Question

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 7.15 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 7.15 km mark was the same as his overall average speed up to that time. _____m/s^2

Answer 1

Answer:

  0.2583 m/s²

Step-by-step explanation:

The relationship between speed, distance, and time is ...

  speed = distance/time

Of course, the relationship between the various units of measure is ...

• 1 km = 1000 m
• 1 min = 60 s

The average speed for the first 7.15 km was ...

  (7150 m)/(1500 s) = 4 23/30 m/s

The acceleration is considered to be uniform over the 60-second period. The distance traveled during that time will be the same as the distance traveled at the initial speed for 30 s plus that traveled at the final speed for 30 s. That is, we can compute the final speed as though "Pre" ran 25.5 minutes (1530 s) at his initial speed and the remaining time (133.6 s) at his final speed.

The distance for 25.5 minutes at 4 23/30 m/s is ...

  (1530 s)(143/30 m/s) = 7293 m

The remaining distance is then ...

  10,000 -7,293 = 2,707 . . . meters

And the final speed is ...

  (2707 m)/(133.6 s) ≈ 20.262 m/s

The acceleration is the change in speed divided by the time period:

  (20.262 m/s -4.767 m/s)/(60 s) ≈ 0.2583 m/s²

_____

Comment on these numbers

According to this scenario, Pre's speed for the last 2000 meters, sustained for more than a minute, was more than 20 m/s -- about double the world record speed for a 100 meter sprint. It was a little more than 4 times his speed for the first 25 minutes.