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Constant acceleration muon problem

  1. Sep 9, 2012 #1
    A muon enters a region with a speed of 5.00 x 10^6 m/s and then is slowed at a rate of 1.25 x 10^14 m/s^2. A) How far does the muon take to stop?

    After some research on the internet I have found an answer to the problem but that does not help me enough. Here is what I have found:

    Im suppose to use this equation: V^2 = Vo^2 + 2a(X-Xo)

    As im sure you can see however I don't know what these symbols in this equation mean. In my book it has the second Vo^2 part with the 2 directly over the o which says to me that it's not saying the square of o. So that confuses me. Secondly here is the answer I found:

    x = -1/2(Vo^2/a) = -1/2((5.00 x 10^6)^(2)/(-1.25 x 10^14))

    This just confuses me further because they didn't show the algebra steps which I kind of need in order to understand how they got the solution. If someone could please help explain this equation to me and show me the steps in the algebra that would be amazing. I mean knowing the answer isn't going to help me on the test.
     
  2. jcsd
  3. Sep 9, 2012 #2
    First of all, how long would it take for a particle going initially with speed v to stop under constant deceleration a? Express t in terms of a and v.

    Then, when you know a, v and t, what is the distance the particle covers?
     
  4. Sep 9, 2012 #3

    Doc Al

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    Staff: Mentor

    The '0' is just a subscript that stands for the initial speed or position at the start of the motion (t = 0).

    So V0 is the same as Vi. And V is the final speed at the end of the motion.

    Another way to write that standard kinematic formula is:
    [tex]V_f^2 = V_i^2 + 2a\Delta x[/tex]

    Since the final speed is 0 ([itex]V_f = 0[/itex]), how might you rewrite the equation and solve for Δx?
     
  5. Sep 9, 2012 #4

    Doc Al

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    FYI, that equation is one of a small set of kinematic equations that you would be wise to become very familiar with. Here's a list: Basic Equations of 1-D Kinematics
     
  6. Sep 9, 2012 #5
    Well if I set the final velocity to 0 then divide by the change in x i would get:

    Change in x = velocity initial^2 + 2a

    but if i plug in the values in at this point i get 2.75x10^14
     
    Last edited: Sep 9, 2012
  7. Sep 9, 2012 #6

    Doc Al

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    No you wouldn't. You'd start with:
    [tex]0 = V_i^2 + 2a\Delta x[/tex]
    And then:
    [tex]- V_i^2 = 2a\Delta x[/tex]
    You do the rest.
     
  8. Sep 9, 2012 #7
    Oh I get it sorry messed up so now I end up with -Vo^2/2a or -1/2(Vo^2/a) right?

    So I think my biggest problem with these types of equations is just going to be figuring out what variables I have and which ones I do not have, any tips?
     
  9. Sep 10, 2012 #8

    Doc Al

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    Right. And remember that since your acceleration is negative (it's slowing down) your answer for Δx will be positive.
    One thing I can recommend is to do as many practice problems as possible. That way you will get intimately familiar with the kinematics equations and how they are used.

    And it would probably be wise to review your algebra as well.
     
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