A particle is moving in xy-plane in circular path with centre at origin.If at an instant the position of particle is given by `(1)/(sqrt(2))(hat i+hat j)` Then velocity of particle is along

A particle is moving in xy-plane in a circular path with centre at origin. If at an instant the position of particle is given by (1)/(sqrt(2)) ( hat(i) + hat(j)) , then velocity of particle is along

A particle moves in x-y plane according to the equation bar(r ) = (hat(i) + 2 hat(j)) A cos omega t the motion of the particle is

The position of a particle is given by vec(r) = 3that(i) - 4t^(2)hat(j) + 5hat(k). Then the magnitude of the velocity of the particle at t = 2 s is

A particle is moving with speed 6m/s along the direction of 2hat(i)+2hat(j)-hat(k) find the velocity vector of a particle ?

The position of a particle at time moving in x-y plane is given by vec(r) = hat(i) + 2 hat(j) cos omegat . Then, the motikon of the paricle is :

The position of a particle at time moving in x-y plane is given by vec(r) = hat(i) + 2 hat(j) cos omegat . Then, the motikon of the paricle is :

A particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are a=(4 hat(i)+3hat(j))m//s^(2) and p=(8 hat(i)-6hat(j))"kg-m/s" . The motion of the particle is