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Find acceleration given initial and final velocity, and displacement

  1. Jul 28, 2013 #1
    1. An electron in the cathode ray tube of a television set enters a region where it accelerates uniformly from a speed of 62300 m/s to a speed of 3.86 × 106 m/s in a distance of 3.17 cm.
    What is its acceleration?

    2. a = [Vf2 - Vi2] / [2d]

    3. a = [(3.86 × 106)m/s2 - 62300m/s2] / [2 * 0.0317m]
    I'm getting the answer 2.35(rounded to hundredth place)*1014m/s2 , but it doesn't seem to be correct when I put that answer on my homework. Please help.
    Last edited: Jul 29, 2013
  2. jcsd
  3. Jul 29, 2013 #2
    Acceleration is rate of change of velocity. It's (vf-vi)/t . Don't square the velocities. And don't divide by the distance.

    More applicable formula here would be
    v^2=u^2 + 2as (s is the distance)
  4. Jul 29, 2013 #3
    Wouldn't you still divide by (2s) anyways based on algebra. To get (a) alone you would subtract u^2 from both sides of the equation and have v^2-u^2= 2as. Then you would divide both sides by (2s) to get (a) alone. Isn't that basically the same as the original equation I had? I'm also confused about what you mean by not squaring the velocities.
    Last edited: Jul 29, 2013
  5. Jul 29, 2013 #4


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    The result is correct.

  6. Jul 29, 2013 #5
    That's what I thought too, I guess the only explanation to why I'm getting this wrong on my homework website is that I am entering it in an incorrect format.
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