Question

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.

Answer 1

0.18 m/s²

Step-by-step 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