00:01
In this question, both observers measure the speed of light to be seized.
00:05
In part a, call the total travel time which is delta t .s and observer at rest relative to the mirror, see the light travel at a distance d1, which is equal to d from the spacecraft of two to the mirror.
00:21
But a distance which is d2 is equal to d minus v delta t s, d minus v delta t s from the mirror back to the spacecraft.
00:30
Because the spacecraft has traveled the distance, which is v delta t s forward, therefore the total distance traveled by the light is d is equal to d1 plus d2, which is equal to d plus d minus v delta ts is equal to c delta ts.
00:50
Hence delta ts is equal to 2d divided by c plus v.
00:57
In part b, we have the observer in the spacecraft measure a length contracted initial distance to the mirror of, which is l is equals to d under root 1 minus v square upon c square.
01:11
And the mirror moving toward the ship at speed v...