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Mechanics 3 - Further kinematics

  1. Mar 3, 2006 #1
    I'm studying M3 as one of the three modules for my further maths alevel, and have just started the first chapter. I'm fine with acceleration as a funtion of time, and integrating to find velocity and displacement, and most of acceleration as a function of displacement, but have come stuck on one of the questions:-

    A particle p moves along the positive x axis. Its acceleration is (x + 3)ms^-2 when its displacement from the origin O is x meters. Given that initially, when t = 0, the velocity of P is 3ms^-1 in the direction Ox and x = 0. Obtain:-
    a) the speed Vms^-1 of P as a funtction of x
    b) x as a function of t


    I started by trying (1/2)v^2 = INT(X+3)
    therefore getting
    v^2 = x^2 + 6x + C
    where C = 9
    therefore:
    v^2 = x^2 + 6x + 9

    The book's answer however is different, v = x + 3
    I dont know how they got that answer, and as for part b) i'm completely lost as i cant figure out how to bring t into the equation.

    Any help would be greatly apprectiated,
    Thanks,
    Bob
     
  2. jcsd
  3. Mar 3, 2006 #2

    qtp

    User Avatar

    how can the velocity = x+3 if the acceleration is x+3 ?
     
  4. Mar 3, 2006 #3
    I don't know, that's the books answer. It could be wrong though. I know a few were in M1.
     
  5. Mar 3, 2006 #4
    You can find da/dx.
    da/dx*v = da/dx*dx/dt = da/dt.
     
  6. Mar 3, 2006 #5
    Your answer is not differant anyway. v^2 = x^2 + 6x + 9 = (x + 3)^2.
    Thus v = x + 3. So why should you say that your answer is wrong according to the book? Use the same concept I have shown dor the second question to solve the first question. It is tricky but solve it properly.
     
  7. Mar 3, 2006 #6
    Ah of course. Silly mistake there, just wasnt thinking. Thank you.
     
  8. Mar 27, 2009 #7
    Here is the answer:

    (a) d/dx(0.5v2) = ( x + 3 )
    => v2/2 = x2/2 + 3x + C
    => v2 = x2 + 6x + C

    when t=0, x=0 and v=3
    Therefore, c = 9

    v2 = x2 + 6x + 9
    => v2 = (x+3)2
    => v = (x+3)

    (b) v = x+3
    a = x+3

    a = dv/dx * dx/dt
    => (x+3) = 1 * (dx/dt)
    Integrating both sides.... ( replace "{" as integration sign... (im a noob in this forum))

    => {dx/x+3 = { dt
    => ln|x+3| = t + c .eqn 1

    when t=0, x=0

    => c = ln3

    therefore... from eqn 1...

    => ln|x+3| = t + ln3
    => ln|x+3| - ln3 = t
    => ln|(x+3)/3| = t
    => (x+3)/3 = expt
    => x + 3 = 3expt
    => x = 3expt - 3
     
    Last edited: Mar 27, 2009
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