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P3 Differential Equation watch

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    A girl live 500 metres from school. She sets out walking at 2ms , but when she has walked a distance of x metres her speed has dropped to ( 2 - 1/400 x ) ms. How long does she take to get to school?
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    have u got an answer, before i scare you with a mahoosive wrong solution?
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    about 6.5 minutes

    thanks for helping!
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    No i got something like 9.5 minutes.. grr.

    To be honest with you though, i'm using M2 methods.. so errrr, i'll keep trying and see what i come up with.

    I'm basing it on the fact that INT (velocity) = displacement
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    (Original post by Sang)
    A girl live 500 metres from school. She sets out walking at 2ms , but when she has walked a distance of x metres her speed has dropped to ( 2 - 1/400 x ) ms. How long does she take to get to school?
    when distance = 0 then speed = 2 ms
    when distance = x then speed = 2 - 1/ 400x

    time = distance / speed

    x / 2 - 1/400 x = 400x^2 - 2 = x^2 = 0.005

    so x = square root of 0.005

    (actually thats probably wrong, im not sure.)
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    i got 6.53878 minutes. So i think you're right
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    (Original post by SUKBarracuda)
    i got 6.53878 minutes. So i think you're right
    Explain how please, genius
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    dx/dt

    dt/dx

    => x=


    in that method please? im stuck, similar questions after this in book that i need to do
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    Please tell me this isn't edexcel, i've never seen anything like this on all the P3 papers i've done. Variable accleration, speed-time graphs etc come up in Mechanics but i'v never had to apply M2 to P3.. weird.
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    negative this is OCR P3 - but it should br totally unrelated you should be able to answer this with pure 3 knowledge only so im told. Also I do s2 not m2.
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    (Original post by BlueAngel)
    when distance = 0 then speed = 2 ms
    when distance = x then speed = 2 - 1/ 400x

    time = distance / speed

    x / 2 - 1/400 x = 400x^2 - 2 = x^2 = 0.005

    so x = square root of 0.005

    (actually thats probably wrong, im not sure.)



    Or dx/ ds = k(2 - 1/ 400x)

    dx / 2 - 1/ 400x = k.ds

    { 1/2 - 1/ 400x .dx = { k. ds

    = LN 400X = ks

    time = e^ 400x - ks + c
    = e^ A400x

    Then I think you have to find what C is and what A is by putting the following in:
    when distance = 0 then speed = 2 ms
    when distance = x then speed = 2 - 1/ 400x

    Please tell me if im right or wrong. Ive got P3 next week too and Im dreading it.
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    v = 2 - (x/400)

    dx/dt = 2 - x/400

    dx/dt = (800 - x)/400

    400int{1/(800 -x)} dx = int{t} dt

    u = 800 -x
    du = -dx

    -400int{1/u}du = t + C

    -400.ln(800 - x) = t + C

    when x = 0 t = 0

    C = -400.ln(800) = -2673.8

    -400.ln(800 - x) = t -2673.8

    so when x = 500

    -400.ln(300) + 2673.8 = t

    t = 392.287 seconds
    = 6.53 minutes.

    :cool:
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    ive never come across du = -dx can elaborate on that please barrucuda?
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    i'm sorry this is just me skipping a step through habit.

    u = 800 -x
    du/dx = -1

    and

    400int{1/(u)} dx

    in order to be integrating with respect to u, you need to remove the dx.

    since du/dx = -1

    dx can be considered -1.du

    and so replaced:

    -400int{1/u}du

    and you can then integrate wrt u.
 
 
 
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