A particle moving on x - axis has potential energy `U = 2 - 20x + 5x^(2)` joule along x - axis. The particle is relesed at `x = -3`. The maximum value of `x` will be (`x` is in metre)

The potential energy U of a particle oscillating along x-axis is given as U=10-20x+5x^2 . If it is released at x=-2 , then maximum value of x is

A particle free to move along the X- axis has potential energy given by U(X)=k[1-exp(-X^(2))] of appropriate dimensions. Then-

A particle of mass 2 kg moving along x-axis has potential energy given by, U=16x^(2)-32x ( in joule), where x is in metre. Its speed when passing through x=1 is 2ms^(-1)

A particle moves along the X-axis as x=u(t-2 s)+a(t-2 s)^2 .

A particle moves along the X-axis as x=u(t-2 s)+a(t-2 s)^2 .

A particle moves along the X-axis as x=u(t-2s)+a(t-2s)^2 .

A particle moves along the X-axis as x=u(t-2s)=at(t-2)^2 .

A particle moves along the X-axis as x=u(t-2s)=at(t-2)^2 .

A particle can move along x axis under influence of a conservative force. The potential energy of the particle is given by U = 5x^2 – 20x + 2 joule where x is co-ordinate of the particle expressed in meter. The particle is released at x = –3 m (a) Find the maximum kinetic energy of the particle during subsequent motion. (b) Find the maximum x co-ordinate of the particle.

A particle is moving along the x-axis with an acceleration a = 2x where a is in ms^(–2) and x is in metre. If the particle starts from rest at x=1 m, find its velocity when it reaches the position x = 3.