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[Quantum Mechanics]Uncertainty principle

  1. Mar 11, 2005 #1
    I have a question for Heisenberg's uncertainty principle. Here is the question:
    You are driving your car when you race thru a stop sign and get subsequently pulled over. The cop measured your speed within a meter of the stop sign. You decide to use the uncertainty principle as your defense, claiming that your measured speed differed from zero by less than the uncertainty principle would allow for. Therefore, your speed could well have been zero and you should not be ticketed. How fast could you have gone for this to be a valid defense? (Mass of the car=1000kg, h=6x10^-34)
    I used m delta(v) x delta(s) >or= h/(4pi)
    solving for delta(v), i got delta(v) >= 4.77x10^-38 m/s. What does this value represent? How do I go from here to find the maximum speed the car could have gone under the valid defense condition? I guess I am not exactly grasping the meaning of this equation. Please let me know.
    Also, another question:
    Not having evaded a ticket in the past problem, you switch to a super-light quantum car wtiht eh mass of an electron. The cop now measures your speed when you are within a micron of the stop sign. How fast can you go now, in km/hr? (mass of electron = 10^-30kg, 1 micron = 10^-6m)
    Again, I used the uncertainty principle equation to solve for delta(v), and I got delta(v) >= 1.33x10^-5 km/hr. Does this mean that the maximum is 1.33x10^-5km/hr? Please please help this stupid person out. Thanx :smile: :!!)
  2. jcsd
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