# Velocity is proportional to distance

## 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!

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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).

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:
$$v^2 = v_{i}^2 + 2a_c(x - x_{i})$$
initial velocity is zero, initial displacement is zero, therefore:
$$v^2 = 2a_cx$$
which gives:
$$v = \sqrt{2a_cx}$$
since 2 and acceleration are both constants:
$$v = C * \sqrt(x)$$
Can this be considered proportional? Perhaps at a stretch?

haruspex
Homework Helper
Gold Member
2020 Award
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?

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:
$$v^2 = v_{i}^2 + 2a_c(x - x_{i})$$
initial velocity is zero, initial displacement is zero, therefore:
$$v^2 = 2a_cx$$
which gives:
$$v = \sqrt{2a_cx}$$
since 2 and acceleration are both constants:
$$v = C * \sqrt(x)$$
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 $$v^2 = v_{i}^2 + 2a_c(x - x_{i})$$. Is it possible that there is a misprint on the textbook?

mfb
Mentor
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.

in my opinion, v^2 should be proportional to displacement, if acceleration is constant, as shown by you just now, which is $$v^2 = v_{i}^2 + 2a_c(x - x_{i})$$. 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."

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.