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Velocity is proportional to distance

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  1. Dec 14, 2014 #1
    1. The problem statement, all variables and given/known data
    An object starts from rest at the origin and moves in the positive x direction. The velocity is proportional to the displacment. I am trying to find the equation linking velocity to displacement.

    2. Relevant equations
    Since the object starts at rest and then moves in the positive direction, acceleration is non-zero. The chapter in the textbook concerns constant acceleration. So, I think v2= vi2+ 2a[x-xi] would be appropriate. vi is initial velocity and xi is initial displacement.

    3. The attempt at a solution
    Setting both the initial velocity and initial displacement to zero gives v = sqrt(2ax). Does this count as velocity proportional to displacement? I found online this type of thing being referred to as proportional to the square root.
    Any help appreciated. Apologies for lack of latex. This isn't on a computer!
     
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  3. Dec 14, 2014 #2

    Orodruin

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    If velocity is proportional to displacement, then acceleration is not constant. In stead you would have v = kx for some constant k (with units 1/time). This is a differential equation that can be solved (note that if x=0, the object will not move, you need some initial displacement).
     
  4. Dec 20, 2014 #3
    Hi Orodurin, thanks for the reply.
    The chapter concerns kinematics with either zero or constant acceleration. It doesn't even assume much familiarity with calculus so I can't see that the question would require one to solve a differential equation. That's why I thought acceleration would have to be constant and non-zero.
    The only solution I can find is the one from my original post:
    [tex]v^2 = v_{i}^2 + 2a_c(x - x_{i})[/tex]
    initial velocity is zero, initial displacement is zero, therefore:
    [tex]v^2 = 2a_cx[/tex]
    which gives:
    [tex]v = \sqrt{2a_cx}[/tex]
    since 2 and acceleration are both constants:
    [tex]v = C * \sqrt(x)[/tex]
    Can this be considered proportional? Perhaps at a stretch?
     
  5. Dec 20, 2014 #4

    haruspex

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    That seems pretty simple. If y is proportional to x, what equation can you write for y as a function of x? Are you sure that's a precise statement of the problem?
     
  6. Dec 21, 2014 #5
    in my opinion, v^2 should be proportional to displacement, if acceleration is constant, as shown by you just now, which is [tex]v^2 = v_{i}^2 + 2a_c(x - x_{i})[/tex]. Is it possible that there is a misprint on the textbook?
     
  7. Dec 21, 2014 #6

    mfb

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    If acceleration is constant - but here, it is not.
    We can see this in post 1 already, where velocity would follow the square root of displacement.
     
  8. Jan 4, 2015 #7
    Thanks for the replies. I am not sure if it is an error in the book, but perhaps the acceleration is not constant and the relationship is simply |v| = kx and I do not need anything more "fancy" to solve the problem (such as differential equations), I might be neglecting something obvious.
    The question is as follows: "... tell whether the object's position x doubles as the clock reading t doubles... The object starts from rest at the origin at t=0 and travels in the positive x direction. Its speed is proportional to its distance from the origin."
     
  9. Jan 4, 2015 #8
    Don't worry: If the distance doubles as time doubles then the acceleration would have to be zero, since the average velocities would have to be the same. But the problem involves non-zero velocity, and so the distance does not double as the time doubles. Kind of obvious and I completely missed it. Thanks for all the replies.
     
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