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Title: Linear motion problem? Post by: Fightboy on Oct 16, 2014 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 9.00-km mark at a time of 25.0 min even. If he accelerates for 60.0 s and maintains the increased speed for the duration of the race, calculate the acceleration that he had. Assume his instantaneous speed at the 9.00-km mark was the same as his overall average speed at that time.
I kept getting very confused on this problem because there were too many unknowns. I don't know his final speed after acceleration, the distance covered during his acceleration, or the distance covered after he was done accelerating. I've tried solving by using a system of equations but i kept getting to many unknown variables which led me too not get any answers. The only thing i pretty much got is that his instantaneous speed at the 9 km mark would be 9000m/1500 s= 6m/s. But i cant use that to solve for acceleration since we don't know the distance traveled. Can anyone help? Title: Re: Linear motion problem? Post by: bio_man on Oct 16, 2014 Similar 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 7.99-km mark at a time of 25.0 min. If he accelerates for 60 s and maintains the increased speed for the duration of the race , calculate the acceleration that he had. Assume his instantaneous speed at the 7.99-km mark was the same as his overall average speed up to that time. Number m/s2 Answer is in the second attachment. Title: Re: Linear motion problem? Post by: bio_man on Oct 16, 2014 Also, another one:
Quote In April 1974, Steve Prefontaine completed a 10-km race in a time of 27 min, 43.6 s. Suppose Steve was at the 8.27-km mark at a time of 25.0 min. If he accelerates for 60 s and maintains the increased speed for the duration of the race, calculate the acceleration that he had. Assume his instantaneous speed at the 8.27-km mark was the same as his overall average speed up to that time. Speed at 8.27 - km mark = (10 km)/(27 min*(60) + 43.6 s) = 25/4159 km/s. Now if he accelerates at x km/s^2, his speed after 60 seconds will be (25/4159 + 60x) km/s. Now if it takes Steve 27 min - 25 min = 2 mins to run the remaining (10 km - 8.27 km) = 1.73 km of the race at a rate of (25/4159 + 60x) km/s then it must be true that: 1.73/(25/4159 + 60x) = 60(2) 1.73 = 3000/4159 + 7200x x = 1.401 * 10^-4 km/s^2 Title: BFSF: Linear motion problem? Post by: joethefoe123 on May 29, 2023 TY :)
Title: BFSF: Linear motion problem? Post by: Addy Hickey on Sep 25, 2023 Thanks!
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