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

If the position vector of a particle is given by vec(r ) = (cos 2t) hat(i) + (sin 2 t) hat(j) + 6 t hat(k) m . Calculate magnitude of its acceleration (in m//s^(2) ) at t = (pi)/(4)

If the velocity is vec(v) = 2 hat(i) + t^(2) hat(j) - 9 vec(k) , then the magnitude of acceleration at t = 0 . 5 s is :

The time dependence of the position of a particle of mass m = 2 is given by vec(r) (t) = 2t hat(i) - 3 t^(2) hat(j) Its angular momentum, with respect to the origin, at time t = 2 is :

The time dependence of the position of a particle of mass m = 2 is given by vec(r) (t) = 2t hat(i) - 3 t^(2) hat(j) Its angular momentum, with respect to the origin, at time t = 2 is :

The time dependence of the position of a particle of mass m = 2 is given by vec(r) (t) = 2t hat(i) - 3 t^(2) hat(j) Its angular momentum, with respect to the origin, at time t = 2 is :

The position of a particle is given by vec r= 3.0 t hat i - 2.0 t^2 hat j + 4.0 hat k m , wher (t) in seconds and the coefficients have the proper units for vec r to be in metres. what is the magnitude of the velocity of particle at t=2 sec ?

If the position vector of a particle is given by vec r= 5t^(2)hat i +7t hat j +4 hat k , then its velocity lies in: