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-2s)=at(t-2)^2 .

Position of particle moving along x-axis is given as x=2+5t+7t^(2) then calculate :

The displacement of a particle moving along x-axis is given by : x = a + bt + ct^2 The acceleration of the particle is.

A particle moves along x-axis whose potential energy versus x-coordinate graph is shown in the given figure. At x = 0, the particle is released form rest. Two potential wells A and B are shown in the figure. Then,

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.

Velocity of particle moving along x-axis is given as v = ( x^(3) - x^(2) + 2) m // sec. Find the acceleration of particle at x=2 meter.