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Work debate

  1. Sep 16, 2005 #1
    disclaimer: we're not studying physics nor do we remember very much about highschool/college physics so please dont flame if this is a dumb question.

    Does it take the same amount of time for a car traveling at 65mph to pass a car going 60mph, as it does for the same car traveling at 5mph to pass the other car while stationary?

    If yes, why? If no, why?

    Thanks very much we appreciate you guys who spend the time to fully answer this for us. (try to keep the explanation simple) :rofl:

    - Cam
  2. jcsd
  3. Sep 16, 2005 #2


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    Hey cam, must be a watercooler question. Think of it this way, the person in the slower car that's being passed is watching the other car pass by at 5 mph in both cases. So relative to the slower moving (or not moving) car, the other car is going the same speed and passing takes the same amount of time.

    Mathematically, the time it takes to go a distance is:

    d = v t
    where d = distance
    v = velocity
    t = time

    So if the velocity difference is the same for both cars (5 mph difference) then given some time t, the distance the other vehicle travels with respect to the slower moving car is the same.

    PS: Hope you won your bet! LOL
  4. Sep 16, 2005 #3


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    it depends on how "anal" you are (and you can't spell "analysis" without "anal"). from a classical mechanics POV, they're the same as Q has said above. from a POV of relativistic physics, not quite the same, but the difference would be unmeasurable.
    Last edited: Sep 16, 2005
  5. Sep 16, 2005 #4
    Actually, according to Special Relativity, there is an ever-so-slight time difference. If the passing time in the reference frame of the stationary car is given by [itex]t_0[/itex], then according to an observer on the ground in the 60/65 mph case, the passing time is given by

    [tex]t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    (where c is the speed of light). However, in this case, v^2/c^2 is about 9.4*10^-15. For small values of this quantity, the above equation is approximately

    [tex]t = t_0 \left ( 1+ \frac{v^2}{2c^2} \right )[/tex],

    and the percent difference between the two is [itex]50 \frac{v^2}{c^2}[/itex], or about 4.967*10^-13 %. Therefore, the difference is so small that it can't be measured, and is probably overcome by thermal effects. So, technically, it would seem that no one should win.
    Last edited: Sep 16, 2005
  6. Sep 17, 2005 #5
    Actually, there is no difference at all.

    In relativity there is no third observer. The time as measured by the occupant
    of either car would be the same in the two situations.
  7. Sep 17, 2005 #6


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    space and time aren't consistant and absolute really. Classical mechanics summed things up nicely and made sense. It was like a machine. And as far as you're concerned cam, at such low speed with such huge objects, there is no difference between the two.

    In the last hundred years, however, it's pretty much been determined that on a super tiny scale, there isn't so much of a sensible, concrete way things work. In fact, particle physics is determined purely by statistical analysis (doing experiments over and over and over and over and over again and forming a table of what particles are most likely to do in certain situations). That is, we only know the tendencies of particles. Not only that, but (as I opened with) space and time aren't consistant. Space is kind of stretched (inconsistantly) by gravity, depending on different densities around the Earth, and time is so bizarre that I don't understand it well enough to explain it.

    So while you and your car, and they and their car have a 5mph difference in both cases, the particles that make you and your cars up aren't behaving in such a consistant way, and the space that your traveling through may be warped in (very slightly!) different ways by the mass and geometry of the Earth.

    disclaimer: i've only completed two semesters of classic physics courses. I'm only two weeks into modern elementary physics.
  8. Sep 17, 2005 #7
    It's funny watching you guys take a simple regular old laymen's question and turn it into a spur of special relativity and quantum mechanics :D
  9. Sep 17, 2005 #8


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    modern physics rocks
  10. Sep 17, 2005 #9


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    In this situation, there is a third observer - the ground. And speeds are measured in the ground's frame of reference (the spedometer is mechanically linked to the wheels (or, at least, they used to be)).
  11. Sep 19, 2005 #10
  12. Sep 19, 2005 #11


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    Interesting forum. Sorta like a cultural cesspool.
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