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

Initial velocity and acceleration of a particles are as shown in the figure. Acceleration vector of particle remain constant. Then radius of curvature of path of particle :

Path traced by a moving particle in space is called trajectory of the particle. Shape of trajectiry is decided by the forces acting on the particle. When a coordinate system is associated with a particle motion, the curve equation in which the particle moves [y=f(x)] is called equation of trajectory. It is just giving us the relation among x and y coordinates of the particle i.e. the locus of particle. To find equation of trajectory of a particle, find first x and y coordinates of the particle as a function of time eliminate the time factor. The position vector of car w.r.t. its starting point is given as vec(r)=at hat(i)- bt^(2) hat(j) where a and b are positive constants. The locus of a particle is:-

The coordinates of a moving particle at time t are given by x=ct^(2) and y=bt^(2) . The speed of the particle is given by :-

A particle is projected from the ground at an angle of 60^(@) with horizontal at speed u = 20 m//s. The radius of curvature of the path of the particle, when its velocity. makes an angle of 30^(@) with horizontal is : (g=10 m//s^(2)

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)

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 the xy-plane with constant acceleration a directed along the negative y-axis. The equation of path of the particle has the form y= bx - cx^2 , where b and c are positive constants. Find the velocity of the particle at the origin of coordinates.

A particle moves in the x-y plane according to the law x=t^(2) , y = 2t. Find: (a) velocity and acceleration of the particle as a function of time, (b) the speed and rate of change of speed of the particle as a function of time, (c) the distance travelled by the particle as a function of time. (d) the radius of curvature of the particle as a function of time.

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.

In the above question, the radius of path of the particle will be