Secretariat ran the 1.25 mi long Kentucky Derby with timesof 25.1 s, 23.6 s, 23.

Question

Secretariat ran the 1.25 mi long Kentucky Derby with timesof 25.1 s, 23.6 s, 23.7 s, 24.0 s, and 22.9 s for each of the quarter mile segments. (a) Find his average speed during each quarter-mile segment. first segment ft/s second segment ft/s third segment ft/s fourth segment ft/s fifth segment 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,anduse vf = vi + at.)
ft/s2 (a) Find his average speed during each quarter-mile segment. first segment ft/s second segment ft/s third segment ft/s fourth segment ft/s fifth segment 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,anduse vf = vi + at.)
ft/s2 first segment ft/s second segment ft/s third segment ft/s fourth segment ft/s fifth segment ft/s

Explanation / Answer

(a)The total distance raned by Secretariat is S = 1.25 mi =1.25 * 5280 ft = 6600 ft The length of each quarter-mile segment is S1 =(S/5) = (6600/5) ft = 1320 ft The average speed duing first quarter-mile segment is u1 = (S1/t1) t1 = 25.1 s or u1 = (1320/25.1) = 52.5 ft/s The average speed duing second quarter-mile segmentis u2 = (S2/t2) t2 = 23.6 s or u2 = (1320/23.6) = 55.9 ft/s The average speed duing third quarter-mile segmentis u3 = (S3/t3) t3 = 23.7 s or u3 = (1320/23.7) = 55.7 ft/s The average speed duing fourth quarter-mile segmentis u4 = (S4/t4) t4 = 24.0 s or u4 = (1320/24.0) = 55 ft/s The average speed duing fifth quarter-mile segmentis u5 = (S5/t5) t5 = 22.9 s or u5 = (1320/22.9) = 57.6 ft/s (b)The average speed for the first four quarter-mile segmentsis uavg = (u1 * t1 +u2 * t2 + u3 * t3 +u4 * t4 /t1 + t2 +t3 + t4) or uavg = (52.5 * 25.1 + 55.9 * 23.6 + 55.7 * 23.7+ 55 * 24.0/25.1 + 23.6 + 23.7 + 24.0) = 54.7 ft/s Let the average acceleration for the entire race be atherefore we get u5 = uavg + at where t = t1 + t2 + t3 +t4 + t5 = 25.1 + 23.6 + 23.7 + 24.0 + 22.9 =119.3 s or a = (u5 - uavg/t) = (57.6 -54.7/119.3) = 0.0243 ft/s2 The average speed duing second quarter-mile segmentis u2 = (S2/t2) t2 = 23.6 s or u2 = (1320/23.6) = 55.9 ft/s The average speed duing third quarter-mile segmentis u3 = (S3/t3) t3 = 23.7 s or u3 = (1320/23.7) = 55.7 ft/s The average speed duing fourth quarter-mile segmentis u4 = (S4/t4) t4 = 24.0 s or u4 = (1320/24.0) = 55 ft/s The average speed duing fifth quarter-mile segmentis u5 = (S5/t5) t5 = 22.9 s or u5 = (1320/22.9) = 57.6 ft/s (b)The average speed for the first four quarter-mile segmentsis uavg = (u1 * t1 +u2 * t2 + u3 * t3 +u4 * t4 /t1 + t2 +t3 + t4) or uavg = (52.5 * 25.1 + 55.9 * 23.6 + 55.7 * 23.7+ 55 * 24.0/25.1 + 23.6 + 23.7 + 24.0) = 54.7 ft/s Let the average acceleration for the entire race be atherefore we get u5 = uavg + at where t = t1 + t2 + t3 +t4 + t5 = 25.1 + 23.6 + 23.7 + 24.0 + 22.9 =119.3 s or a = (u5 - uavg/t) = (57.6 -54.7/119.3) = 0.0243 ft/s2 The average speed duing third quarter-mile segmentis u3 = (S3/t3) t3 = 23.7 s or u3 = (1320/23.7) = 55.7 ft/s The average speed duing fourth quarter-mile segmentis u4 = (S4/t4) t4 = 24.0 s or u4 = (1320/24.0) = 55 ft/s The average speed duing fifth quarter-mile segmentis u5 = (S5/t5) t5 = 22.9 s or u5 = (1320/22.9) = 57.6 ft/s (b)The average speed for the first four quarter-mile segmentsis uavg = (u1 * t1 +u2 * t2 + u3 * t3 +u4 * t4 /t1 + t2 +t3 + t4) or uavg = (52.5 * 25.1 + 55.9 * 23.6 + 55.7 * 23.7+ 55 * 24.0/25.1 + 23.6 + 23.7 + 24.0) = 54.7 ft/s Let the average acceleration for the entire race be atherefore we get u5 = uavg + at where t = t1 + t2 + t3 +t4 + t5 = 25.1 + 23.6 + 23.7 + 24.0 + 22.9 =119.3 s or a = (u5 - uavg/t) = (57.6 -54.7/119.3) = 0.0243 ft/s2 The average speed duing fourth quarter-mile segmentis u4 = (S4/t4) t4 = 24.0 s or u4 = (1320/24.0) = 55 ft/s The average speed duing fifth quarter-mile segmentis u5 = (S5/t5) t5 = 22.9 s or u5 = (1320/22.9) = 57.6 ft/s (b)The average speed for the first four quarter-mile segmentsis uavg = (u1 * t1 +u2 * t2 + u3 * t3 +u4 * t4 /t1 + t2 +t3 + t4) or uavg = (52.5 * 25.1 + 55.9 * 23.6 + 55.7 * 23.7+ 55 * 24.0/25.1 + 23.6 + 23.7 + 24.0) = 54.7 ft/s Let the average acceleration for the entire race be atherefore we get u5 = uavg + at where t = t1 + t2 + t3 +t4 + t5 = 25.1 + 23.6 + 23.7 + 24.0 + 22.9 =119.3 s or a = (u5 - uavg/t) = (57.6 -54.7/119.3) = 0.0243 ft/s2

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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