In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.71 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 7.71 km mark was the same as his overall average speed up to that time.

0.18 m/s²

Explanation:

The total time taken to cover a 10 km race is 27 min, 43.6 seconds.

At 25 min, Pre was at 7.71 km mark. Therefore the average speed = 7.71 km / 25 min = 7710 m / (25 * 60) s = 5.14 m/s

The distance remaining = 10 km - 7.71 km = 2.29 km = 2290 m

The remaining time = 27 min, 43.6 seconds - 25 min = 2 min 43.6 second = 163.6 seconds

She accelerates for 60 seconds, therefore the distance covered (S) during the acceleration (a) is:

S₁ = 5.14(60) + 0.5a(60)² = 308.4 + 1800a

She maintains the speed for the remaining distance (S₂). The remaining time = 163.6 seconds - 60 seconds = 103.6 seconds. The final speed after the acceleration = (5.14 + 60a) m/s

S₂ = (5.14 + 60a)* 103.6 = 532.5 + 6216a

S₁ + S₂ = 2290 m

(308.4 + 1800a) + (532.5 + 6216a) = 2290

8016a + 840.9 = 2290

8016a = 1449.1

a = 0.18 m/s²

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)

there are no contact forces mentioned on the list of choices that you've provided.