Jake decided to walk from his room to his class 1.0 miles away. The class will s

Question

Jake decided to walk from his room to his class 1.0 miles away. The class will start in 13.5 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 3.2 mm/s^2. 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? In this case, what is the smallest magnitude of acceleration that will still get him to class on time? (Answer either in mm/s^2, or m/s^2)

Explanation / Answer

here

the distance to cover is d = 1 miles = 1609.34 m

time t= 13.5 min = 810 sec

and the acceleration is a = 3.2mm/s^2 = 0.0032m/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 * 0.0032 * 810^2

d = 1049.76 m

so the Jake will be late because he has to cover the distance of 1609.34 m but in covers only 1049.76 m

then by using the same equation

d = 0.5 * a * t^2

1609.34 = 0.5 * 0.0032 * t^2

t = sqrt ( 1609.34 / 0.5 * 0.0032 )

t = 1003 sec

so the time he will be late is 1003 - 810 =193 sec = 3.21 min

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