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 moving in x-y plane. At an instant, it has velocity (4 hat (i) + 4 hat(j)) m//s and acceleration (3 hat(i) + 5 hat(j)) m//s^(2) At that instant, the radius of curvature of its path will be :

A particle is moving in x-y plane. At an instant, it has velocity (4 hat (i) + 4 hat(j)) m//s and acceleration (3 hat(i) + 5 hat(j)) m//s^(2) At that instant, the radius of curvature of its path will be :

A particle is moving in such a way that at one instant velocity vector of the particle is 3hati+4hatj m//s and acceleration vector is -25hati - 25hatj m//s ? The radius of curvature of the trajectory of the particle at that instant is

A particle is moving in such a way that at one instant velocity vector of the particle is 3hati+4hatj m//s and acceleration vector is -25hati - 25hatj m//s ? The radius of curvature of the trajectory of the particle at that instant is

A particle has an initial velocity 3hati+3hatj and acceleration of 0.41hati+0.3hatj . Its speed after 10s is

A particle has an initial velocity 3hati+4hatj and an accleration of 0.4hati+0.3hatj . Its speed after 10s is

A particle P is at the origin starts with velocity u=(2hati-4hatj)m//s with constant acceleration (3hati+5hatj)m//s^(2) . After travelling for 2s its distance from the origin is

A particle P is at the origin starts with velocity u=(2hati-4hatj)m//s with constant acceleration (3hati+5hatj)m//s^(2) . After travelling for 2s its distance from the origin is