[" The speed "v" of a particle moving along the axis of "x" given by the relation "],[v^(2)=n^(2)(8bx-x^(2)-12b^(2))" .Then the minimum time taken by the particle to go "],[" from "x=5b" to "x=6b" is (approximately) "]

The speed v of a particle moving along a straight line is given by a+bv^(2)=x^(2) , where x is its distance from the origin. The acceleration of the particle is

The speed v of a particle moving along a straight line is given by a+bv^(2)=x^(2) where x is its distance from the origin. The acceleration of the particle is

The speed v of particle moving along x-axis is given by v^(2) = 8bx - x^(2) - 12 b^(2) , where 'b' is a constant. Find amplitude, maximum velocity and time period of oscillation.

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

A particle is performing SHM along x-axis with amplitude 6.0 cm and time period 12 s. What is the minimum time taken by the particle to move from x=+3cm to x=+6cm and back again?

The acceleration of a particle moving along x -axis is a= -100x .It is released from rest at x=2. Here a and x are in S.I.units.The minimum time taken by particle to go from x=2 to x=0 is (pi)/(alpha) seconds.Find the value of (alpha)/(4) .

The potential energy U(x) of a particle moving along x - axis is given by U(x)=ax-bx^(2) . Find the equilibrium position of particle.

The speed v of a particle moving along a straight line. When it is at distance x from a fixed point on the line is v^(2)=144-9x^(2) . Select the correct alternatives

A particle is moveint along the x-axis whose instantaneous speed is given by v^(2)=108-9x^(2) . The acceleration of the particle is.