If such drivers are traveling on a straight road at 30 mph (44 ft/s) and their cars can decelerate at 27 ft/s², determine the shortest stopping distance d for a normal driver from the moment he or she see the pedestrians.

35.85 ft

Explanation:

initial velocity, u = 44 ft/s

acceleration, a = - 27 ft/s²

Let the stopping distance is d. the final velocity, v = 0 ft/s

Us third equation of motion

v² = u² + 2ad

0 = 44 x 44 - 2 x 27 x d

1936 = 54 d

d = 35.85 ft

Thus, the stopping distance is 35.85 ft.