Instantaneous velocity of a particle moving in `+x` direction is given as `v = (3)/(x^(2) + 2)`. At `t = 0`, particle starts from origin. Find the average velocity of the particle between the two points `p (x = 2)` and `Q (x = 4)` of its motion path.

If the velocity of a particle moving along x-axis is given as v=(3t^(2)-2t) and t=0, x=0 then calculate position of the particle at t=2sec.

If the velocity of a particle moving along x-axis is given as v=(3t^(2)-2t) and t=0, x=0 then calculate position of the particle at t=2sec.

If the velocity of a particle moving along x-axis is given as v=(3t^(2)-2t) and t=0, x=0 then calculate position of the particle at t=2sec.

The positon of an object moving along x-axis is given by x(t) = (4.2 t ^(2) + 2.6) m, then find the velocity of particle at t =0s and t =3s, then find the average velocity of particle at t =0 s to t =3s.

The velocity of a particle moving in the x direction varies as V = alpha sqrt(x) where alpha is a constant. Assuming that at the moment t = 0 the particle was located at the point x = 0 . Find the acceleration.

The velocity of a particle moving in the positive direction of the x axis varies as v=5sqrtx , assuming that at the moment t=0 the particle was located at the point x=0 , find acceleration of the particle.

The instantaneous velocity of a particle moving in xy-plane is vecV=(ay)hati+(V_(0))hatj where y is the instantaneous y co-ordinat of the particle and V_(0) is the particle starts from origin then its trajectory is

A particle moves along x-axis and its displacement at any time is given by x(t) = 2t^(3) -3t^(2) + 4t in SI units. The velocity of the particle when its acceleration is zero is

The motion of a particle, along x-axis, follows the relation x = 8t- 3t^(2) . Here x and t are expressed in metre and second respectively . Find (i) the average velocity of the particle in time interval 0 to 1 s and (ii) its instantaneous velocity at t = 1 s.