**The fifth scene: 2020.3.9 **

Given "u" the velocity of A relative to the earth, "w" the velocity of spaceship B relative to A, you should find "v" the velocity of B relative to the earth.

------->velocity positive direction

Earth ……………………………….A…………………………………….B

……………………………………....u...…………………........…......…..v

u is the velocity of A relative to the Earth

v is the velocity of B relative to the Earth

x is the velocity of Earth relative to B

w is the velocity of B relative to the A

Presume: u=0.2C, w=0.8C

w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.8C

0.2-v=(1-0.2v)*0.8

0.84v = -0.6

v = -0.714C, the velocity v of B relative to the earth is -0.714C, velocity direction is opposite to A.

you can seee that v and u are in the different direction.But A sees B moving away at 0.8C, what happened?

Presume: u=0.2C, w=0.1C

w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.1C

0.2-v=(1-0.2v)*0.1

0.98v = 0.1

v = 0.102C, v and u are in the same direction.

You can see different u and different w, the direction of v is different, which is very interesting.

Deduction:

w = (u+x)/(1+u*x/C^2)

if u = 0.2C, put it into the formula. w = (0.2+x)/(1+0.2x) ==> w = (1+5x)/(5+x) ==> x=(5w-1)/(5-w),

so if x>0, then (5w-1>0 and 5-w>0) or (5w-1<0 and 5-w<0), simplify this inequality (w>0.2 and w<5) or (w<0.2 and *w>5, it is not correct*). So we get:

....if 0.2<w<1, then x>0.

....if w<0.2 or *w>1(it is not correct)*, then x<0.

Finally, we can get:

....if 0.2C<w<1C, then x>0, v<0; this mean u and v have the contrary velocity direction.

....if -1C<w<0.2C, then x<0, v>0; this mean u and v have the same velocity direction.

....if w=0.2C, then x=0, v=0;

It is so interesting, the value of w will determine the direction of v.

？？？

We need to think: is this a Digital Game?

**Edited by TonyYuan2020, 19 March 2020 - 06:56 AM.**