Answer:
(a) [tex]\Delta t'=1.38\mu s[/tex]
(b) [tex]\Delta x'=-389.46m[/tex]
Explanation:
We use Lorentz transformations, since they relate the measures of a physical magnitude obtained by two different observers, these are:
[tex]\Delta x'=\frac{\Delta x-u\Delta t}{\sqrt{1-\frac{u^2}{c^2}}}\\\Delta t'=\frac{\Delta t-\frac{u\Delta x}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]
Here [tex]\Delta x[/tex] is the spatial separation according to O, [tex]\Delta x'[/tex] is the spatial separation according to O', [tex]\Delta t[/tex] is the time interval according to O, [tex]\Delta t'[/tex] is the time interval according to O', [tex]u[/tex] is the relative speed between the two observers and [tex]c[/tex] is the speed of light. All we do now is write the quantities we were given, recall that [tex]1\mu s=10^{-6}s[/tex]
(a)
[tex]\Delta t'=\frac{0.48*10^{-6}s-\frac{0.97c(45m)}{c^2}}{\sqrt{1-\frac{(0.97c)^2}{c^2}}}\\\Delta t'=\frac{0.48*10^{-6}s-\frac{0.97(45m)}{(3*10^{8}\frac{m}{s})}}{\sqrt{1-0.97^2}}\\\Delta t'=1.38*10^-6 s=1.38\mu s[/tex]
(b)
[tex]\Delta x'=\frac{45m-(0.97*3*10^8\frac{m}{s})0.48*10^{-6}s}{\sqrt{1-\frac{(0.97c)^2}{c^2}}}\\\Delta x'=-389.46m[/tex]
The minus sign means that the second event is closer for one observer and the first is closer for the other.
