00:01
So here we have a diagram in the situation.
00:04
S prime is moving away from observer in s at a speed of u is equal to 0 .6 times c relative to an observer in the stationary frame at s.
00:18
A particle is moving to the right and away from observer in the frame s prime at speed b prime.
00:28
So we need to calculate what speed is observed by the observer in the rest frame and that speed is v for three different cases.
00:39
So for the first case, and in this case, since the motion is all in the positive x direction, we let vx prime just be v prime and vx just be v.
00:52
So firstly, we can decalculate v and we know v from the lorence transformation is v prime, plus.
01:02
U all over 1 plus u times v prime over c squared and these values are all known so v prime is given to be 0 .4c and u we know it's 0 .6 times c so we would assume by looking at these values that v should be the speed of light we'll find out now that's over 1 plus u times v prime over c squared is just u 0 .4 without the c times 0 .6 without the c since the c's cancelled and we get the observed speed by someone in frame s of the particle moving would speed v prime in in frame s prime 0 .806 times c so it's actually less than the speed of life.
02:34
Next, we'll do the same, use the same equation, and find v for a different value of v prime...