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Current time:0:00Total duration:6:45

Video transcript

all right we can now do the math to solve for V so let me just simplify the right-hand side of this equation V - negative V well that's just going to be 2 V + 1 minus negative of V squared over C squared well that's just 1 plus positive V squared over C squared and let's see what can we do next well we can multiply we're going to want to solve for V we can multiply both sides of this equation by 1 plus V squared over C squared so let's do that 1 plus V squared over C squared on the right hand side those cancel on the left hand side we can distribute the 0.8 and we will get 0.8 or we can distribute the 0.8 C I should say 0.8 C plus 0.8 C V squared over C squared well the C of the numerator is going to cancel with one of these C's in the denominator so it's going to be plus 0.8 V squared over C and then is going to be equal to 2 V and remember we want to solve for V so we're setting up essentially a quadratic in V so let's now move let's now find ourselves give ourselves some real estate so let's go right over there let's subtract 2v from both sides and actually I'm going to write it in order of degree so on the left hand side I have 0.8 I'll write that over C V squared and then minus 2 V I just subtracted 2v from both sides plus 0.8 C plus 0.8 C is equal to 0 I guess we want to simplify a little bit we can multiply both sides by C I don't that'll well let me just go with that if we multiply both sides by C this will become 0.8 V squared minus 2 C V plus 0.8 C squared is equal to 0 and we could keep trying to algebraically manipulate this but we could just go straight to the quadratic formula here to solve 4v I'll do that in a different color just for kicks V is going to be equal to negative B so this right over here is B and write this down this is our B this is our a and this is our C we're talking about the coefficient or we're talking about the different variables using the quadratic formula so V is going to be negative B so the negative of negative 2 C so it's going to be 2 C plus or minus the square root of b squared so negative 2 C squared is positive 4 C squared minus 4 times a so minus 4 times 0.8 times C so times 0.8 C squared 0.8 C squared all of that over 2a so 0.8 times 2 is 1.6 let's see this is going to be equal to 2 C plus or minus the square root now let's see I can factor out a 4 C squared 4 C squared times 1 minus 0.8 times 0.8 is 0.6 for all ideas I factored out a 4 C squared from both terms of course all of that over 1.6 scroll down a little bit and this is going to be equal to it this is all algebra at this point to see plus or minus if I fact if I take the 4 C squared out of the radical it's going to be it's going to be 2 C so 2 C plus or minus 2 C times the square root of 1 minus 0.64 well that's 0.36 things are getting nicely simple now all over 1.26 the square root of 0.36 the principal root of that is 0.6 so that's 0.6 and move down a little bit so this is going to be equal to 2 C and I can I can see the end plus or minus two times zero point six is 1.2 C all of that over one point six so we've got two possible values for V what we can tell if we add if we do the plus up here you're going to end up with 3.2 C divided by one point six which is going to be a speed a velocity of greater than the speed of light so we can rule out the positive version of it so we know the answer is going to have to be the negative version of it so to see minus 1.2 C over one point six which is equal to 0.8 C over one point six well zero point eight is half of one point six so this is zero point five see that was really really really cool because what's what what do we know now we know now that if a in a frame of reference a feels like it is stationary we have A's friend moving with a relative velocity of eight tenths of the speed of light there could be a third party see that defines a frame of reference moving away from a with the velocity of half the speed of light half the speed of light and from a frame of reference it looks like C's velocity is closer to bees than it is to ace but we're dealing in this relativistic this this this this Einstein Ian's world because if we then go into C's frame of reference it actually looks like a and B are both leaving C with a speed of 0.5 C zero so here you can view the velocity is positive 0.5 C and here you could say the velocity is negative 0.5 negative 0.5 C this was really cool we're able to find a frame of reference that you could think of it is right in between and what's going to be really cool about this is that we can actually you know what what some people don't like about Minkowski space-time diagrams is that it looks asymmetric even though if B is moving in a stream of reference with a velocity of positive eight tenths of the speed of light well then if you were in B's frame of reference a is moving to the left with a speed of 0.8 times C but now we can define another I guess you could say a neutral frame of reference where they're both moving in different directions with the same speed which may actually makes the the the interpreting the space-time diagram a little bit easier we will do that in future videos but this is a fun problem in a it's in an on itself