# Relative Speed vs speed of light [duplicate]

I've got two questions which are related and I've always wondered about and got no answer for. (I'm not a physics student and know only basic physics, so the easier explanation the better)

1. Two cars traveling at 60km/h in opposite direction. The speed observed by an onlooker is 60 km/h for each car. If you are in the car, the observed speed of the other car is 120km/h. Correct?

We say that speed of light is the fastest you can travel. In the car example (ok, maybe spaceships to be a little more realistic) if they both travel in the speed of light or near. They will observe the speed of light * 2 Is this possible? or is the actual highest velocity speed of light /2

1. Similar reasoning, if you are in a train travelling in the speed of light, then you are unable to walk forward in that train, since you can't travel faster than the speed of light?
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## marked as duplicate by Qmechanic♦, Manishearth♦, John Rennie, dmckee♦Jul 16 '12 at 13:33

Possible duplicate: physics.stackexchange.com/q/7446/2451 – Qmechanic Jul 16 '12 at 8:45
youtube.com/watch?v=KZ0gQevrPTo – ZolaKt Jul 16 '12 at 10:32
I guess it somewhat answers my question. But I still have a problem really grasping the answer :) – Vixen Jul 16 '12 at 12:41

The non-mathematical answer is that, in our universe, it turns out that velocities don't add by simple (vector) addition. In your two-cars-scenario, your measured speed from one car (of the other car) is only approximately 120 km/h, although it's a very good approximation. John Rennie's possible duplicate in the comments of the OP gives the actual formula. When you get near the speed of light, the error in the "simple addition" method of summing velocities becomes large enough to be noticed.

As to your second question, it doesn't matter how fast the train is going (it'll have to be slightly less than light speed, but that's not important for our purposes), you can still walk forward. In the frame of the train (which your traveler is in), the train isn't moving at all. Someone outside the train will measure the velocity of the person walking on the train to be less than the speed of light, no matter what, because the correct formula for measuring velocities applies equally well in this case.

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Thanks, that was what I was expecting as an answer – Vixen Jul 16 '12 at 14:29
possible pre-duplicate with physics.stackexchange.com/questions/186418/… . I think that the answer is not complete. In the initial frame of the speeds ( observer standing on the road ) , you don't need special relativity. Transformations are needed only in the other frames ( the fast cars ). – igael May 28 '15 at 10:21