Question

Jake decided to walk from his room to the class 1.2 miles away. The class will start in 14.3 minutes. As he is walking, he is getting more and more concerned about making it on time. So he keeps walking ever a little faster, maintaining a tiny acceleration of 3.9 mm/s2. Is he going to be late after all? If you find that he will be late, answer by how many minutes. Perhaps he will arrive early? Then he did not need to exert himself quite so much! In this case, what is the smallest magnitude of acceleration that would still get him to class in time (the answer may be in either mm/s2, or in m/s2)

Answer 1

the distance to cover is d = 1.2 miles = 1931.21 m

time t= 14.3 min = 858 sec

and the acceleration is a = 3.9 mm/s^2 = 0.0039m/s^2

using the kinematic equation S = ut + 0.5 at^2

u = 0 m/s, S = 0.5 * a * t^2

S = 0.5 * 0.0039 * 858^2 = 1435.52 m, so the Jake will be late

then by using the same equation

d = 0.5 * a * t^2

1931.21 = 0.5 * 0.0039 * t^2

t = sqrt ( 1931.21 / (0.5 * 0.0039 )

t = 1081.56 sec

the time taken by the Jack is 1081.56 sec = 16.58 min

smallest accleration a = 2s/t^2

a = 2* 1435.52/1081.56^2

a = 2.45 mm/s^2