The position vector of a particle is r =2t2x + 3ty + 4z Then the instantaneous velocity v and acceleration a respectively lie

The position vector is given as r = 2t²i + 3tj + 4k. To find the instantaneous velocity (v), we take the first derivative of r with respect to time: v = dr/dt = 4ti + 3j + 0k. Since the z-component is zero, the velocity vector lies entirely in the xy-plane. To find the acceleration (a), we take the derivative of the velocity: a = dv/dt = 4i + 0j + 0k [1]. The acceleration vector has only an x-component, meaning it is directed along the x-axis. Therefore, the velocity lies on the xy-plane and the acceleration is along the x-direction. This follows the standard kinematic procedure where velocity and acceleration are successive derivatives of the position vector.

Sources

1. [1] https://www.maths.dur.ac.uk/users/W.J.Zakrzewski/dynamics/problems_solns.pdf