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Differentiation Problem

  1. May 15, 2004 #1
    I’m having some difficulty with this question:

    (b) A particle starts with a speed of 20 m/s and moves in a straight line. The particle is subjected to a resistance which produces a retardation which is initially 8 m/s^2 and which increases uniformly with the distance moved, having a value of 9 m/s^2 when the particle has moved a distance of 5 m.
    If v m/s is the speed of the particle when it has moved a distance x m:

    (i) prove that, while the particle is in motion,

    v dv/dx = - ( 8 + x/5 )

    (ii) Calculate the distance moved by the particle in coming to rest


    I can’t seem to get out part (i), which is probably key to solving part (ii). I’m having problems with the force equation, more specifically with the resistance increasing with time/distance.

    Thanks in advance for any help.
     
  2. jcsd
  3. May 15, 2004 #2

    jcsd

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    express v in terms of x.

    edited: sorry that's a rather silly way to go about it. What does d2v/dx2 equal? so what does dv/dx equal?
     
    Last edited: May 15, 2004
  4. May 15, 2004 #3
    The way that I usually do it is to get a force equation for the question first... then work out dv/dx etc. Any help with actually getting a force equation?
     
  5. May 15, 2004 #4

    jcsd

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    I'll give you a start: d2v/dx2 is going to be linear (the question tells you this), therefore d2v/dx2 = k (k = a constant), what is k?
     
  6. May 16, 2004 #5
    I’m still not getting it out… am I supposed to be using equations of motion to get a, which is equal to v(dv/dx) and (d^2x)/(dx^2)? I’ve been trying to do it that way, but I’m still getting nowhere. I still can't figure out where the 8m/s^2 and 9m/s^2 retardations come in. I’d be grateful for a little more help.
     
  7. May 16, 2004 #6

    arildno

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    Hint:
    The acceleration of an object is given by [tex]\frac{dv}{dt}[/tex].
    Assuming that the velocity may be written as a function of the position x(t), we have, by the chain rule:
    [tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/tex], since by definition of velocity, we have [tex]v=\frac{dx}{dt}[/tex]
     
  8. May 16, 2004 #7

    jcsd

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    Okay, a = dv/dx,

    As I siad before we know d2v/dx2 = a constant as the question tells us this:, therefore from the question we can work out that: d2v/dx2 = [(-9) - (-8)]/5 = -1/5

    So now integrate to find dv/dx
     
  9. May 16, 2004 #8

    arildno

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    [tex]a\neq\frac{dv}{dx}[/tex], since the dimensions are unequal.
    Rather, we have:
    [tex]v(x(t))=\frac{dx}{dt}\rightarrow{a}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/tex]
     
  10. May 16, 2004 #9

    jcsd

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    D'oh, yes your're right of course acceltrtion doesn't equal dv/dx
     
  11. May 16, 2004 #10
    Ok, so now I'm just as lost. a = v(dv/dx), right? But I still dont know what I should be doing next...
     
  12. May 16, 2004 #11

    arildno

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    Well, we're halfway, so you'r just halfway as lost.
    By Newton's 2. law, you have:
    [tex]a=\frac{F}{m}[/tex]
    Now, you have been given that the resistance(force) induces a retardation of absolute value (8+x/5) (right?), so by assuming v is in the positive direction, the induced acceleration (F/m) must be -(8+x/5)
     
  13. May 16, 2004 #12
    Yes, I know how to apply the force equation (F=Ma, etc), but what I dont understand is where the (8 + x/5) part comes from - I don't know how to firgure out that. That's my original question, and I still don't know how to do it.
     
  14. May 16, 2004 #13

    arildno

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    Initially (that is, when you have moved zero distance!), the resistance force induces an acceleration of -8 m/s^(2) (that is, a retardation by assuming v positive)
    This is given in the exercise.
    You are then informed that the resistance increases in a uniform manner with the distance travelled so far.
    Now uniform manner means that the resistance must increase in a linear relation with x (the distance travelled).
    Hence, the induced acceleration must be on the form: -(8+sx), where s is a constant.

    You are then informed that when you have traveled 5 meters (x=5), the induced acceleration is -9m/s^(2).

    Hence, you may conclude that s=1/5
     
  15. May 17, 2004 #14
    Thanks, that's really all I wanted to know.
     
  16. May 17, 2004 #15

    arildno

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    Well, I had to take my time, you know..:wink:
     
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