Recent content by greenpeace16

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

    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...
  2. G

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

    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?
  3. G

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

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

    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...
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