The position vector of the particle is given by r = 2t^2x + 3ty + 4z. To find the instantaneous velocity and acceleration, we need to differentiate the position vector with respect to time.

Taking the derivative of r with respect to t, we get the velocity vector v = d/dt (2t^2x + 3ty + 4z).

The derivative with respect to t of 2t^2x is 4tx, the derivative of 3ty with respect to t is 3y, and the derivative of 4z with respect to t is 0.

So, the velocity vector v = 4tx + 3ty.

From the velocity vector, we can see that the velocity lies on the xy-plane, as there is no component in the z-direction (since 4tx and 3ty).

Now, to find the acceleration vector, we differentiate the velocity vector v with respect to t.

The derivative with respect to t of 4tx is 4x, and the derivative of 3ty with respect to t is 3y.

So, the acceleration vector a = 4x + 3y.

From the acceleration vector, we can see that the acceleration