Jake decided to walk from his room to the class 1.4 miles away. The class will s

Question

Jake decided to walk from his room to the class 1.4 miles away. The class will start in 14.2 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 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).

Explanation / Answer

Given,

Distance = S = 1.4 miles = 2253 meters.

time = T = 14.2 minutes = 852 sec ; a = 9 mm/s2 = 0.009 m/s2

Speed of jack will be:

v = u + at = 0 + 0.009 x 852 = 7.67 m/s

We now need to know the acceleration :

v2 = u2 + 2 a S

a = 7.67 x 7.67 / 2 x 2253 = 0.013 m/s2

We need to ckeck if, with this acceleration jack can reach on time or not:

S = ut + 1/2 at2

t = sqrt (2253 x 2 / 0.013) = 588.7 sec

Hence t = 588.7 sec < T = 852 sec.

Hence, Jack needs the acceleration of a' = 0.013 m/s^2 = 13 mm/s^2 to reach before time in the class.

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