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 accelerated for 60 s and then maintained his increased speed for the remainder of the race. calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 8.13 km mark was the same as his overall average speed up to that time.

Answer 1

Answer:$0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance$=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$