ch37-p067 - 67. Interpreting vAB as the x-component of the...

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(1 )(1 ) ) AC AB BC AB BC AC β ββ βββ −++ = −−+ and expand: 1 – β AC + β AB + β BC β AC β AB β AC β BC + β AB β BC β AB β BC β AC = 1 + β AC β AB β BC β AC β AB β AC β BC + β AB β BC + β AB β BC β AC We note that several terms are identical on both sides of the equals sign, and thus cancel, which leaves us with β AC + β AB + β BC β AB β BC β AC = β AC β AB β BC + β AB β BC β AC which can be rearranged to produce 22 AB BC AC AB BC AC + =+ . The left-hand side can be written as 2 β AC (1 + β AB β BC ) in which case it becomes clear how to obtain the result from part (a) [just divide both sides by 2 (1 + β AB β BC ) ]. 67. Interpreting v AB as the x -component of the velocity of A relative to B , and defining the corresponding speed parameter AB = v AB / c , then the result of part (a) is a straightforward
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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