Respuesta :
Answer:
[tex]x'_t=-5880\ m[/tex]
[tex]x'_p=-1680\ m[/tex]
[tex]t'_t=28\ \mu s[/tex]
[tex]t'_p=18.2\ \mu s[/tex]
Explanation:
Given:
- position of tree on the earth, [tex]x_t= 0[/tex]
- position of pole on the earth, [tex]x_p=30\ km[/tex]
- time at which the event occurs, [tex]t=20\ \mu s[/tex]
- velocity of observer in the rocket with respect to the earth, [tex]v=0.7c[/tex]
Now the space coordinate x for tree and pole as observed from the rocket:
[tex]x'_t=\gamma (x_t-v.t)[/tex]
where:
[tex](\rm lorentz\ factor)\ \gamma= \frac{1}{\sqrt{1-\frac{v^2}{c^2} } } = \frac{1}{\sqrt{1-\frac{(0.7c)^2}{c^2} } }=1.4[/tex]
[tex]x'_t=1.4\times (0-(0.7\times 3\times 10^8)\times (20\times 10^{-6}))[/tex]
[tex]x'_t=-5880\ m[/tex]
and
[tex]x'_p=\gamma (x_p-v.t)[/tex]
[tex]x'_p=1.4\times (3000-(0.7\times 3\times 10^8)\times (20\times 10^{-6}))[/tex]
[tex]x'_p=-1680\ m[/tex]
Now, time coordinates:
[tex]t'_t=\gamma (t-\frac{v.x_t}{c^2} )[/tex]
[tex]t'_t=1.4\times (20\times 10^{-6}-\frac{0.7c\times 0}{c^2} )[/tex]
[tex]t'_t=28\ \mu s[/tex]
and
[tex]t'_p=\gamma (t-\frac{v.x_p}{c^2} )[/tex]
[tex]t'_p=1.4\times (20\times 10^{-6}-\frac{0.7c\times 3000}{c^2} )[/tex]
[tex]t'_p=18.2\ \mu s[/tex]
