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| | From:  Vork- (Original Message) | Sent: 12/21/2006 5:44 PM |
Problem to solve, created by Vork: For his birthday, young Geekfurd Whiz received a digital car clock and a speedometer calibrated in feet per second. He rushed out to the garage to install his gifts in his homemade gocart. During a test run, Geekfurd accelerated at a constant rate along a trail parallel to the back of his dad's property. He noticed that as he came level with one corner of the property his velocity was 30 feet per second. Five seconds later he passed the other corner doing 40 feet per second. Hey, thought Geekfurd, now I know the width of the back of our property. Show the logic steps Geekfurd used to calculate the width of the property. |
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This one's gonna require some serious thought.... Problems like this have always been tough for me  ! |
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| | | From: Vork- | Sent: 12/22/2006 10:46 PM |
Hint: Find the average velocity across the back of the lot in feet per second. If you know average velocity and how many seconds elapse, you can calculate how many feet are covered. |
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| | | From: Vork- | Sent: 12/22/2006 11:00 PM |
Just so I don't give the answer I will use beans as an example. If I want to find the avearage length of several beans of varying lengths, I measure all of them, add their lengths together and divide by how many there are. That finds the same number of beans all the same length laid end to end that would equal the original beans' total length. |
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<Badger reaches for his calculator> |
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| | | From: Vork- | Sent: 12/24/2006 2:16 PM |
Everyone thinks differently, and visualization of how to solve problems comes more easily to some than others. I was lucky to have an excellent math and physics teacher in high school. I seem to have a knack for solving these problems. If I can show someone how to approach such problems it gives me a good feeling. I am not trying to show off my math skills to be king of the castle, so to speak. So, back to the average velocity. It is not the fastest nor the slowest speed, but the one exactly half way between. This is only for CONSTANT acceleration problems. If Geekfurd sped up, stopped part way, or went sideways, the problem would be insoluble unless you knew the details. |
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The back side is 200 ft? I can't remember how I came to that conclusion lol ... --L.B. |
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| | | From: Vork- | Sent: 12/29/2006 7:57 PM |
Close, Badger, and thanks for trying. The average velocity (speed in a certain direction) is the sum of the start and end velocities divided by 2 thus: Average velocity is (30 +40)/2= 35 feet per second.
The average velocity X the total time = the length, thus: 35 feet per second times 5 seconds is 175 feet. |
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Np Vork  ... It looks like I tried every way but the right way to solve that one! |
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| | | From: Vork- | Sent: 1/14/2007 9:36 PM |
Ok, I know this is not a high point for most people, but in the interest of completing the post: Geekfurd's acceleration was (40-30)/5 which is 2 feet per second per second. |
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Ty for the mindbender Vork  ! I sat there for quite a while drawing diagrams and trying different math solutions to try ta figure that thing out lol... --B. |
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