Answer to Question #96308 in Mechanics | Relativity for Umaima

Question

Prove that VcA = vccb +VBA when a,b,c don't pass through same event

Answer 1

The objects A, B, C move with regard to each other. After time t after the objects started moving, the distance between A and B (yellow) was $V_{BA}t.$ The distance between B and C was $-V_{BC}t$ (light blue), it is negative because C has lower velocity than B (C has less incline). And the distance between A and C was $-V_{CA}t$ (blue) for the same reason. Therefore, from the drawing we have

$$d(C;B)=d(B;A)+d(C;A).$$

Now substitute what these distances are in terms of relative velocities:

$$-V_{CB}t=V_{BA}t-V_{CA}t, \\V_{CA}=V_{CB}+V_{BA}.$$