Answer:
a) 0.94 C
b) 7.14 m
c) 3.11 m
d) 1.40 s
e) 2.93 s
Explanation:
First we need to set up a coordinate system. This will have the positive X axis pointing north. So spaceship A has positive speed, and spaceship B has negative speed.
The Lorentz transformation for speed is:
[tex]u' = \frac{u - v}{1 - \frac{u*v}{c^2}}[/tex]
u: speed of spaceship A as observed by you
v: speed of spaceship B as observed by you
In the case of the speed of spaceship A as observed by spaceship B:
[tex]u' = \frac{0.7c - (-0.7c)}{1 - \frac{0.7c*(-0.7c)}{c^2}} = 0.94c[/tex]
The transform for lengths is:
[tex]L = L0 * \sqrt{1 - \frac{v^2}{c^2}}[/tex]
For the case of spaceship A as observed by you:
[tex]L = 10 m * \sqrt{1 - \frac{(0.7c)^2}{c^2}} = 7.14 m[/tex]
For the case of spaceship A as observed by spaceship B:
[tex]L = 10 m * \sqrt{1 - \frac{(0.94c)^2}{c^2}} = 3.12 m[/tex]
The time dilation equation is:
[tex]T = \frac{T0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
For the case of the event as observed by you:
[tex]\frac{1 s}{\sqrt{1-\frac{(0.7c)^2}{c^2}}} = 1.40 s[/tex]
For the case of the event as observed by spaceship B:
[tex]\frac{1 s}{\sqrt{1-\frac{(0.94c)^2}{c^2}}} = 2.93 s[/tex]