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Thread: Maximum Velocity

  1. #1
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    Maximum Velocity

    Hi everyone. I'm currently working on a velocity problem. Here's what I did:

    A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t)=12t^2-36t+15. At t = 1, the particle is at the origin

    Find the maximum velocity of the particle for 0≤t≤2.

    V'=24t-36
    24t-36=0
    6(4t-6)=0
    t=3/2

    I drew a sign chart, and I found that 3/2 is actually the minimum. How can I get the maximum velocity?

    Thanks a lot.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You might want to make a sketch of that vt graph to help you.

    $\displaystyle v(t)=12t^2-36t+15 = 3(4t^2 - 12t +5) = 3(2t - 1)(2t - 5)$

    The x intercepts are at t = 0.5 and t = 2.5

    Draw the graph and you'll see that the maximum will occur at t = 0.
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  3. #3
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    Thanks a lot.
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  4. #4
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    Hello, zhangxupage2!

    $\displaystyle \text{A particle moves on the }x\text{-axis so that its velocity at any time }t \ge 0$
    $\displaystyle \text{is given by: }\: v(t)\:=\:12t^2-36t+15$
    $\displaystyle \text{At }t = 1\text{, the particle is at the origin.}$
    $\displaystyle \text{Find the maximum velocity of the particle for }0 \le t \le 2.$

    $\displaystyle v' \:=\:24t-36 $
    $\displaystyle 24t-36\:=\:0 \quad\Rightarrow\quad t \,=\,\frac{3}{2}$

    $\displaystyle \text{I drew a sign chart, and I found that }\frac{3}{2}\text{ is actually the minimum.}$ . Right!
    $\displaystyle \text{How can I get the maximum velocity?}$

    We know that the velocity function is a parabola.

    And you found its vertex (minimum velocoty).

    . . $\displaystyle f\left(\frac{3}{2}\right) \:=\:12(\frac{3}{2})^2 - 36(\frac{3}{2}) + 15 \:=\:-12 $


    So test the endpoints of the given interval.

    . . $\displaystyle v(0) \:=\:12(0^2) - 26(0) + 15 \:=\: 15$

    . . $\displaystyle v(2) \:=\:12(2^2) - 36(2) + 15 \:=\:-9$

    Therefore . . .

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