Answer :
Answer:
a) [tex] x=2*10^{9} m[/tex] and [tex] t=8.35 s[/tex]
b) t = γt', so it is 8.35 s.
Explanation:
a) The equation of Lorentz transformations is given by:
[tex] x=\gamma(x'+ut')[/tex]
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
- x' = 0
- t' = 5.00 s
- u =0.800 c, c is the speed of light 3*10⁸ m/s
[tex]\gamma=\frac{1}{\sqrt{1-(u/c)^{2}}}[/tex]
[tex]\gamma=\frac{1}{\sqrt{1-(0.800c/c)^{2}}}[/tex]
[tex]\gamma=\frac{1}{\sqrt{1-(0.800)^{2}}}[/tex]
[tex]\gamma=1.67[/tex]
[tex] x=1.67(0+0.800c*5.00)[/tex]
[tex] x=2*10^{9} m[/tex]
Now, to find t we apply the same analysis:
[tex] t=\gamma(t'+\frac{ux'}{c^{2}})[/tex]
but as x'=0 we just have:
[tex] t=\gamma(t')[/tex]
[tex] t=1.67*5.00=8.35 s[/tex]
b) Here, Mavis reads 5 s on her watch and Stanley measured the events at a time affected by the Lorentz factor, in other words t = γt', if we see it is the same a) part. So the time interval will be equal to 8.35 s.
I hope it helps you!
Answer:
a) x = 2 × 10⁹m and t = 8.34s
b)t = 8.34s
Explanation:
Part a
The Lorentz factor is given by
[tex]\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2} } }[/tex]
[tex]\gamma = \frac{1}{\sqrt{1-(\frac{0.8c}{c})^2 } } \\\\ = 1.667[/tex]
From Lorentz transformation we have
[tex]x = \gamma (x'+ut')\\\\ x = \gamma (0 + ut') \\\\ x= \gamma ut'[/tex]
substitute the values
[tex]x = (1.667)(0.8c)(5.0s)\\\\x=(1.667)(0.8\times 3.0 \times 10^8m/s)(5.0s)\\\\x=2.0\times10^9m[/tex]
From Lorentz transformation we have
[tex]t = \gamma (t'+\frac{ux'}{c^2} ) \\\\t = \gamma(t'+\frac{u(0)}{c^2} )\\\\t= \gamma t'[/tex]
[tex]t = (1.667)\times 95.0s)\\\\t=8.34s[/tex]
Part b
From the concept of timr dilation we have,
[tex]\Delta T = \gamma \Delta T_0[/tex]
[tex]= 1.667\times 5.0\\\\=8.34s[/tex]
The value of time found is same as part a