Post a New Question

Physics Problem

posted by .

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 -

    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
    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.

  • Physics Problem -

    shut up!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question