The position of a particle is given as
`vecr=at hati+bt^(2)hatj`
Find the equation of trajectory of the particle.

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

Velocity vector of a particle is given as vecv=4thati+3hatj At t=0 position of the particle is given as vecr_(0)=2hatj Find the position vector of the particle at t=2 sec and te average acceleration of the particle for t=0 to t=2sec.

If the position vector of the particle is given by vecr = 3 t^2 hati + 5t hatj + 4hatk . Find the velocity of the particle at t = 3s.

The position vector of a particle vecr =(t^2-1)hati +2thatj . Determine the trajectory of the particle in the xy plane.

The position of a particle is given by vecr = 3t hati + 2t^(2) hatj + 5hatk , where t is in seconds and the coefficients have the proper units for vecr to be in meters. The direction of velocity of the particle at t = 1 s is

The position of a particle is given by vecr = 3t hati + 2t^(2) hatj + 5hatk , where t is in seconds and the coefficients have the proper units for vecr to be in metres. The direction of velocity of the particle at t = 1 s is

The position vector of a particle is vecr = (a cos omega t)hati + (a sin omegat)hatj . the velocity of the particle is

The position of a particle is given by vecr = 2t^(2) hati + 3t hatj + 4hatk where t is in second and the coefficients have proper units for vecr to be in metre. The veca(t) of the particle at t = 1 s is