00:01
Now in part e, both observes measure the speed of light to be seen.
00:07
In part a we have call the total travel time which is delta t s, an observer at rest relative to the mirror sees the light travel at a distance of d1, which is equal to d from the spacecraft to the mirror.
00:24
But a distance d2 which is equals to d minus d delta ts, distance d2 d2 is equals to d minus v delta t s from the mirror back to the spacecraft because the spacecraft has traveled the distance v t delta v delta t s forward therefore the total distance traveled by the light is d is equal to d1 plus d2 hence d plus d minus v delta t s is equal to c delta ts hence, delta t s is equals to 2d divided by c plus v, which is equal to 2d divided by a c plus 0 .65c.
01:12
We get 2 multiply with 5 .66 into 10, 10 meter, divided by a c, which is 1 .650 multiply 3 into 10 ,000 to power 8 meter per seconds, is equal to 2 to 9 seconds.
01:29
Now in part b, the observer in the spacecraft measure a length contracted initial distance to the mirror of l is equal to d under root 1 minus v square by c square and the mirror moving toward the ship at the speed of v.
01:50
Consider the motion of light toward the mirror and time interval delta t1.
01:54
Light travel toward the mirror at speed c while the mirror travel to be...