Final answer:
The student's physics problem involves finding the velocity and acceleration vectors for a particle on a circular path, demonstrating centripetal acceleration, and calculating the centripetal force vector as a function of time.
Step-by-step explanation:
The student's question pertains to the kinematics of a particle in motion in the xy-plane, specifically asking for the velocity and acceleration vectors as functions of time, the nature of the acceleration with respect to the circular path (centripetal acceleration), and the corresponding centripetal force vector as a function of time.
Answer (a): Velocity and Acceleration as Functions of Time
For the given position function r(t) = (4.0 cos 3t)i + (4.0 sin 3t)j, the velocity v(t) is the first derivative of r(t) with respect to t, and the acceleration a(t) is the derivative of v(t).
V(t) = dr/dt = -12.0 sin(3t)i + 12.0 cos(3t)j
A(t) = dv/dt = -36.0 cos(3t)i - 36.0 sin(3t)j
Answer (b): Centripetal Acceleration
The acceleration points towards the center of the circular path, which is a characteristic of centripetal acceleration.
Answer (c): Centripetal Force as a Function of Time
The centripetal force vector F(t) can be found using Newton's second law, F = ma, where m is the mass of the particle.
F(t) = ma(t) = (-36.0 m)(0.50 kg) cos(3t)i - (36.0 m)(0.50 kg) sin(3t)j