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silverxx

  • 8 months ago

kinamatics

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  1. silverxx
    • 8 months ago
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    the dog from rest accelerates 5m/s^2 instantly at the sight of the boy and runs after a very scared boy. since the dog chases him, he accelerates 2 m/s^2 running from 10m/s seped. how far a distance the dog can overtakes the boy?

  2. silverxx
    • 8 months ago
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    ow thats speed not seped.

  3. theEric
    • 8 months ago
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    This one is not ambiguous, thankfully. It looks like it was translated from another language, though. Do you understand the situation?

  4. silverxx
    • 8 months ago
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    yes but no idea how to answer this one.

  5. theEric
    • 8 months ago
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    Actually, there is missing information. How far apart are the boy and the dog in the beginning? Without this, we can't find an exact answer. The answer would have a variable in it to account for this distance.

  6. theEric
    • 8 months ago
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    Is there any more information in that problem?

  7. silverxx
    • 8 months ago
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    yay. thats the only infos on the problem.

  8. theEric
    • 8 months ago
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    What I would do is come up with two equations: one to describe the position of the dog, one to describe the position of the boy. Then set them equal to each other, which makes the equation true for only the one time at which the dog catches the boy. When you find that time, you can plug it into the equation for either the dog's or boy's position equation to find out the position where they meet. Then you can find how far that is from where the dog started.

  9. theEric
    • 8 months ago
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    But time will depend on how far apart the dog is from the boy in the beginning. So you start of with your basic \(x=x_0+v_0t+\frac12at^2\) equation. One for the dog, one for the boy.

  10. theEric
    • 8 months ago
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    So you'll plug in the values from the problem. Can you do that?

  11. theEric
    • 8 months ago
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    Well, \(x_0\) for the boy will be unknown, so you let it be a variable. I think you'll be able to use the quadratic formula to get \(t\). And then you can plug that \(t\) into the equation for the dog's position. I have to go. Take care!

  12. silverxx
    • 8 months ago
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    thank you so much =)

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