In April 1974, Steve Prefontaine completed a 10-km race in a time of 27 min, 43.
ID: 1525526 • Letter: I
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 8.13-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 8.13-km mark was the same as his overall average speed up to that time.
PS: Please double check your work guys, I've been having trouble with this one and would like to see the process behind solving this. It should be noted that some of the other posts with answers on here have been wrong.
Explanation / Answer
Speed at 8.13 Km
u=8130/(25*60) =5.42 m/s
Lets us suppose final velocity after 60 sec from 8.13 km is v
v-u=at v=5.42 + a*60
v=5.42 + 60a
also S1=ut+(1/2)at2
= 5.42*60 + (0.5)a*3600
=1800a + 325.2 m
Now the remaining time to complete rest race t=(27*60 + 43.6)-60 -(25*60) =103.6 sec
now distance travelled during ths time will be S2=v*103.6 m =561.51 + 6216a m
now total distance travelled is S1 +S2 +8130 =10000
1800a + 325.2 +561.51 + 6216a + 8130 = 10000
8016a =983.29
a = 0.123 m /s2
=1590-917.21 a=0.084 m/s2