Calculate Distance Travelled by Einsteinian Meter Rod in 1.9245-09 Seconds

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The discussion centers on calculating the distance traveled by an Einsteinian meter rod moving at 0.5c in a specified time frame. The calculation presented indicates that the rod travels approximately 0.2887 meters in 1.9245 x 10^-9 seconds. There is confusion regarding the notation used for scientific values, particularly distinguishing between "6.66667-09" and "6.66667 x 10^-9," which represent different values. The importance of using consistent reference frames in relativistic calculations is emphasized, as well as the need to apply special relativity principles when dealing with high-speed scenarios. Overall, clarity in notation and understanding of relativistic effects are critical for accurate calculations.
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I have an Einsteinian meter rod traveling at .5c which is 1 meter per 6.6667-09 seconds.

How do I discover how far the rod will travel in 1.9245-09 seconds?
 
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(meters/second) X seconds = meters

so 1.9245/6.6667 =
 
Ivan Seeking said:
(meters/second) X seconds = meters

so 1.9245/6.6667 =
.288673557 meters in 1.9245-09 seconds.

Thanks, Ivan. I'm just beginning to calculate with sci notation and get dazzled by the new appearance of things (very slow brain for math).
 
Hold on for a moment: is that 6.66667-09 (which would be a peculiar calculation) or 6.66667x10-9 ("scientific notation"). Those are very different numbers!
 
HallsofIvy said:
Hold on for a moment: is that 6.66667-09 (which would be a peculiar calculation) or 6.66667x10-9 ("scientific notation"). Those are very different numbers!
A couple of years ago Edexcel wrote in the exams book that the gravitational constant = 6.67300 × 11-11
 
HallsofIvy said:
Hold on for a moment: is that 6.66667-09 (which would be a peculiar calculation) or 6.66667x10-9 ("scientific notation"). Those are very different numbers!
It would look like this written out:

.0000000066667

I've been using 6.66667-09 to indicate this in a thread and no one has said anything about it, although everyone else is writing it the way you do. I just figured they were interchangable.

If not, then I'm curious to find out what interesting thing I have been indicating without realizing it.
 
zoobyshoe said:
It would look like this written out:

.0000000066667

I've been using 6.66667-09 to indicate this in a thread and no one has said anything about it, although everyone else is writing it the way you do. I just figured they were interchangable.

If not, then I'm curious to find out what interesting thing I have been indicating without realizing it.

6.66667-09 ≈ 0.0000000384432

6.66667 x 10-09 = 6.66667e-09 = 0.00000000666667
 
zoobyshoe said:
It would look like this written out:

.0000000066667

I've been using 6.66667-09 to indicate this in a thread and no one has said anything about it, although everyone else is writing it the way you do. I just figured they were interchangable.

If not, then I'm curious to find out what interesting thing I have been indicating without realizing it.

You have been indicating the reciprocal of the 9th power of 20/3 (or 6.66667)...but that would be a very unusual representation, given the context.

So, everyone understood what you meant, or what you meant to mean.
 
I think you need to use SR in this question, since distance contracts when traveling at high speed.

d=vt

d\sqrt {1- \frac{(0.5c)^2}{c^2}}=(0.5c)(6.66667 x 10^(-9))

and solve for d from then on.
 
  • #10
Nenad said:
I think you need to use SR in this question, since distance contracts when traveling at high speed.

d=vt

d\sqrt {1- \frac{(0.5c)^2}{c^2}}=(0.5c)(6.66667 x 10^(-9))

and solve for d from then on.

Yes, using relativist speeds in this problem (to confuse the issue) didn't seem like a particularly good idea for someone struggling with basic scientific notation.

But if this is indeed meant to be a relativistic problem then it is inadaquately stated. The speed of 0.5c is measured relative to who? This time of 1.9245 x 10^(-09) is measured relative to who? If the speed, time and distance are all measured in a consistant reference frame then the distance traveled (in that same reference frame) will still be sucessfully computed by d = vt.
 
  • #11
uart said:
If the speed, time and distance are all measured in a consistant reference frame then the distance traveled (in that same reference frame) will still be sucessfully computed by d = vt.
I can state the same problem in terms of non-relativistic speeds:

If I have a dragonfly which is flying straight and steady at a rate of 1 meter per .0493 seconds, how do I determine how far it will fly in 1.9245x10-9 seconds?
 
  • #12
d= vt=1 1 m/(0.0493 sec)*1.9245x10-9= (1.9245/0.0493)x10-9= 25.4349 x10-9.

Since the "least accurate" number (0.0493) had 3 significant figures, I would write that as 2.54x10-8 meters.
 
  • #13
HallsofIvy said:
(1.9245/0.0493)x10-9
Thanks HallsofIvy. I'm interested in the way you expressed this, with the single exponent outside the parenthesis. It makes me wonder how you would put it if both had an exponent, say if .0493 had been 4.93x10-7.
 
  • #14
<br /> \frac{1.9245 \times 10^{-9}}{4.93 \times 10^{-7}}\ =<br /> \ \frac{1.9245}{4.93} \times \frac{10^{-9}}{10^{-7}}\ =<br /> \ (1.9245/4.93) \times 10^{-2}<br />
 
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