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One more

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A muon (an elementary particle) enters a region with a speed of 4.20 106 m/s and then is slowed at the rate of 1.10 1014 m/s2.
    (a) How far does the muon take to stop?
    ________ m


    2. Relevant equations

    x(t)= initial position + final velocity * time

    v(t)= (acceleration * time) + initial velocity

    x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position

    x= initial position * (average velocity * time)

    average velocity= (final velocity - initial velocity) / (2)

    (final velocity^2) - (initial velocity^2) = 2 * acceleration * change in position




    3. The attempt at a solution

    I came out with this as my seconds, t, .00000003818181818. Is this correct for the t value and where should I put this in at?
     
  2. jcsd
  3. Feb 5, 2009 #2

    Nabeshin

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    Your time looks to be about correct. You can substitute it into your equation number three.

    Alternatively, you can solve this one without time with another one of the equations you mentioned =)
     
  4. Feb 5, 2009 #3
    Some of the equations you typed are wrong. Acceleration and initial velocity are given. What is the final velocity of the particle (hint: read the problem)? Which kinematical equation should be used to calculate the distance traveled when acceleration and the velocities are known?
     
  5. Feb 5, 2009 #4
    By using the last equation I listed, I came out with -1.90909091 x 10^-8. This doesn't seem correct? Is this value correct and should it be a positive number?
     
  6. Feb 5, 2009 #5

    Nabeshin

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    Umm, no that is not correct. Perhaps you substituted something in wrong?

    As far as signs are concerned remember acceleration opposes the velocity, so its sign is negative.
     
  7. Feb 5, 2009 #6
    Hummm... this is what I did:

    0^2 - (4.20 x 10^6)^2 = 2 * (-1.1 x 10^14)* change in X

    {4200000/((-1.4 x 10^14) * 2)}

    Change in X = -1.90909091 x 10^ -8.

    Maybe I entered it into my calculator incorrectly?
     
  8. Feb 5, 2009 #7

    Nabeshin

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    Looks like you forgot to square the initial velocity. You have the ^2 written in your first step, but you drop it in the 2nd. Also, you dropped the - sign from steps 1 to 2.

    Looks like those are your problems,
    Cheers!
     
  9. Feb 5, 2009 #8
    Let's try this again.....


    -.0801818182?
     
  10. Feb 5, 2009 #9

    Nabeshin

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    Again, you forgot the sign, but yes.
     
  11. Feb 5, 2009 #10
    +.0801818182 m Thanks so much for your help!!
     
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