Suppose that Secretariat ran the Kentucky Derby with times of 25.4 s, 24.0 s, 24
ID: 2026217 • Letter: S
Question
Suppose that Secretariat ran the Kentucky Derby with times of 25.4 s, 24.0 s, 24.0 s, and 22.5 s, for the quarter mile.
(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
(b) Assuming that Secretariat's instantaneous speed at the finish line was the same as the average speed during the final quarter mile, find his average acceleration for the entire race. (Hint: Recall that horses in the Derby start from rest.)
_____ ft/s2
Explanation / Answer
time taken for first quater mile is t1=25.4s time taken for first quater mile is t2=24.0s time taken for first quater mile is t2=24.0s time taken for first quater mile is t3=24.0s time taken for first quater mile is t3=24.0s time taken for first quater mile is t4=22.5s (a) 1 mile =5280 ft quarter mile = 1/4 mile = 5280/4 ft =1320 ft average speed during first quarter is v =distance/time =1320 /25.4 =51.968 ft/s average speed during second quarter is v =distance/time =1320 /24 =55 ft/s average speed during third quarter is v =distance/time =1320 /24 =55 ft /s average speed during fourth quarter is v =distance/time =1320 /22.5 =58.666 ft/s (b) from the equation of motion v = u +at..................(1) 'a' is acceleration throughout the journey v is final velocity = average velocity in fourth quarter =58.666ft /s u is the intial velocity horses starting from the rest u =0ft/s the total time taken to reach finish line is t =25.4+24+24+22.5 =95.9 s from equation (1) 58.666=0+a (95.9) acceleration a = 58.666/95.9 =0.611741 ft /s^2 time taken for first quater mile is t4=22.5s (a) 1 mile =5280 ft quarter mile = 1/4 mile = 5280/4 ft =1320 ft average speed during first quarter is v =distance/time =1320 /25.4 =51.968 ft/s average speed during second quarter is v =distance/time =1320 /24 =55 ft/s average speed during third quarter is v =distance/time =1320 /24 =55 ft /s average speed during fourth quarter is v =distance/time =1320 /22.5 =58.666 ft/s (b) from the equation of motion v = u +at..................(1) 'a' is acceleration throughout the journey v is final velocity = average velocity in fourth quarter =58.666ft /s u is the intial velocity horses starting from the rest u =0ft/s the total time taken to reach finish line is t =25.4+24+24+22.5 =95.9 s from equation (1) 58.666=0+a (95.9) acceleration a = 58.666/95.9 =0.611741 ft /s^2 average speed during second quarter is v =distance/time =1320 /24 =55 ft/s average speed during third quarter is v =distance/time =1320 /24 =55 ft /s average speed during fourth quarter is v =distance/time =1320 /22.5 =58.666 ft/s (b) from the equation of motion v = u +at..................(1) 'a' is acceleration throughout the journey v is final velocity = average velocity in fourth quarter =58.666ft /s u is the intial velocity horses starting from the rest u =0ft/s the total time taken to reach finish line is t =25.4+24+24+22.5 =95.9 s from equation (1) 58.666=0+a (95.9) acceleration a = 58.666/95.9 =0.611741 ft /s^2 average speed during third quarter is v =distance/time =1320 /24 =55 ft /s average speed during fourth quarter is v =distance/time =1320 /22.5 =58.666 ft/s average speed during fourth quarter is v =distance/time =1320 /22.5 =58.666 ft/s (b) from the equation of motion v = u +at..................(1) 'a' is acceleration throughout the journey v is final velocity = average velocity in fourth quarter =58.666ft /s u is the intial velocity horses starting from the rest u =0ft/s the total time taken to reach finish line is t =25.4+24+24+22.5 =95.9 s from equation (1) 58.666=0+a (95.9) acceleration a = 58.666/95.9 =0.611741 ft /s^2Related Questions
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