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 8.27 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 8.27 km mark was the same as his overall average speed up to that time. acceleration: 0.181 m/s2 contact us help 8

Answer 1

Speed =Distance/time

Initial speed (Speed at 8.27km mark) =8.27km/25min

$u=0.3308km/min= 5.513m/s$

Accelerated for 60 s to achieve increased speed

$v=u+at$

$v=5.513m/s+a*60s$

${\color{Red} v=5.513+60a}$ -----------------(1)

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Distance covered while accelerating

$s=ut+1/2at^{2}$

$s=5.513m/s*60s+1/2a(60s)^{2}$

${\color{Red} s=330.78+1800a}$ -----------(2)

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Time remaining = 27min 43.6s - 25min -60s =103.6s
Remaining distance =10km -8.27km - s=d

$1730m-s=d$

Remaining distance is covered  using increased speed

$v=d/103.6$

$v=(1730-s)/103.6$

From (1)   $5.513+60a =(1730-s)/103.6$

$(5.513+60a) 103.6=1730-s$

$s=1730-(5.513+60a) 103.6$ -------------(3)

Put (3) in (2)

$1730-(5.513+60a) 103.6=330.78+1800a$

$1730-571.147-6216a=330.78+1800a$

$1730-571.147-330.78=1800a+6216a$

$828.073=8016a$

ANSWER: ${\color{Red} a=0.1033m/s^{2}}$

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