Two vehicles travel towards each other, one at 75% of the speed of light, the other at 50% the speed of light?

Question

How would they percieve each other. The combined relative speed would be greater than the speed of light. Would they meet or pass each other before seeing each other?

Answer 1

The tau factors are:
1 / sqrt( 1 - 0.75^2) = 1.511858
1 / sqrt( 1 - 0.50^2) = 1.154701

These sum to 2.666558, which works out to
1 / sqrt( 1 - 0.9270187^2) = 2.666558

So the apparent combined speed is 92.7% of light speed.

They would see each other, of course. Since each vehicle is traveling slower than light, the light from each goes out ahead of each vehicle. The other vehicle would be able to see this light.

Answer 2

The main principle of relativity is you can't tell if you're moving smoothly without looking outside.

Based on that, the drivers should be able to see normally even moving at the speed of light. Because if your image disappeared when you were moving at the speed of light, you could tell you were moving at the speed of light just by looking in a mirror, right? You wouldn't need to look outside, right? Which would violate the principle of relativity.

That's half the problem solved. Next, speed is distance divided by time (as in miles/hour). So if speed were to be the same then the distance and time would have to be different. Which meant that there must be something suspect with time.

So a moving observer and a stationary observer would observe different times.

What Albert Einstein said (from the above) is that light propagates the same for everyone. So if you stand still or move, light is always relative to you, i.e it still moves 186,000 miles per second.

What these two drivers are doing is increasing their energy and mass and the linkage between these two is 'c', the speed of light. So as their energy and mass increase towards the speed of light, time to them slows. They will still see images at the speed of light.

Answer 3

The speed of light is a constant regardless of the motion of the observer;
I.e an astronaut travelling at 50 % of the speed of light would still observe the other vehicle as travelling at 75% of the speed of light relative to him; The displacement with respect to velocity is not the issue - spatial displacement is observed and temporal (time) displacement is observed all other factors are constant. No need to do tricky mathematical equations to derive this - it has already been determined by einstein and the scientific community for almost a century.

Answer 4

i dont really understand all this working out of number because im not exactly an expert...but i would have thought by what i already know about relativity and things(not a lot) that both cars would indeed see each other because neither are going at the speed of light however their combined speed would equal greater than the speed of lightwouldnt they percieve each other at slower rates...but at equal ones because its their combined speed which equals more than the speed of light?

Answer 5

The speed with which the distance between the
ships shrinks from the point of view of
"motionless observer" is indeed v1 + v2 = 1.25c,
that is greater than c.

The relative speed of the spaceships with respect
to each other, as seen by their crews is not simply
v1 + v2. The formula is more delicate and is in fact
v = (v1 + v2)/(1 + v1v2/c²) =
(3c/4 + c/2) / (1 + 3/4 * 1/2) =
(5c/4) / (11/8) =
10/11c ~ 0.909 c

This formula can be easily derived using nothig
else than simple geometry.
Draw a line a and choose points -c, -3c/4, 0, +c/2 +c on it.
Choose any projection center point A.
Draw three lines u(A,-c), v(A,+c/2), w(A, +c).
Now choose another line b carefully, such that
b is divided in two equal parts by u,v, and w at points
|UV| = |VW|.

The projection Z of point -3c/4 from point A onto line b
represents velocity of the -3/4 ship with respect to
+c/2 ship represented by point V, that is v = |VZ|/|VU| c.

Answer 6

According to relativity nothing can travel faster than light. As the two vehicles approached each other they would perceive each other as travelling at no more than the speed of light. If they collided they would be turned into elementary particles.

Answer 7

No. The time derivative of their separation would exceed speed of light c to one at rest, but that's not any one object's velocity, so there is no violation of special relativity. In the reference frame either observer, the other ship would be approaching at a speed less c. I won't do the Lorentz transform calculation, but it would be something like 85 to 90 % of c. Keep in mind that relativity plays strange games with space and time.

Answer 8

hi Pete - the concern is which you replaced your inertial physique interior the midst of the question. You began via asserting that each and every traveller replaced into shifting at 0.9c in opposite guidelines - yet relative to what inertial physique? then you fairly stated that the inertial physique replaced into linked to traveller A. as quickly as you probably did that, the courting between the two visitors replaced, given so you might state relative velocities in terms of one physique of reference. once you do this and persist with the Lorentz remodel equations as defined above, you detect that no merchandise can attain the cost of light relative to a distinctive observer in any physique of reference. issues like measured term and measured length interior the inertial physique exchange as a manner to make this all genuine.

Answer 9

I don't know the answer to this question, but was wondering how fast the speed cop would have to go in order to catch them ?

Answer 10

Pas mal pour apprendre l'anglais !
En progrès je dois dire ...