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Use parametric equations to find x for a given value of y. Part of HELM Book 2.2.2.

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A particle with mass $m$ falls under gravity so that at time $t$ its distance from the $y$-axis is $2t$ and its distance from the $x$-axis is $−mg\\dfrac{t^2}{2}+ 3$ where $g$ is a constant (the acceleration due to gravity). Find the value of $t$ when the particle crosses the $x$-axis and, at this time, find the distance from the $y$-axis.

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(a) $x=2t$, $y=-mg\\dfrac{t^2}{2}+3$

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(b) \\begin{align} 0&= -mg\\frac{t^2}{2}+3\\\\ -3 &= -mg\\frac{t^2}{2}\\\\ \\frac{3}{mg} &= \\frac{t^2}{2}\\\\ \\frac{6}{mg}&=t^2\\\\\\therefore\\, t &= \\sqrt{\\frac{6}{mg}}\\qquad\\textrm{since } t>0 \\end{align}

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(c) $x = 2\\sqrt{\\dfrac{6}{mg}}$

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Begin by obtaining the parametric equations of the path of the particle:

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(Note, to enter $−mg\\dfrac{t^2}{2}+ 3$, type -m*g*t^2/2+3)

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$x=$ [[0]], $\\qquad y = $ [[1]]

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Now find the value of $t$ when $y=0$

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(To enter a square root, type sqrt() e.g. sqrt(6) for the positive square root of 6)

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Finally, obtain the value of $x$ at this value of $t$:

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