a. For the first 2 seconds, while acceleration is constant, d = ½at2 Substituting the given values d = 10 meters, t = 2 seconds gives a = 5 m/s2
b. The velocity after accelerating from rest for 2 seconds is given by v = at, so v = 10 m/s
c. The displacement, time, and constant velocity for the last 90 meters are related by d = vt. To cover this distance takes t = d/v = 9s. The total time is therefore 9 + 2 = 11 seconds
