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Thread: velocity acceleration problem

  1. #1
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    velocity acceleration problem

    velocity acceleration problem-van.jpg
    The van travels over the hill described by y = (-0.0015x^2 + 15) ft. If it has a constant speed of 75 ft/s, determine the x and y components of the van's velocity and acceleration when x = 50 ft.

    This is what I found:

    v_y = -0.15v_x

    v = \sqrt{v_x^2 + v_y^2}

    v_x = 74.17 ft/s \leftarrow

    v_y = 11.13 ft/s \uparrow

    a_y = -16.504 - 0.15a_x

    Now, how to complete the answer?
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  2. #2
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    Re: velocity acceleration problem

    Starting from scratch I get the following.

    let $c=-0.0015$

    let $\dot{u} = \dfrac{du}{dt}$, i.e. a dot above a variable represents it's time derivative.

    Note that $x$ is moving in the negative direction.

    $y = c x^2 + 15$

    $\dfrac{dy}{dx} = 2cx$

    $\dot{y} = \dfrac{dy}{dx}\dot{x} = 2 c x \dot{x}$

    $75^2 = \dot{x}^2 + \dot{y}^2 = (1+4 c^2 x^2)\dot{x}^2$

    $v_x = \dot{x} = -\dfrac{75}{\sqrt{1+4c^2 x^2}}$

    $v_y = \dot{y} = \dfrac{150 c x}{\sqrt{1+4 c^2 x^2}}$

    $a_x = \dot{v_x} = \dfrac{dv_x}{dx}v_x$

    $a_y = \dot{v_y} = \dfrac{d v_y}{dx} v_x$

    $a_x = -\dfrac{600 c^2 x}{\left(4 c^2 x^2+1\right)^2} \dfrac{75}{\sqrt{1+4c^2 x^2}}=-\dfrac{45000 c^2 x}{\left(4 c^2 x^2+1\right)^3}$

    $a_y = -\dfrac{150 c}{\left(4 c^2 x^2+1\right)^{3/2}}\dfrac{75}{\sqrt{1+4c^2 x^2}}= -\dfrac{11250 c}{\left(4 c^2 x^2+1\right)^{5/2}}$

    plugging in $c$ and $x=50$ we obtain

    $\vec{a} = (a_x,a_y) = \{-4.7356,-15.9619\}~ft/s^2$
    Thanks from HallsofIvy
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  3. #3
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    Re: velocity acceleration problem

    ignore this post, it seems to be incorrect.
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  4. #4
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    Re: velocity acceleration problem

    $\vec{v} = \left<|v|\cos{\theta},|v|\sin{\theta}\right>$, where $\theta$ is the angle relative to the positive x-axis of the line tangent to the path at $x=50$ ...

    $y' = -0.003x \implies y'(50) = -0.15 \implies \theta = \pi + \arctan(-0.15)$

    $\vec{v} \approx \left<-74.17,11.13\right>$


    $\vec{a} = \left<\dfrac{v^2}{r}\cos{\phi},\dfrac{v^2}{r}\sin{ \phi}\right>$, where $r = \dfrac{\left[1+(y')^2\right]^{3/2}}{|y''|}$ and $\phi = \theta + \dfrac{\pi}{2}$

    $r = \dfrac{(1+0.0225)^{3/2}}{.003} \approx 344.64 \, ft$

    $\vec{a} = \dfrac{75^2}{r} \approx 16.32 \, ft/s^2$

    $a_x = \dfrac{75^2}{r} \cos{\phi} \approx -2.42 \, ft/s^2$

    $a_y = \dfrac{75^2}{r} \sin{\phi} \approx -16.14 \, ft/s^2$
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  5. #5
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    Re: velocity acceleration problem

    Quote Originally Posted by romsek View Post
    ignore this post, it seems to be incorrect.
    actually the correction was correct.

    velocity acceleration problem-clipboard01.jpg
    Thanks from skeeter
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  6. #6
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    Re: velocity acceleration problem

    Thanks romsek, fantastic explanation (I always love to solve using this way)

    thanks skeeter, beautiful explanation by using the radius of curvature.
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