# Boundaries

• January 8th 2011, 11:17 AM
Natasha1
Boundaries
I never understand boundaries and would need some help on this question

In a race, Paula runs 25 laps of a track.
Each lap of the track is 400m, correct to the nearest metre.
Paula's average speed is 5.0 m/s, correct to one decimal place.

Calculate the upper bound for the time that Paula takes to run the race.
Give your answer in mins and seconds, correct ot the nearest second!
• January 8th 2011, 11:34 AM
snowtea
Upper bound for time is found using the lower bound on average speed and upper bound on track length.

I'm not sure what correct to one decimal place means.
If this means the digit is correct, then the average speed lower bound is 5.0m/s.
If it means correct rounded to 1 decimal place, then the average speed lower bound is 4.95 m/s.
If it means with an error of 0.1 m/s, then 4.9 m/s.

Upper bound on track distance is 401m.
• January 8th 2011, 11:38 AM
Natasha1
so I do 401 divide by 4.95 which gives me 81.01010101 mins/sec is this correct? And are the units correct too?
• January 8th 2011, 11:41 AM
snowtea
I think you have the right idea.
$time = distance / (avg\, speed)$, so you want to maximize distance and minimize average speed to get maximum time.
Units for time should be seconds.
• January 8th 2011, 11:43 AM
Natasha1
so time = 81 seconds
• January 8th 2011, 11:48 AM
snowtea
Quote:

Originally Posted by Natasha1
so time = 81 seconds

Close. Is 81 an upper bound of 81.010101... ?
• January 8th 2011, 11:50 AM
Natasha1
I don't know
• January 8th 2011, 12:09 PM
snowtea
Is $81 \geq 81.010101...$?
• January 8th 2011, 12:11 PM
Natasha1
strickly smaller
• January 8th 2011, 12:25 PM
snowtea
The question says compute an upper bound for time.
How can 81 seconds be an upper bound, when you computed something larger than 81 seconds for your answer.
• January 8th 2011, 12:27 PM
Natasha1
Got it! It's 82 dooooooooohhhhhh!