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quickquestion

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« on: August 06, 2014, 06:47:01 pm »
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My two dogs were running on the beach when I called them back. The faster dog was 100m away and the slower dog was 70 m away. The faster dog runs twice as fast as the slower dog. How far away was the second dog when the first dog reached me?

keltingmeith

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« Reply #1 on: August 06, 2014, 07:17:08 pm »
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Let's call the speed of the slower dog s, and the distance they are from you d. It then follows that the speed of the faster dog is 2s, however we can still the distance d.

This means we have the equations $d=2s + a$ and $d=s + b$, where a and b refer to the dog's initial position. The faster dog is 100m away, and the slower dog is 70m away. Since we know that these linear equations are RISING, and we want them to get closer to 0, we will take these as negative values. This gives us equations $d=2s-100$ and $d=s-70$. Now, we want to know where the second dog is when the first dog reaches us.

When the first dog reaches us, it will have a distance of 0m - so, we solve this for s: $0=2s-100 \implies s=50m/s$. This means that the second dog has a speed of 50m/s when the first dog has reached this. Subbing this in, we get $d=50-70=-20m$, which means that the dog is 20m away.

I know that all the negative signs are very confusing, however it's basically impossible to understand why we use negative symbols in these situations with having at least a basic knowledge of vectors, so you'll have to just take this as gospel for now.

smile123

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« Reply #2 on: November 20, 2016, 09:25:06 pm »
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RuiAce

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