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Magnitude of Acceleration

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data
    electron starting from rest and moving with constant acceleration travels 4.7cm in 8.2ms. What is the magnitude of this acceleration? Answer in km/s^2

    2. Relevant equations
    X=Xi + (Vi)t + 1/2at^2

    3. The attempt at a solution
    Vi = 0 so the equation becomes X = Xi + 1/2at^2
    t = 8.2ms and X = 4.7 so 4.7 = Xi + 1/2a(8.2)^2

    I feel like I did everything right up to this point but all I need is acceleration to plug in but I'm not sure how to get that.
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 7, 2010 #2


    User Avatar
    Homework Helper
    Gold Member

    You do not need to plug in acceleration. You are trying to solve for a.

    Thus, what you need to plug in is xi, the starting position of the particle.
  4. Sep 8, 2010 #3
    4.7cm = .000007km and 8.2ms = .0082 seconds

    electron is from rest so Vi=0 so the equation is 2(X-Xi)/t^2, I have time and I have X which equals .000047km. How to I find Xi, or is that also zero?
    Last edited: Sep 8, 2010
  5. Sep 8, 2010 #4
    problem statement says that: "starting from rest", which means that initial velocity is zero. For initial position Xi you can chose whichever value you like. What is given - the distance moved, that is X-Xi. You can chose X = 4.7 cm only if your Xi = 0.
  6. Sep 8, 2010 #5
    thanks guys, yeah I got it, it was 1.398 km/s^2
  7. Sep 8, 2010 #6
    your welcome
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