Solving for relativistic velocity using Newtonian physics

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gtguhoij
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Homework Statement
Here is another exercise focusing on how the distance, or gap, between two objects changes over time. If a taxi is racing toward you from the north at 3/4 light speed, and another taxi is racing toward you from the south at 1/2 light speed, how quickly from your perspective is each taxi approaching the other?
Relevant Equations
d = vt


The answer is tD = [D_0 - 3/4ct - 1/2ct] I just have 2 questions.

I realize for 2 vectors approaching it is negative for distance and for velocity positive. What be the rule for time? How do I find vector answers for velocity and distance and time?

I am confused why I have "td = ..." ? Can someone explain? "td" doesn't even make sense.
 
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Leave vectors aside. You are one single observer; for you, the “mutual “ speed (wording used by W. Rindler in his book on relativity) by which the given distance between cars ( consider them as material points) decreases is simply :

##v = 0.75c +0.5c =1.25c##

no wonder that this result is greater that ##c##. No relativistic composition of speeds is to be made here. This is the answer to “how quickly...”.
So, if you are given the initial distance of the cars , simply divide it by that mutual speed, and this is the time for them to crash into each other, according to your wrist watch.
Be careful, don’t stay on their trajectory, if you don’t want to be reduced into a bloody steak.
But the point where they meet isn’t probably your original position .
 
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I ‘ ll add another clarification. The relativistic composition of velocities applies here if taxi driver A wants to know what is the speed of taxi B relative to him. The result is less than ##c##, as can be easily verified: 0.909c (approx).
Which is the wrist watch time for collision, for both A and B?
Determine the ##\gamma ## factor using the speed just found .
Determine the contracted distance ##\frac{D}{\gamma}## between cars.
Divide this contracted distance by the relativistic speed, and you are done.
 
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