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Homework Help: Equations of motion problem

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    An air track glider is placed on a linear air-track which is slightly tilted. It is given a velocity of 1.5 metres per second up the track. If its acceleration is 2 metres per second squared down the track, find the time at which it is 1m below its starting point.

    Please explain how to solve this problem. Thanks

    2. Relevant equations



    3. The attempt at a solution

    s=(u+v)/2 xt


    1= (1.5+v)/2 x t

    t=1-(1.5+v)/2
     
    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 26, 2010 #2
    try using the equation final position= initial position + (initial velocity)(time) +1/2(acceleration)(time)^2. also try making the final position negative because it is below the starting point
     
  4. Sep 26, 2010 #3
    Thanks for the help. I have got to t+1/t=1.5
    Can you help me from here to find t.
     
  5. Sep 26, 2010 #4
    if i did the problem, it would become -1m= 0m + 1.5t + (1/2)2t^2. by moving the -1 over to the other side, i would use the quadratic formula dueto the equation becoming, t^2 +1.5t+1. (i just re-arranged it after moving the one over.
     
  6. Sep 26, 2010 #5
    I have tried to use the quadratic formula but there is a negative discriminant so so it won't work. Are there other ways to solve the equation?
     
  7. Sep 26, 2010 #6
    i see. the acceleration is negative. i did not see that. so the equation should be -(t)^2 +1.5t +1 . quadratic should give you two answers, one of which should work.
     
  8. Sep 26, 2010 #7
    Hi, if my understanding towards your question is correct, that the initial velocity of the glider is 1.5 m s-1 upwards, acceleration is 2 ms-2 downwards, and that you need to find the time at which displacement is 1m downwards,

    then I'll be using the same method as jmb88korean:

    s = ut + 0.5at2

    1 = 1.5t + 0.5(2)t2
    t2 - 1.5t -1 = 0
    t = ( -(-)1.5 +/- root(1.52 + 4)) / 2
    t = 2 or -0.5

    Since by convention, there is no such thing is negative t, we take t = 2.
     
  9. Sep 26, 2010 #8
    I calculated the same answers as Ambidext when I did the problem.
     
  10. Sep 26, 2010 #9
    Thankyou jmb88korean and Ambidext for the help, I agree that t=2 is the answer.
     
  11. Sep 26, 2010 #10
    Actually, mathematically, we know how to omit -0.5.

    Can anyone explain though, why, in newtonian mechanics concept, WHY the -0.5 is irrelevant?
     
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