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anonymous

  • 5 years ago

question is, When is the particle moving forward (in the positive direction)? equation: v(t)= 3t^2-12t+9>0 answer = positive direction when t<1 and t>3 negative direction when 1<t<3 i cant understand how the math book gets those answers!!! please help (its a rate of change question)

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  1. anonymous
    • 5 years ago
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    Factor that polynomial you got there, to =3(t-1)(t-3)- then you find that when t<1 and when t>3, your equation is positive. this function is stating that when v(t) is postive, the particle is moving in the "positive" direction and when the function is negative, it's moving in the "negative" direction.

  2. anonymous
    • 5 years ago
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    factor out the 3 from the polynomial, then factor it like any other polynomial. Or use the AC method if you're comfortable with it.

  3. anonymous
    • 5 years ago
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    maybe im factoring wrong, but can someone go through the factoring for me?

  4. anonymous
    • 5 years ago
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    v(t) = 3t^2 - 12t + 9 v(t) = 3 * (t^2 - 4t + 3) v(t) = 3 * (t - 1)(t - 3)

  5. anonymous
    • 5 years ago
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    i got to that, but then wouldnt it be t>1 and t>3 ?

  6. anonymous
    • 5 years ago
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    not t<1 and t>3

  7. anonymous
    • 5 years ago
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    It's when the quantities (t-1) and (t-3) are both either negative or positive. Because then you have 3 * a positive scalar quantity. So when t > 3, both (t-1) and (t-3) will be positive. When t < 1 both will be negative. It can't be when t>1 and t>3, because that would be the same as saying t>1. You only have one t, it can't be two things at once. Plus, think about when t=2. (t-2) is positive but (t-3) is negative. It will be traveling in the negative direction. Hope that helps clear things up a bit. :)

  8. anonymous
    • 5 years ago
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    oh alright, i got it!! thanks soo much!

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