Question

In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min, 43.6 s43.6 s. Suppose "Pre" was at the 8.13 km8.13 km mark at a time of 25.0 min25.0 min. If he accelerated for 60 s60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s60 s interval. Assume his instantaneous speed at the 8.13 km8.13 km mark was the same as his overall average speed up to that time.

Answer 1

Total distance covered in the race $d=10\,km$

Total time taken to complete the race $t=27\,min\,43.6\,s$

Time taken to complete distance d1 = 8.13 km is $t_1=25\,min$

Average speed during this time is , u = d1/t = 8.13*10^3/(25*60) = 5.42 m/s

Remaining distance is d2+d3 = 10-8.13 = 1.87 km = 1870 m -----(1)

Distance $d_2$ is completed with an acceleration $a$ in a time $t_2=60\,s$

$d_2=ut_2+\frac{1}{2}at_2^2$ ----(2)

$a=\frac{v-u}{t_2}$ ---(3)

Distance $d_3$ is completed with average speed $v$ in a time $t_3=27min\,43.6\,s-25min-60\,s=1min43.6\,s=103.6\,s$

$d_3=v \,t_3$ ---(4)

using (2) and (4) in (1)

ut2+(1/2)at2^2+vt3 = 1870.....(5)

using equations (3) and (4)

ut2+(1/2)at2^2+(u+at2)t3 = 1870

a = (2*1870-2*5.42*60-2*5.42*103.6)/(60^2+2*60*103.6) = 0.1835 m/s^2 ....Answert