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Thread: particle motion

  1. #16
    Grand Panjandrum
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    I have moved this thread to calculus, since no one is interested in the per-calculus solution.

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  2. #17
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    Re: particle motion

    Quote Originally Posted by euclid2 View Post
    a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
    Can u do the problem


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  3. #18
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    Re: particle motion

    Quote Originally Posted by abdurrahmanzafar View Post
    Can u do the problem


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    why are you reviving a thread that is over 7 years old?
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  4. #19
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    Re: particle motion

    Quote Originally Posted by skeeter View Post
    why are you reviving a thread that is over 7 years old?
    I'm new to this app honestly i don't really no what to do i just needed help on that problem


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  5. #20
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    Re: particle motion

    The derivative of the position function $d=t^2-8t+15$ is velocity ... the particle is at rest when velocity is zero.

    Find $v(t)=d(t)$, set equal to zero, and solve for $t$.
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  6. #21
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    Re: particle motion

    Quote Originally Posted by skeeter View Post
    The derivative of the position function $d=t^2-8t+15$ is velocity ... the particle is at rest when velocity is zero.

    Find $v(t)=d(t)$, set equal to zero, and solve for $t$.
    Got it thanks


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  7. #22
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    Re: particle motion

    Without using Calculus- since the position function $x= t^2- 8t+ 15$, is quadratic complete the square:
    $x= t^2- 8t+ 16- 16+ 15= (t- 4)^2- 1$. That is a parabola, opening upwards, with vertex at (4, -1). If t< 4, as t increases, x goes downward. The velocity is negative. If t< 4, as t increases, x goes upward. The velocity is positive. At the vertex, t= 4, the velocity is 0.
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  8. #23
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    Re: particle motion

    Quote Originally Posted by HallsofIvy View Post
    Without using Calculus- since the position function $x= t^2- 8t+ 15$, is quadratic complete the square:
    $x= t^2- 8t+ 16- 16+ 15= (t- 4)^2- 1$. That is a parabola, opening upwards, with vertex at (4, -1). If t< 4, as t increases, x goes downward. The velocity is negative. If t< 4, as t increases, x goes upward. The velocity is positive. At the vertex, t= 4, the velocity is 0.
    Thanks a bunch for the answer


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