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    A particle starts with speed 20ms^-1 and moves in a straight line. The particle is subject to a retardation which is inititally 5ms^-2 and which increases uniformly with the distance moved, having a value of 11ms^-2 when the particle has moved 12m. Given that the particle has speed vms^-1 and has moved a distance of x, show that, while the particle is in motion

    v.dv/dx = -(5+x/2) <=== easy peasy

    Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

    Answer is 20m and I dont get that answer :'(
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    I also have another question.

    A rock climber of mass 80kg is attached to one end of a flexible rope of modulus 45000N and length 20m. She slips from an overhang which is 10m vertically below the point to which the other end of the rope is attached.
    Calculate: the speed of the climber when the rope becomes taught (14ms^-1 and easy )
    the maximum distance of the climber below the point where she slipped. 13m
    the time from slipping to reaching her lowest point. 1.75s
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    v.dv/dx = -(5+x/2)
    ∫ v dv = - ∫ 5 + 0.5x dx
    0.5v² = -5x - 0.25x² + C
    v=20 when x=0:
    200 = 0 + C
    So:
    0.5v² = -5x - 0.25x² + 200
    v=0:
    -0.25x² - 5x + 200 = 0
    x² + 20x - 800 = 0
    (x-20)(x+40) = 0
    x=20m
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    (Original post by dvs)
    v.dv/dx = -(5+x/2)
    ∫ v dv = - ∫ 5 + 0.5x dx
    0.5v² = -5x - 0.25x² + C
    v=20 when x=0:
    200 = 0 + C
    So:
    0.5v² = -5x - 0.25x² + 200
    v=0:
    -0.25x² - 5x + 200 = 0
    x² + 20x - 800 = 0
    (x-20)(x+40) = 0
    x=20m

    tx..:'(

    ∫ 0.5x dx does not equal x^2........but x^2/4
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    (Original post by lgs98jonee)
    A rock climber of mass 80kg is attached to one end of a flexible rope of modulus 45000N and length 20m. She slips from an overhang which is 10m vertically below the point to which the other end of the rope is attached.
    Calculate: the speed of the climber when the rope becomes taught (14ms^-1 and easy )
    the maximum distance of the climber below the point where she slipped. 13m
    When she's at the point when the string is taught:
    KE=0.5mv²=0.5(80)(14)²=7840
    PE=mgh=80(9.8)x=784x
    EPE=0

    When she's at the lowest point:
    KE=0
    PE=0
    EPE=(X/2l)e²=1125x²

    Conservation of energy gives:
    1125x²-784x-7840=0
    x=3.01m [Using the quad. formula]

    You want the distance from the point she fell, so add this to 10 (since that's when the string becomes taught) to get 13m.

    the time from slipping to reaching her lowest point. 1.75s
    I've been trying for quite some time now, but I can't seem to get 1.75...
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    (Original post by dvs)


    I've been trying for quite some time now, but I can't seem to get 1.75...
    Thanks

    Ne1 else able to get the 1.75s?
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    Climber falls naturally for 13m, until when the rope just becomes taught, Then falls for a distance of x metres while the rope gets stretched.

    Naturall fall.
    s = ½gt²
    13 = 4.9t² => t = 1.6288
    t1 = 1.63 s
    ========

    Constrained fall
    Tension in rope is T = lambda*x/l

    T = md, where d is the deceleration provided by the rope on the climber.
    Effective acceleration on the climber is (g - d)

    g - d = v.dv/dx
    v.dv.dx = (g - d) = (g - T/m)
    v.dv.dx = 9.8 - lambda.x/(lm)
    v.dv/dx = 9.8 - 45000.x/(20.80)
    v.dv/dx = 9.8 - 28.125x

    integrating,

    ½v² = 9.8x -14.0625x² + c
    v² = 19.6x - 28.125x² + c

    at x=0, v = 14 => c = 196

    v² = 19.6x - 28.125x² + 196
    v = rt(196 + 19.6x - 28.125x²)

    now to simplify the expression a bit,

    196 + 19.6x - 28.125x² = -28.125(x² - 0.6969x - 6.9288)
    196 + 19.6x - 28.125x² = 28.125{7.0902 - u²}, where u = x - 0.3484 giving du = dx

    we can now write,

    dx/dt = v = rt{k²(a² - u²)}, where a = rt(7.0902) = 2.6627 and k = rt(28.125) = 5.3033
    du/dt = k.rt(a² - u²)
    du/{k.rt(a² - u²)} = dt
    (1/k)(1/a)arcsin(u/a) = t + c'

    at t = 0, x = 0, giving u = -0.3484 => c' = -9.2925*10^(-3)

    at x = 3.0112m, u = 2.6628 giving,

    t = (1/5.3033)(1/2.6627).arcsin(2.6628/2.6627) + 9.2925*10^(-3)
    t = (1/5.3033)(1/2.6627).arcsin(1) + 9.2925*10^(-3)
    t = (1/5.3033)(1/2.6627).(pi/2) + 9.2925*10^(-3)
    t = 0.1112 + 9.2925*10^(-3)
    t = 0.1205
    t2 = 0.12s
    ========

    total time of fall is t = t1 + t2
    t = 1.63 + 0.12
    t = 1 75s
    =======
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    Hmm, nicely done Fermat.

    I approached it by first finding the time taken for free fall, then I went to see if the climber moves in SHM at some point and what the amplitude and period of that motion was. I obviously messed up somewhere!
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    tx every1. i have another question. I can do a,b,c but not d :'(

    A particle P starts from rest at a fixed point O nad moves on a horizontal place in a straight line Ox. At time ts after leaving O, the acceleration of P is 15+4t-3t^2. The particle comes to instantaneous rest at the point A.
    a. Show that P takes 5s to reach O.
    b. Calculate the distance between O and A giving your answer to the nearest m.

    The mass of P is 0.2kg. As P moves during the interval 0<=t<=5, calculate:

    c. The greatest kinetic energy of P
    d. the greatest value of the magnitude of the force exerteed on P giving your answer in newtons



    b. 115m
    c. 130J
    d. 8N <=====this is the bit i cant do :'(
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    (Original post by dvs)
    Hmm, nicely done Fermat.

    I approached it by first finding the time taken for free fall, then I went to see if the climber moves in SHM at some point and what the amplitude and period of that motion was. I obviously messed up somewhere!
    Ta, but it's still a heck of a lot of work. I don't think this question could ever be described as an exam-type question :eek:
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    What about me newest question
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    Lol, i saw the mistake Fermat.....
    You deleted before I could quote it though

    4-6t=0
    so t=1.5s? I dont think so
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    DRAT

    Force exerted on P = F,say where F = ma
    F = m(15 + 4t - 3t²)
    dF/dt = m(4 - 6t) = 0 for a turning point
    4 - 6t = 0
    t = 2/3 s
    ======

    d²F/dt² = m(-6) = -ve => a max

    @ t = 2/3

    F = 0.2(15 + 8/3 - 3*(4/9))
    F = 0.2(17.666 - 1.333)
    F = 3.267 N
    =========
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    (Original post by lgs98jonee)
    What about me newest question
    Max force => max acceleration

    Oh well a bit late
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    (Original post by Fermat)
    DRAT

    Force exerted on P = F,say where F = ma
    F = m(15 + 4t - 3t²)
    dF/dt = m(4 - 6t) = 0 for a turning point
    4 - 6t = 0
    t = 2/3 s
    ======

    d²F/dt² = m(-6) = -ve => a max

    @ t = 2/3

    F = 0.2(15 + 8/3 - 3*(4/9))
    F = 0.2(17.666 - 1.333)
    F = 3.267 N
    =========
    That is what I got


    Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

    Heinamann just dont realise how much student's time their mistakes cost.

    (Original post by JamesF)
    Max force => max acceleration
    Yeh, thanks...I did do GCSE Physics...can you get the 8N answer though, as opposed to 3.3?
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    (Original post by lgs98jonee)
    That is what I got


    Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

    Heinamann just dont realise how much student's time their mistakes cost.



    Yeh, thanks...I did do GCSE Physics...can you get the 8N answer though, as opposed to 3.3?
    I think it's a mistake in the book.
    Like jamesf said, max force => max accln

    and max accln is 16 1/3, if you look at the eqn 15 + 4t - 3t².
    You would need an accln of 40 m/s² to get a force of 8 N.
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    (Original post by Fermat)
    I think it's a mistake in the book.
    Like jamesf said, max force => max accln

    and max accln is 16 1/3, if you look at the eqn 15 + 4t - 3t².
    You would need an accln of 40 m/s² to get a force of 8 N.
    Yeh I worked backwards and couldnt work out where they got 40 from as a=49/3 . Tx
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    (Original post by lgs98jonee)
    That is what I got


    Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

    Heinamann just dont realise how much student's time their mistakes cost.



    Yeh, thanks...I did do GCSE Physics...can you get the 8N answer though, as opposed to 3.3?
    When t = 5
    a = 15 + 20 - 75 = -40

    ma = - 8N
    |ma| = 8N

    Aitch
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    (Original post by Fermat)
    DRAT

    Force exerted on P = F,say where F = ma
    F = m(15 + 4t - 3t²)
    dF/dt = m(4 - 6t) = 0 for a turning point
    4 - 6t = 0
    t = 2/3 s
    ======

    d²F/dt² = m(-6) = -ve => a max

    @ t = 2/3

    F = 0.2(15 + 8/3 - 3*(4/9))
    F = 0.2(17.666 - 1.333)
    F = 3.267 N
    =========
    In the given interval, the acceleration of greatest magnitude isn't at a turning point.

    [See post above this one]

    Aitch
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    (Original post by Fermat)
    DRAT

    Force exerted on P = F,say where F = ma
    F = m(15 + 4t - 3t²)
    dF/dt = m(4 - 6t) = 0 for a turning point
    4 - 6t = 0
    t = 2/3 s
    ======

    d²F/dt² = m(-6) = -ve => a max

    @ t = 2/3

    F = 0.2(15 + 8/3 - 3*(4/9))
    F = 0.2(17.666 - 1.333)
    F = 3.267 N
    =========
    I have to admit that when I did this question a couple of months ago, I did exactly this, looked in the back of the book and found 8N.

    Then the word "magnitude" poked me in the eye. Looking at the curve, it's suddenly clear that, within the limits, the curve is furthest from y=0 at x=5.

    Aitch
 
 
 
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