A runner hopes to complete the 10000-m run in less than 30min. After running at constant speed for exactly 27min, there are still 1200m to go. The runner must then accelerate at 0.20m/s^2 for how many seconds in order to achieve the desired time?

Answer 1

Here is a solution from eight years ago by Mathmate. Note in his problem the distance remaining was 1100 meters, so adjust that for your 1200.

<< A runner hopes to complete the 10,000m run in less than 30 minutes. After exactly 27 minutes, there are still 1100m to go. The runner must then accelerate at 0.20 m/s^2 for how many seconds in order to achieve the desired time?

I get 80.9 but I know that the answer is 3.1 seconds. How do you get 3.1 seconds? (What formula/processes?) Thanks.

Physics Problem - MathMate Monday, August 31, 2009 at 1:04am

Assuming he ran at uniform speed for the first 27 minutes. Then initial speed,
u = (10000-1100)m/(27*60)sec. = 5.494 m/s
Remaining time is 3 minutes, or 180 s.
Distance to run, S = 1100m
a = acceleration = 0.2 m/s
let t=time of acceleration, then
new speed=u+at
time to run at new speed=(180-t) sec.
Distance run during acceleration
=ut+(1/2)at²
Summing up distance run in 3 minutes,
1100 = ut+(1/2)at² + (u+at)*(180-t)
which simplifies to
t² -360t +10000/9 = 0
From which t can be solved using the quadratic formula as
t=3.113 sec. or t=356.887 sec.
We reject the second solution because it exceeds the 180 sec. limit.