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# Linear motion problem?

wrote...
Posts: 29
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9 years ago
 Linear motion problem? 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? Read 2794 times 4 Replies

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wrote...
Educator
9 years ago
 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/s2Answer is in the second attachment. Attached file(s)  Thumbnail(s): You must login or register to gain access to these attachments.
wrote...
Educator
9 years ago
 Also, another one:QuoteIn 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
wrote...
5 months ago
 TY :)
wrote...
2 months ago
 Thanks!
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