5. The joint angles as a function of time of a person performing a squat exercise for the ankle, knee, and trunk are as follows: (t) = 90t2 - 12t + 1.5 0k (t) = 16t - 24t + 1.5 OT(t) = -535t24t + 1.6 These expressions are valid for 0 < t < 0.05s and the angles are in radians. The length of the feet are 0.2m, lower leg is 0.4m, of the upper leg is 0.5m, and the trunk is 0.85m. The mass of the feet are 1kg, of the lower legs is 6 kg, of the upper legs is 13, and of the trunk (which also includes the head and arms) is 45kg. The center-of-mass of each segment is located halfway along its length. a) What is the center-of-mass position at t=0.02s? b) What is the velocity of the lower leg center of mass at t=0.02s? [Answer: (a) 0 + 1.0 m, (b) 1.6 - 0.5 m/s] Figure for Problem 6