A particle whose velocity is given as `vecv=hati+6thatjm//s` is moving in x-y plane. At t-0, particle is at origin. Find the radius of curvature of path at point `((sqrt(2))/(3)m,(2)/(3)m)`

If equation of path of moving particle is given by y=2x^(2) ,radius of curvature of path of particle at origin is

A particle starts moving with velocity "vecV=hat i(m/s)", and acceleration "vec a=(hat i+hat j)(m/s^(2))".Find the radius of curvature of the particle at "t=1sec".

A particle is moving in an isolated x-y plane. At an instant, the particle has velocity (4hati+4hatj)m//s and acceleration (3hati+5hatj)m//s^(2) . At that instant what will be the radius of curvature of its path ?

A particle is given an initial velocity of vecu=(3 hati+4 hatj) m//s . Acceleration of the particle is veca=(3t^(2) +2 thatj) m//s^(2) . Find the velocity of particle at t=2s.

The velocity of an object is given by vecv = ( 6 t^(3) hati + t^(2) hatj) m//s) . Find the acceleration at t = 2s.

The velocity of an object is given by vecv = ( 6 t^(3) hati + t^(2) hatj) m//s) . Find the acceleration at t = 2s.

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.

The velocity of an object is given by vecv = ( 6 t^(3) hati + t^(2) hatj) m//s . Find the acceleration at t = 2s.

A particle moves in plane with acceleration veca = 2hati+2hatj . If vecv=3hati+6hatj , then final velocity of particle at time t=2s is

Motion of Mass Center in Vector Form A 2.0 kg particle has a velocity of Vecv_1=(2.0hati-3.0hatj)m /s, and a 3.0 kg particle has a velocity vecv_2=(1.0hati+6.0hatj)m/s. (a) How fast is the center of mass of the particle system moving? (b) Find velocities of both the particles in centroidal frame.