Question

From t = 0 to t = 3.72 min, a man stands still, and from t = 3.72 min to t = 7.44 min, he walks briskly in a straight line at a constant speed of 2.53 m/s. What are (a) his average velocity avg and (b) his average acceleration avg in the time interval 1.00 min to 4.72 min? What are (c) avg and (d) avg in the time interval 2.00 min to 5.72 min?

Answer 1

distance = speed * time
(a) avg speed = total_distance/total_time = (7.44-3.72)s*2.53m/s / (7.44-0)s = 1.265 m/s
(b) avg acceleration = change_in_speed/time
now just plug and chug