Napril 1974, steve prefontaine completed a 10 km race in a time of 27 min , 43.6 s . suppose "pre" was at the 7.57 km mark at a time of 25.0 min . 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 7.57 km mark was the same as his overall average speed up to that time.

Explanation:

Given:

Assume:

Since the person has the instantaneous velocity at 25 min equivalent to the average velocity of the person during this time interval. So, let us find out the average velocity of the person till 25 min time interval.

Since the person moves with a constant acceleration for the first 60 s and then moves with a constant velocity after this instant. This means the final velocity of the person at the end of 60 s is 5.047 m/s.

Now using the equation of constant acceleration, we have

Hence, the acceleration of Steve in the 60 s interval is .

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day