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

magnityde and direction

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.