What is the acceleration of a particle moving along a parabolic path?

[greenpeace16]

Homework Statement

A particle moves uniformly with speed v along the path y=kx^2 .Taking k as a positive constant, acceleration of the particle at x=0 is _________?

Homework Equations

options-a)2kv^2 b)kv^2 c)1/2(kv^2) d)2kv

The Attempt at a Solution

I did- y=kx^2
differentiating we get,
v= 2kx
again differentiating we get,
a=2k
so there'es an uniform acceleration of a=2k at all points.
but none of the options match!

 

[PeterO]

Homework Statement

A particle moves uniformly with speed v along the path y=kx^2 .Taking k as a positive constant, acceleration of the particle at x=0 is _________?

Homework Equations

options-a)2kv^2 b)kv^2 c)1/2(kv^2) d)2kv

The Attempt at a Solution

I did- y=kx^2
differentiating we get,
v= 2kx
again differentiating we get,
a=2k
so there'es an uniform acceleration of a=2k at all points.
but none of the options match!

You have differentiated the equation of the path. Not sure whether that is really the same as differentiating the velocity.

When x = 0 , the parabola in the region can be approximated to a circle, so it amounts to circular motion. [small spherical mirrors focus just like parabolic mirrors for example] The trick is to relate the focal length of the parabola to the radius of curvature of the circle.

 

[greenpeace16]

You have differentiated the equation of the path. Not sure whether that is really the same as differentiating the velocity.

When x = 0 , the parabola in the region can be approximated to a circle, so it amounts to circular motion. [small spherical mirrors focus just like parabolic mirrors for example] The trick is to relate the focal length of the parabola to the radius of curvature of the circle.

Well, isn't finding how y changes with respect to x of a particle, knowing the velocity of the particle?
And anyhow, the focus of this parabola can be found from- if y= ax^2 + bx + c , then the focus is at {-b/2a,(-b^2/4a) + c + 1/4a} and so for this problem we get the focal length to be 1/4k and near x=0 this can be approximated to the radius of the curvature as you mentioned, and so we know, a= v^2/r and hence here, a=v^2/(1/4k) = v^2*kv = 4kv^2.
This is not one of the options either. Did i do anything wrong?

 

[BruceW]

I did- y=kx^2
differentiating we get,
v= 2kx
again differentiating we get,
a=2k
so there'es an uniform acceleration of a=2k at all points.
but none of the options match!

You've differentiated y with respect to x. This doesn't give you velocity. I don't know about tricks with parabolas, but I think the most straightforward way to do this question is to write down the position vector, and then the velocity vector, and acceleration vector, and use the information you are given in the question to solve for the magnitude of acceleration.

 

[PeterO]

Well, isn't finding how y changes with respect to x of a particle, knowing the velocity of the particle?
And anyhow, the focus of this parabola can be found from- if y= ax^2 + bx + c , then the focus is at {-b/2a,(-b^2/4a) + c + 1/4a} and so for this problem we get the focal length to be 1/4k and near x=0 this can be approximated to the radius of the curvature as you mentioned, and so we know, a= v^2/r and hence here, a=v^2/(1/4k) = v^2*kv = 4kv^2.
This is not one of the options either. Did i do anything wrong?

Do you mean f = 1/(4k) or 1/4 of k?

I recall the radius of curvature is 2f?

 

[greenpeace16]

You've differentiated y with respect to x. This doesn't give you velocity. I don't know about tricks with parabolas, but I think the most straightforward way to do this question is to write down the position vector, and then the velocity vector, and acceleration vector, and use the information you are given in the question to solve for the magnitude of acceleration.

there is no more information given other than the path equation and that it moves uniformly with speed v.
the position vector is xi + yj = xi+kx^2j. Now if I am not supposed to differentiate it, how can i get velocity vector?

 

[BruceW]

you've got the position vector correct. Now you differentiate that with respect to time to get the velocity vector. (keeping in mind that the unit vectors i and j don't depend on time). And I know that you don't already have an explicit equation for the function of x with time, but for now you can just use a variable, for example \dot{x} to represent it.

 

[greenpeace16]

I meant 1/(4k) i.e reciprocal of 4k.And yeah, for spherical mirrors, I do know that focal length is twice the radius of curvature but is it the same here?We're approximating the parabola to be a circle so isn't the focal length=to radius of curvature?
And yeah, when we differentiate,
y=kx^2
velocity = v = dx/dt
dy/dt = 2kx(dx/dt)
differentiating again, which gives
d^2y/dt^2 = acceleration = 2k(dx/dt)^2+2kx(d^2x/dt^2)
since u need acceleration at x=0
acceleration is 2k(dx/dt)^2 = 2kv^2
If we took 2f=R, then we would get the same result, but I'm not really convinced whether that's right.And just because the answers match, I can't leave it at that.

 

[BruceW]

And yeah, when we differentiate,
y=kx^2
velocity = v = dx/dt
dy/dt = 2kx(dx/dt)
differentiating again, which gives
d^2y/dt^2 = acceleration = 2k(dx/dt)^2+2kx(d^2x/dt^2)
since u need acceleration at x=0
acceleration is 2k(dx/dt)^2 = 2kv^2

You've got to the right answer. I think the only unfounded assumption you've made here is that the x component of acceleration is equal to zero. But this happens to be true, if you work it out. Or maybe you did work it out, but didn't show your working here? In any case, you got the right answer :)