Jake decided to walk from his room to the class 1.2 miles away. The class will s

Question

Jake decided to walk from his room to the class 1.2 miles away. The class will start in 13.6 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

Here the distance to cover is d = 1.2 miles = 1931.208 m

The time t= 13. min = 816 sec

And the acceleration is a = 9mm/s^2 = 0.009 m/s^2

Here by using the second equation

d = ut + 0.5 * a* t^2

u = 0 m/s

d = 0.5 * a * t^2

d = 0.5 * 0.009 * 816^2

d = 2996.35 m

so the Jake will not be late then by using the same equation

The smallest magnitude of acceleration that took him to class on time is,

a=d/0.5*t^2=1931.208 m/0.5* 816^2=3.6 mm/s^2

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