Two spaceships approach the Earth from opposite directions. According to an obse

Q: Two spaceships approach the Earth from opposite directions. According to an obse

a) Ship A observed from ship B V_ba =(Va+Vb)/[1+VaVb/c^2] V_ba =(0.687c+0.863c)/(1+0.687c*0.863c/c^2] V_ba =0.973c b) Ship B observed from ship A V_ab =V_ba =0.973c (since both of these quantities just represent the relative velocity of the two ships as measured by them.)

ID: 2272480 • Letter: T

Question

Two spaceships approach the Earth from opposite directions. According to an observer on the Earth, ship A is moving at a speed of 0.687c and ship B at a speed of 0.863c. What is the velocity of ship Aas observed from ship B? Of ship B as observed from ship A? (Indicate the direction with the sign of your answer. Assume ship A is moving in the positive direction).

Pleaes don't give me an answer that is higer than the speed of light. This is modern physics, not classical. Thanks!

ship A observed from ship B c ship B observed from ship A c

Explanation / Answer

Ship A observed from ship B

V_ba =(Va+Vb)/[1+VaVb/c^2]

V_ba =(0.687c+0.863c)/(1+0.687c*0.863c/c^2]

V_ba =0.973c

Ship B observed from ship A

V_ab =V_ba =0.973c

(since both of these quantities just represent the relative velocity of the two ships as measured by them.)

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Two spaceships approach the Earth from opposite directions. According to an obse

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