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January 27, 2015

January 27, 2015

Posted by **Zack** on Sunday, August 30, 2009 at 9:50pm.

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 1:04amAssuming 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.

- Physics Problem -
**Jeff**, Sunday, September 12, 2010 at 5:10pmshut up!

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