Question

Answer both for best answer

1. ) In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 7.85 km mark at a time of 25.0 . If he accelerated for 60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 7.85 km mark was the same as his overall average speed up to that time.

2.

You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators.

If state law mandates that elevators cannot accelerate more than 3.90 m/s or travel faster than 14.8 m/s, what is the minimum time in which an elevator can travel the 373 from the ground floor to the observatory floor?

Answer 1

1) At to = 25 min

v = 7850/(25*60) = 5.24 m/s

He had to cover = 10 - 7.85 = 2.15 km = 2150 m

In t = 27 min 43.6 s - 25 min = 163.6 s

After he accelerates for 60 s, he will have velocity V,

which he will run for the remaining (163.6 - 60)s = 103.6 s

We know that V = v + at = 5.24 m/s + a*60s

Distance covered while accelerating is

s = vt + 0.5at^2 = t*(v + at) = 60*(5.24 + 0.5* a*60)

Distance covered while coasting,

S = Vt = (5.24m/s + a*60s)*103.6s

= 542. 864 + a*6216

We know that s + S = 2150 m

2150 = 60*(5.24 + 0.5*a*60) + 542.864 + a*6216

2150 = 314.4 + 1800a + 542.864 + 6216a

a = 0.16 m/s^2