Secretariat ran the 1.25 mi long Kentucky Derby with times of25.2 s, 23.7 s, 24.

Question

Secretariat ran the 1.25 mi long Kentucky Derby with times of25.2 s, 23.7 s, 24.8 s,23.4 s, and 23.5 s for each of the quarter mile segments. (a) Find his average speed during eachquarter-mile segment. first segment 1 ft/s second segment 2 ft/s third segment 3 ft/s fourth segment 4 ft/s fifth segment 5 ft/s
(b) Assuming that Secretariat's instantaneous speed at the finishline was the same as the average speed during the final quartermile, find his average acceleration for the entire race.(Hint: Recall that horses in the Derby start from rest,and use vf = vi +at.)
6 ft/s2 (a) Find his average speed during eachquarter-mile segment. first segment 1 ft/s second segment 2 ft/s third segment 3 ft/s fourth segment 4 ft/s fifth segment 5 ft/s
(b) Assuming that Secretariat's instantaneous speed at the finishline was the same as the average speed during the final quartermile, find his average acceleration for the entire race.(Hint: Recall that horses in the Derby start from rest,and use vf = vi +at.)
6 ft/s2 first segment 1 ft/s second segment 2 ft/s third segment 3 ft/s fourth segment 4 ft/s fifth segment 5 ft/s

Explanation / Answer

Average speed = distance/time A mile is 5280 feet, so a quarter mile is 5280/4 = 1320 feet. Thus,the average speeds per segment are: 1. 1320/25.2 = 52.38 ft/s 2. 1320/23.7 = 55.70 ft/s 3. 1320/24.8 = 53.23 ft/s 4. 1320/23.4 = 56.41 ft/s 5. 1320/23.5 = 56.17 ft/s The entire race took him 25.2 + 23.7 + 24.8 + 23.4 + 23.5 = 120.6seconds. It took him 120.6 seconds to accelerate from 0 to 56.17ft/s. Using vf = vi + at, we have 56.17 = 0 + a*120.6 a = 56.17/120.6 = .466 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.