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Calculus problem concerning acceleration

  1. Aug 11, 2012 #1
    1. The problem statement, all variables and given/known data
    a particle of mass accelerates at rest from the origin . The acceleration of the particle 'a' is given by a=k/(x+4)^4 , where x is the distance from the origin . given data : when x=5 , velocity 'v'=8

    what is the maximum velocity that can be obtained ?

    3. The attempt at a solution

    I even got stuck when I thought about it briefly. When the particle's velocity reaches maximum , the acceleration must be 0 , and either k=0 or x must be at infinity . After that , I have no idea what i can do about it . Shall I use definite integration (from 0 to infinity) or indefinite integration (which leaves the integration constant C )? what I am sure is merely that k has to be obtained by integration and the given numerical data.
     
    Last edited: Aug 11, 2012
  2. jcsd
  3. Aug 11, 2012 #2
    It looks like you can use definite integration (you know when x goes from 0 to 5, v goes from 0 to 8) and then solve for k. After that, you should have a velocity equation in terms of x and you just have to find it's maximum value. EDIT: Actually, I don't think that would work, sorry. Since the 'a' is given in terms of x, integrating it won't give a formula for velocity because you would be integrating with respect to position rather than time. Hrm. I'll try to think about it some more and see if I can come up with something.
     
    Last edited: Aug 11, 2012
  4. Aug 11, 2012 #3
  5. Aug 11, 2012 #4

    AGNuke

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    Actually, you need to know one more definition of acceleration, Newton's Third Equation of motion. Here's its differential equation.[tex]a=\frac{dv}{dt}=\frac{v.dv}{v.dt}=\frac{v.dv}{({dx}/{dt})dt}=v\frac{dv}{dx}[/tex]

    Now, put [itex]a = \frac{k}{(x+4)^4}[/itex] and by workabouts, you get

    [tex]\int_{0}^{v}vdv=k\int_{0}^{x}\frac{dx}{(x+4)^4}[/tex]

    Solve for x=5; v=8, find k. Then by calculus, find the maximum value.
     
    Last edited: Aug 11, 2012
  6. Aug 11, 2012 #5

    Ray Vickson

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    The force F = a/m is derivable from a potential V(x): [itex] F = -dV(x)/dx,[/itex] so
    [tex] V(x) = \frac{k/m}{3(4+x)^3}.[/tex] Since the force is "conservative", the total energy is conserved; that is,
    [tex] E \equiv \frac{1}{2} m v^2 + V(x) = \text{const},[/tex]
    where v = velocity. We have v = 0 when x = 0 and v = 8 when x = 5, so we can find k (assuming a given value of m, such as m = 1).

    RGV
     
  7. Aug 11, 2012 #6

    AGNuke

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    But it isn't necessary to have mass given in Kinematics > Rectilinear Motion. The Question can be solved using Newton's Third Equation of Motion.
     
  8. Aug 12, 2012 #7
    I just worked it out with indefinite integrals on both sides...the max. velocity should be the square root of the difference of the integration constants : C1(from v*dv)-C2( from k*(x+4)^-4*dx
     
  9. Aug 12, 2012 #8

    AGNuke

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    Why indefinite integral!? I even gave you limits of integration. You have been provided that the body was at rest at origin, so v=0 at x=0. Just integrate from 0 to v and 0 to x.

    Then to find k, put v=8, x=5.
     
  10. Aug 13, 2012 #9
    the value of k has been obtained with that method , but in order to find the max. velocity of the particle , the integrated equation must have a constant , C1(from v*dv)-C2( from k*(x+4)^-2*dx) . The value of C1-C2 can be obtained by substituting v=0 and x=0 into the integrated equation , although I don't know if this method makes sense

    By the way , I gave a wrong equation and wrong data when I started this thread . The correct version is updated here:
     

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    Last edited: Aug 13, 2012
  11. Aug 13, 2012 #10

    gabbagabbahey

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    You mean [itex]V(x) = \frac{k \cdot m}{3(4+x)^3}[/itex], right? :wink:

    (No need to assume a value for the mass!:smile:)
     
  12. Aug 13, 2012 #11

    gabbagabbahey

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    I'm not sure what you are doing here. You can find the velocity as a function of position [itex]v(x)[/itex] using either AGNuke's method or Ray Vickson's. What do you get when you do that?

    Afterwards, you can find the maximum velocity the same way you find the maximum value of any function - first look for places where [itex]v'(x)=0[/itex] and [itex]v''(x)<0[/itex], and if you don't find any, see where the function is unbounded (i.e. where [itex]v(x) \to \infty[/itex]).
     
  13. Aug 13, 2012 #12
    as v'(x)=0 , x tends to infinity . Unless there is a constant in v(x), substituting x tends to infinity into v(x) gives v(x)=0
     
  14. Aug 13, 2012 #13

    gabbagabbahey

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    How about you start by showing us what you got for v(x)...
     
  15. Aug 13, 2012 #14
    oh..i just found out what's wrong with me . The integral i obtained is (v^2)*0.5=-k[1/(x+4)]+(k/4) . previously i missed the constant part of this equation , k/4 . Now that when i substitute x tends to infinity I will get v=[(k/4)*2]^0.5 , v=24^0.5 ms^-1
     
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