A particle moves in space such that its position vector varies as `vec(r)=2thati+3t^(2)hatj`. If mass of particle is 2 kg then angular momentum of particle about origin at `t=2` sec is
A particle moves on the xy-pane such that its position vector is given by vec(r)=3t^(2) hati-t^(3) hatj . The equation of trajectory of the particle is given by
A particle move in x-y plane such that its position vector varies with time as vec r=(2 sin 3t)hat j+2 (1-cos 3 t) hat j . Find the equation of the trajectory of the particle.
A particle of mass 2 kg is moving such that at time t, its position, in meter, is given by vecr(t)=5hati-2t^(2)hatj . The angular momentum of the particle at t = 2s about the origin in kg m^(-2)s^(-1) is :
A particle moves so that its position vector varies with time as vec(r )= A cos omegathat(i)+A sin omega t hai(j) . The initial velocity of the particel the particle is
A particle moves in xy plane with its position vector changing with time (t) as vec(r) = (sin t) hati + (cos t) hatj ( in meter) Find the tangential acceleration of the particle as a function of time. Describe the path of the particle.
If postion vector of a particle is given by vec(r) = 10 alpha t^2 hati +[5 beta t- 5]hatj . Find time when its angular momentum about origin is O.
A particle of mass m moves in the xy-plane with velocity of vec v = v_x hat i+ v_y hat j . When its position vector is vec r = x vec i + y vec j , the angular momentum of the particle about the origin is.
A body is projected up such that its position vector varies with time as r = { 3thati + (4 t - 5t^2)hatj} m. Here, t is in seconds. Find the time and x-coordinate of particle when its y-coordinate is zero.
A particle moves in xy-plane. The position vector of particle at any time is vecr={(2t)hati+(2t^(2))hatj}m. The rate of change of theta at time t = 2 second. ( where 0 is the angle which its velocity vector makes with positive x-axis) is :
