<p>blue light has higher frequency</p>
<p>Did we ever establish the answer to the relativistic motion one, with the .6cs?</p>
<p>i thought it was somewhere between .6c and c, cause objects were moving faster relatively to each other, but nothing can move faster than light (theoreticaly)</p>
<p>This test is a lot easier than the one I took in October.
I think the curve would just be the same to compensate for the level of difficulty.</p>
<p>For the chart one with heat, was the ans E?</p>
<p>^^^Yeah, it's between .6c and c.</p>
<p>it is less clear if u assume that the two objects at 0.6c are moving towards each other and have not met.</p>
<p>but if u assume they had met and pass each other after 1 second. it is undisputable that their distance apart is 1.2c.</p>
<p>so if i blast a beam of light at the object at that time from the other object and wait for the light to return in order to observe and measure it's speed, the light after 1 second will be at 1c and the object will be at 1.8c, after 2 seconds, at 2c and object at 2.4c. so after 3 seconds at 3c and the object will be at 3c. so it finally gets reflected. but after 3 seconds the the other object will be 4.8c away so its another catch-up game again. so after 4 seconds (from the start) its 1c, 5.4c. after 5 seconds, 2c, 6c, after 6 seconds 3c, 6.6c, after 7 seconds 4c 7.2c, after 8 5c 7.8c, after 9 6c 8.4 after 10 7c 9c after 11 8c 9.6c, after 12 9c 10.2c after 13, 10c 10.6c after 14, 11c 11.2c, after 15, 12c, 11.8c. so it takes somewhere between 14 to 15 seconds for the light to be reflected. so relatively speaking, i will observe that the object is moving much more quicker than light (though impossible) because light takes between 14 to 15 second to reflect back.</p>
<p>but question was what is observed speed. Everything we observe, is light, so how can we see something to move faster than light?</p>
<p>speed = distance over time.</p>
<p>so one way of observing speed for unrealistically fast objects is to shoot a damn light at it and wait for it to return.</p>
<p>Nothing can go faster than c due to "time dilation". We need to remember that everything is relative with respect to c. You can not just add up the individual velocity. And no, it is not possible to even "observe" that something is going quicker than the speed of light. While it is true that the objects will cover a certain distance in a certain amount of time, the "observed time" would be much higher due to time dilation (thus, yielding an observed velocity that is less than c).</p>
<p>Hmm I think the formula is this thinggg:</p>
<p>(u + v) / (1 + uv/c^2)</p>
<p>So... 1.2 / (1 + .36)
...Or ~0.88.</p>
<p>
<p>The simple addition of velocities does not, however, extend to light or even to objects moving at speeds close to that of light ... The correct, relativistic formula for the "addition" of velocities -- that is, the one that follows from the theory of relativity -- looks like this. Imagine that a reference frame, Sme, is moving with velocity u past you. Now if an object moves at velocity v (parallel to u), as measured by me in my reference frame, then its speed as measured by you would not be vyou = u + v, but would instead be:</p>
<p>vyou = (u + v) / (1 + uv/c^2)
</p>
<p>The only thing is that this formula applies to objects moving in the same direction as observed by a third person. So it's not the exact same thing. But the same principle should apply...? Since it doesn't really matter if the frame of reference is moving or not...?</p>
<p>Yeah, it's that thing.</p>
<p>please do enlighten me. time dilation was not part of my syllabus and i took my a levels 2 years ago (had conscription). can someone work out the answer? and give explanations. thank you.</p>
<p>That question would be stupid if the answer were 1.2c, so the answer is .6c to c.</p>
<p>when you got speed vs observed speed graph, its domain is <c</p>
<p>^
^
what were the options? i forgot.
poseur, i see what you mean by the "objects moving in the same direction as observed by a third person".
we agree that velocity is a not a scalar quantity. so u + v = 0.6 + - 0.6 = 0 so the formula no longer yields the answer.
and since velocity is moving in seperate directions, the denominator becomes 1 -0.36 which is less than 1. so anything divided by less than 1 gives u a "more than" per se.</p>
<p>c, .6c, 1.2c, between .6 and c and something else...
But yeah, the equation posted by Poseur is right (I just checked with my AP Physics book). The exactly velocity is .882c => the right answer is "between .6c and c".</p>
<p>QUOTE: Imagine that a reference frame, Sme, is moving with velocity u past you. Now if an object moves at velocity v (parallel to u), as measured by me in my reference frame, then its speed as measured by you would not be vyou = u + v, but would instead be:</p>
<p>vyou = (u + v) / (1 + uv/c^2)</p>
<p>Usme = 0.6
v= -1.2</p>
<p>so vyou = -0.6/(1+ (0.6*-1.2)/ 1) = -2.143 ??</p>
<p>lol.. the you and me thing is so confusing.</p>
<p>The equation posted by Poseur is that of relative velocity- the velocity measured by an object of another object moving relative to it (in this case, the spaceship and the particle). The velocity measured by the third observer for each of these objects is still .6c.</p>
<p>but question was about velocity of 2nd object seen by 1st object. So it's faster than .6 (seen by stationary observer).</p>