Upper and Lower Bounds. Need relative help. Thanks.?

Question

Upper and Lower Bounds?

Hey. I've got some questions on the topic of Upper and Lower bounds in which i have some difficulties with. All comments will be so much appreciated. Many thanks!

1. A car travels 5000m, correct to the nearest metre, in 2 minutes 21.7 seconds, correct to the nearest tenth of a second. Find the upper and lower bounds of the average speed in metres per second.

Regarding question 2 i found that at the end it requires you to divide certain upper bounds with certain lower bounds. I did convert the 2 minutes into 160 seconds and added it onto the 21.7. I found the upper and lower bounds by using add and sub and using 10 and 1/10 for the secnd. Why do you this to certain bounds? Also has Speed = Distance/Time got something to do with this?

Many thanks! All comments are so very much appreciated.

Answer 1

I believe this is the deal.

What you're actually given a range of values, right?  So the car travels 5000m, +/- 1 meter.  The time is 21.7, +/- .1 seconds.  So your bounds are:

4999-5001 and 21.6-21.8.  Then you're correct, you use the classic equation distance = rate* time (I just wrote it in the way I memorize it, but your way is fine too).  I'll write it as:

D = R*T

So we're looking for the highest and lowest speed.  Actually your form of R = D/T is more helpful to illustrate what these values will be.

To get the biggest value of R, we'd need a big D and a small T.  So to get the highest R, divide the biggest distance 5001 with the smallest time 21.6.  That's the upper bound.

To get the lowest value of R, we reverse things: we want the smallest D and the biggest T.  Divide 4999 / 21.8 and you get the lower bound.  

Hope this helps.