HPSLe 14.2010 - Lec 13 Continued Solutions to Brain Teasers Involving Time#1 Suppose one has a clock that loses one minute a day You start it on

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Lec.# 13 Continued- Solutions to Brain Teasers Involving Time #1. Suppose one has a clock that loses one minute a day. You start it on time, and suppose each year has 365 days. When will it next read the correct time? Solution: Lets assume an old fashion clock or watch that has 12 hours on it and no pm/am indicator, and forget daylight savings issues. Further suppose the clock loses time at a constant rate. If the clock reads 12:00 today at noon, tomorrow at the same real time it will read 11:59. In 60 days from today it will read 11:00 when the real time is 12:00. So in 12x60=720 days it will read 12:00 when it really is 12:00. Notice a ‘stopped clock’ reads the correct time twice a day so it is accurate much more often.
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#2. You approach a mountain range that is 30 miles to the summit and thirty miles down the other side to a nice restaurant. You go up the mountains at an average of 30 MPH (miles per hour). How fast would you have to average on the way down to the restaurant to average 45 MPH for the whole trip? Solution: Average speed =(Distance Traveled) /(Time Elapsed) Distance Traveled =60 Miles. Average Speed=45MPH Hours for trip Well you have already taken an hour to get up to the top, so you must get down in 1/3 hour, so Speed Down=30/(1/3)=90 MPH. #2B. What if you went up at 20MPH what speed must you go down at to average 60 MPH?!? 45 = 60/ T Þ T = 60/45 = 4 /3
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Solution to Brain Teasers With Time #3. A worm starts crawling on a strange rubber rope that is 1 kilometer long. The worm crawls at a steady pace of 1 centimeter (cm) per second ( there are 100,000 cm in a kilometer). After the first second, the rope stretches uniformly like a rubber band to 2 KM, after the next second to 3 KM, etc. (because the stretching is uniform, the rope stretches behind as well as in front of the worm, for example after the first second the rope is 2 KM and the worm’s 1cm has stretched to 2cm). Will the worm reach the other end? If so, how long will it take? Start End
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Solution to Worm’s Journey Let’s measure the worm’s progress after each sec. as a fraction of the distance needed. After the first he is 1/(100,000) there, after 2 sec. he covers 1/2x(100,000) of what’s needed, after 3 sec. 1/3x(100,000). After N sec. he has covered as a fraction of the required distance F N equal to: Now the problem of whether or not he arrives at the end is decided by whether or not the sum of all the fractions ever gets to 1 corresponding to completion of 100% of the journey. That will happen if there is some N* where ). 1 ( 000 , 100 1 1 = = N n N n F 000 , 100 * 1 ... 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 + + + + + + + + + N
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Solution for Traveling Worm The series in question is known as the harmonic series . You can take out a calculator and start adding up the terms. It turns out that they sum up to infinity, but you would not find that out in your lifetime! Here is a proof.
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This note was uploaded on 12/05/2010 for the course PSYCH Psy Beh F2 taught by Professor Williamh.batchelder during the Fall '10 term at UC Irvine.

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HPSLe 14.2010 - Lec 13 Continued Solutions to Brain Teasers Involving Time#1 Suppose one has a clock that loses one minute a day You start it on

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