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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 7.43 km mark at a time of
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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 7.43 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 7.43 km mark was the same as his overall average speed up to that time.
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2021-09-02T19:33:29+00:00
2021-09-02T19:33:29+00:00 1 Answers
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Answer:
Acceleration, a = 0.101 m/s²
Explanation:
Average speed = total distance / total time.
At the 7.43km mark, total distance = 7.43km or 7430m
Total time = 25 * 60 s = 1500s
Average speed = 7430m/1500s = 4.95m/s
He then covers (10 – 7.43)km = 2.57 km = 2570 m
in t = 27m43.6s – 25min = 2m43.6s = 163.6 s
Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 – 60)s = 103.6 s.
From V = u + at; V = 4.95m/s + a *60s
Distance covered while accelerating is
s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²
Distance covered while at constant velocity, v after accelerating is
D = velocity * time
Where v = 4.95m/s + a*60s
D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²
Total distance covered after initial 7.43 km, S + D = 2570 m, so
2570 m = 297m + a*1800s² + 512.82m + a*6216s²
2570 = 809.82 + a*8016
a = 809.82m / 8016s² = 0.101 m/s²