Question #96308
Prove that VcA = vccb +VBA when a,b,c don't pass through same event
1
Expert's answer
2019-10-11T10:50:57-0400


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 VBAt.V_{BA}t. The distance between B and C was VBCt-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 VCAt-V_{CA}t (blue) for the same reason. Therefore, from the drawing we have


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

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


VCBt=VBAtVCAt,VCA=VCB+VBA.-V_{CB}t=V_{BA}t-V_{CA}t, \\V_{CA}=V_{CB}+V_{BA}.



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Comments

Assignment Expert
28.10.19, 17:40

Dear visitor, please use panel for submitting new questions

Umaima
28.10.19, 16:05

Suppose that earth is moving through luminiferous ether at speed v suppose that you are Michelson & that's u have interferometer with two arms each of length L if one arm is along the direction of earth's motion & one arm is perpendicular direction how long it will take a bit of light to make complete circuit along each arm ?

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