

equation (1) becomes

$0 = \triangle t - \frac{u \triangle x}{c^{2} }$

$0 = (0.014 * 10^{-6} ) - \frac{u * 13.2}{(3*10^{8}) ^{2} }$

$(0.014 * 10^{-6} ) = \frac{u * 13.2}{(3*10^{8}) ^{2} } \\u = 95454545.46 m/s$

u = 95.46 * 10⁶ m/s