91影视

Let Anna's frame be$s^1$and Bob鈥檚 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" ${\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}\mathbf{=}{\mathit{纬}}_{\mathbf{r}}\left[\frac{v}{c^2}\left(x_2^{\text{'}}-x_1^{\text{'}}\right)+\left(t_2^{\text{'}}-t_1^{\text{'}}\right)\right]$

After substituting the values, we get

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

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

$t_2-t_1=40脳{10}^{-0}\mathrm{s}$

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

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

${纬}_{v}v=-6脳{10}^{蟿}\mathrm{m}/\mathrm{s}$

Squaring on both sides and solving further yields

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

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

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

$v=5.88脳{10}^7\mathrm{m}/\mathrm{s}$

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