Re: Is the speed of light really constant ?

From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 09/25/04 

Date: Sat, 25 Sep 2004 18:31:51 GMT

"Pax" <pax1@whitesweb.com> wrote in message
news:rEe5d.987$zc1.913@newssvr12.news.prodigy.com...

"Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message
news:%rc5d.968$ED.330@text.news.blueyonder.co.uk...
>
> "Pax" <pax1@whitesweb.com> wrote in message
> news:ZVa5d.978$zc1.409@newssvr12.news.prodigy.com...
>
> You seem to have failed the challenge and are not believed.

It was a challenge then, was it? Very well.

"Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message
news:6pN4d.425$ED.183@text.news.blueyonder.co.uk...
>

A car leaves San Francisco and travels Interstate 80 to New York at a
constant velocity v1 is passed by another car that leaves San Francisco an
hour later, also travelling at a constant speed v2. How fast are the cars
travelling if they meet 1200 miles from San Francisco, and for how long has
each been travelling?

e) Other - specify.
60mph for 20 hours and 63.158mph for 19 hours. (2nd velocity actually
computes to 1200.002 miles)

Bonus point:
How would you display this on a distance/time graph?

Oh, goodie!

x = 1st car with v1 = 60mph (1200 miles)
y = 2nd car with v2 = 63.158 mph (1200.002 miles)
z = meeting point of x and y

D = Distance (the verticle, progresses from bottom to top, S to N)
T = Time (the horizontal, progresses from left to right, W to E)

D 1200 miles (marked in 20 segments of 60 miles each)
^______________________________________________________z
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
x__y___________________________________________________T 20 hours
 | | | | | | | | | | | | | | |
| | | | | |
                             Time (marked in hours) ------------------->

Now it's your turn.

Ok... Correct diagram (well done), and the correct answer was indeed e), but
unfortunately you gave a specific example so similar to my suggested answers
as makes no difference.
> a) 600 mph for 2 hours and 1200 mph for 1 hour
| > b) 60 mph for 20 hours and 63 mph for 19 hours
| > c) 30 mph for 40 hours and 30.77 mph for 39 hours
| > d) 1200 mph for 1 hour and 'c' for 0 hours.
| > e) Other - specify.
If we take my rounding into consideration (and you also have rounded, giving
1200.002 miles), then rounding IS acceptable, therefore the answer you gave
is b), and not e).
Why would you choose b) in preference to c) ?
The correct answer is calculated as follows.
1200 = v1 * t (You have chosen v1 = 60 and t = 20)
1200 = v2 * (t-1) ( 63.158 * 19 = 1200.002, rounded to 1200)
v1*t = v2 *(t-1)
v2 = v1*t /(t-1)

Hence the correct answer is:
e) v1 mph for t hours and v1*t/(t-1) mph for t-1 hours

Sanity check:
Let v1 = 1
Time to travel 1200 miles = 1200 hours = 1200/24 = 50 days
v2 = v1*t/(t-1) = 1200/1199 = 1.0008340283569641367806505421184 mph
(rounding by my calculator).

Since v2 is marginally greater than v1 the answer appears to be correct.

Considering the diagram above as marking 3 sides of a square sans its
eastern, are you capable of visualizing two diagonal lines, one beginning at
the point indicated by x (located at the SW corner of the diagram),
traveling upward at a 45 degree angle and terminating at z (located at the
NE corner of the diagram), and another beginning at the point indicated by y
(located at the SW corner of the diagram + 1), traveling upward at
(approximately) a
1. ___?___ degree angle and terminating at z (still located at the NE corner
of the diagram)?

Yes I can, that is what I have drawn at
  http://www.androc1es.pwp.blueyonder.co.uk/SekerinTime.htm
where cars are continually released at 1 hour intervals,

and at
  http://www.androc1es.pwp.blueyonder.co.uk/actual_data.htm
If you look very carefully, you'll see an orange band crossing over the
yellow
at about half distance. This is where the later light (or second car if you
like)
has passed the earlier slower light (or first car) on it's way to the finish
line.
The later light arrives BEFORE the earlier light. Hence the very real
observation
by the amateur astronomer of the light curve of V1493 Aql at the top of the
page,
reproduced by my program, the code for which may be found at
 http://www.androc1es.pwp.blueyonder.co.uk/C_Program_for_Copernicus.htm
and an executable copy of which is available at
 http://www.androc1es.pwp.blueyonder.co.uk/virus_alert.htm

2. Will the angle Tyz, representing the 2nd car, be wider or more narrow
than the angle Txz, representing the 1st car?
It will be wider.

3. Is the Pythagorean Theorem helpful for finding any of these angles?

Hmm... Not really.
The angle A = zT/xT radians, or zT/xT * 360/2pi degrees.
Why are you interested in the angle?

4. Employing the Pythagorean Theorem, what is the approximate length, in
number of segments, of the hypotenuse of the right triangle Tyz?

Who cares?
sqrt( zT ^2 + xT ^2), but so what?

5. What degree of angle is bounded at point z of the right triangle Tyz?
Who cares about the angle?

6. Bonus question: Is the last angle more or less acute that the one found
at the top of your head?

I'm not wearing a dunce's cap, but your's appears to be firmly in place.

There are no points to be won, there is no bonus point if you answer the
last question correctly, no penalty if you fail, since I'm not grading you
and really don't care if you get any of it right or not.

And I don't really care if you have more interest in insult than physics
either, but
your response "Oh goodie" above betrays your supposed lack of interest as
FALSE.

Androcles.

Oh, well - Pax

.~*~._.~*~._.~*~._.~*~._.~*~._.~*~._.~*~._.~*~._.~*~.

May people say of you:
"The world is a better place because you are in it."

>From Andromeda:
"Dillon Hunt, there are three kinds of people in this
universe, those who can count, and those who can't."

1. 46.4688 degrees
2. wider
3. yes
4. approximately 27.58623 segments
5. 43.5312 degrees
6. less

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