Question

n April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.57 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 7.57 km mark was the same as his overall average speed up to that time.

Answer 1

Answer:

$0.084\ m/s^2$

Explanation:

Given:

• Length of the race = 10 km
• Distance traveled by Steve in 25 min = 7.57 km
• Time interval for the constant acceleration = 60 s
• Initial velocity of Steve = 0 m/s

Assume:

• u = initial velocity
• v = final velocity
• t = time interval
• a = constant acceleration

Since the person has the instantaneous velocity at 25 min equivalent to the average velocity of the person during this time interval. So, let us find out the average velocity of the person till 25 min time interval.

$v_{avg} = \dfrac{7.57\ km}{25\ min}\\\Rightarrow v_{avg} = \dfrac{7.57\times 1000\ m}{25\times 60\ s}\\\Rightarrow v_{avg} = 5.047\ m/s$

Since the person moves with a constant acceleration for the first 60 s and then moves with a constant velocity after this instant. This means the final velocity of the person at the end of 60 s is 5.047 m/s.

$\therefore v = 5.047\ m/s\\u = 0\ m/s\\t = 60\ s$

Now using the equation of constant acceleration, we have

$v = u +at\\\Rightarrow a = \dfrac{v-u}{t}\\\Rightarrow a = \dfrac{5.047-0}{60}\\\Rightarrow a =0.084\ m/s^2$

Hence, the acceleration of Steve in the 60 s interval is $0.084\ m/s^2$.