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Simple acceleration puzzle I need help with,I forgot how to do.

  1. May 30, 2007 #1

    inv

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    1. The problem statement, all variables and given/known data
    Two markers M1 & M2're set up a vertical distance h apart.
    [​IMG]
    When a steel ball is released from rest from a pt a distance x above M1,it's found that the ball takes time t1 to reach M1 and time t2 to reach M2.
    Which expression gives the acceleration of the ball?
    A.Wrong,B.Wrong,C.Wrong,D.Correct - >[tex]\frac{2h}{t_{2}^2-t_{1}^2}[/tex]
    Was wondering how to get that expression,I've done it b4 but forgot I'm afraid,pls guide?


    2. Relevant equations
    Acceleration is the rate of change of velocity,ie (v-u)/t where v is final velocity,u is initial velocity,t is time taken for the change in velocity to occur.


    3. The attempt at a solution

    Well,I've [tex]\frac{h+x}{t_{2}}[/tex] as v
    [tex]\frac{x}{t_{1}}[/tex] as u
    So I do (v-u)/t but I get something far from what the correct answer is ie
    [tex]\frac{t_{1}(h+x)-xt_{2}}{t_{2}t_{1}(t_{2}-t_{1})}[/tex]
     
    Last edited: May 30, 2007
  2. jcsd
  3. May 30, 2007 #2

    andrevdh

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    Homework Helper

    Use

    [tex]s = ut + \frac{1}{2}gt^2[/tex]
     
  4. May 30, 2007 #3

    inv

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    Solved*!Tq for answering!
     
    Last edited: May 30, 2007
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