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Your college quadrangle is 85 meters long and 66 meters wide. When you are late for class, you can walk at 7 miles per hour. You are at one corner of the quad and your class is directly opposite corner. How much time can you save by cutting across rather than going around the edge.
So i sketched it out
85^2+66^2=x^2 sqrt11581 = x 107.6=x
So in order to find how much time it saves we have to change 7 miles per hour to 6 meters per hour?
Yes, that is the distance which you cut accross. Now, how much distance does save you? You would have walked 85+66 without cutting, so subtract *(85+66) - √[88²+65²]*
how many meters***
this in bold is the distance travelled in miles, and then divide by 7 to find the number of hours saved.
http://www.wolframalpha.com/input/?i=%28%2885%2B65%29-%E2%88%9A%2888%C2%B2%2B65%C2%B2%29%29%2F7 yes just a bit less than 6
So 5.799 hours saved?
(If you would like we can do a by-hand approximation (I mean Taylor Polynomial) of the square root.....)
Wells this is an example problem from my engineering textbook. Was just wondering how to get the time saved, but why do we do *(85+66) - √[88²+65²]*
is it because 88+66 is the walking around the corner
and sqrt88^2+65^2 is the cut through? Also why do we square this?
well, (85+66) miles is the distance that takes you to walk without cuttting across, correct?
And *√[88²+65²] miles* is the distance that takes you when you cut across.
So, the distance you saved by walking across is (85+66) - √[88²+65²]
wait also where did the 88 come from?
88 should be 85
so that means that the hours saved would be 6.285
When you cut, you walk: *√[85²+66²] miles* Without cutting across *(85+66) miles* And then *(85+66) - √[85²+66²] miles* is the distance that you have saved by cutting across. THEN, the time your have saved [knowing that you walk 7 miles per hour] *( (85+66) - √[85²+66²] ) / 7*
Alright That makes a lot more sense putting it that way