Bob is watching Anna fly by in her new high-speed plane, which Anna knows to be $\mathbf{60}\mathit{m}$in length. As a greeting, Anna turns on two lights simultaneously, one at the front and one at the tail. According to Bob, the lights come $\mathbf{40}\mathbf{}\mathit{n}\mathit{s}$apart.

(a) Which comes on first?

(b) How fast is the plane moving?

Let Anna's frame be$s^1$and Bob’s be $s$. And Anna is moving on the positive x-axis at velocity $v$with respect to Bob. Event $1$is the turning on of the front light. According to Anna,$x_2^{\text{'}}-x_1^{\text{'}}=-60\mathrm{m}$and $t_2^{\text{'}}-t_1^{\text{'}}=0$ The time interval between two events according to Bob is as follows:

localid="1659084015792" ${\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}\mathbf{=}{\mathit{\gamma }}_{\mathbf{r}}\left[\frac{v}{c^2}\left(x_2^{\text{'}}-x_1^{\text{'}}\right)+\left(t_2^{\text{'}}-t_1^{\text{'}}\right)\right]$

After substituting the values, we get

localid="1659084075710" $t_2-t_1={\gamma }_v\left[\frac{v}{c^2}(-60\mathrm{m})+0\right]$

As the above expression is negative, event $2$occurs before event $1$and therefore, the tail light turns on $40ns$before the front light.

$t_2-t_1=40\times {10}^{-0}\mathrm{s}$

${\gamma }_v\left(\frac{(-60\mathrm{m})v}{c^2}\right)=40\times {10}^9\mathrm{s}$

${\gamma }_{r}v=-\frac{2\times {10}^3\times {\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}{3\mathrm{m}/\mathrm{s}}$

${\gamma }_{v}v=-6\times {10}^{\tau }\mathrm{m}/\mathrm{s}$

Squaring on both sides and solving further yields

$v^2=\frac{36\times {10}^{14}{\mathrm{m}}^2/{\mathrm{s}}^2}{\left(1+\frac{36\times {10}^{14}}{c^2}\right)}$

$=\frac{36\times {10}^{14}{\mathrm{m}}^2/{\mathrm{s}}^2}{(1+0.04)}$

$=3.46\times {10}^{15}{\mathrm{m}}^2/{\mathrm{s}}^2$

$v=5.88\times {10}^7\mathrm{m}/\mathrm{s}$

Therefore, the plane is moving at a velocity of $0.2c$.