Velocity is proportional to distance

  • #1
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Homework Statement


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.

Homework 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.

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!
 

Answers and Replies

  • #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).
 
  • #3
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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?
 
  • #4
haruspex
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The velocity is proportional to the displacement. I am trying to find the equation linking velocity to displacement.
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?
 
  • #5
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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?
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?
 
  • #6
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in my opinion, v^2 should be proportional to displacement, if acceleration is constant
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.
 
  • #7
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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?
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.
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."
 
  • #8
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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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