0:00
Hi there.
00:01
So for this problem we have to a space shift that have opposite directions and leave the earth and each with a speed of 0 .60 times the speed of light or respect to the earth.
00:19
So we're given that.
00:26
And so for part a of this problem, we need to calculate the velocity of the speed of the speed.
00:32
Spaceship 1 relative to the spaceship 2.
00:38
So we take the positive eruption to be the directions of motion of a spaceship 1.
00:45
So we consider first the spaceship 2 as reference frame as and the earth as the reference frame s prime.
00:55
So the velocity of the earth relative to the spaceship 2, it is given by 0 .6 times the speed of light.
01:07
And the velocity of the spaceship 1 relative to the earth, we are going to call that prime.
01:16
And that has a value of 0 .6 times the speed of light.
01:24
So we need to solve for the velocity of spaceship 1 relative to the spaceship 2 using the following equation.
01:33
So are you going to call the ux is equal to ux prime plus the velocity of the spaceship 2, 1 times the product between these two speeds over the speed of light? so we substitute all of these values in here.
02:03
Oh, square, square, remember that.
02:06
So we will have, in the numerator we will have 0 .6 times the speed of light plus 0 .6 times the speed of light.
02:20
And the denominator, we will obtain that since we multiplied those two velocities and those two velocities has the speed of light, we will obtain a speed of flying squared in the denominator, in the denominator...