For a certain interval of motion, the pin P (see figure below) is forced to move in the

fixed parabolic slot by the vertical slotted guide, which moves in the x-direction at a constant rate *v**x* = 20 mm/s. All measurements are in millimeters and seconds. Calculate the magnitudes of the velocity **v** and acceleration **a** of pin P when x = 60 mm.

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# solution:-y=\frac{x^{2}}{160}Given, velocity in x-direction \left(V_{x}\right)=20 \mathrm{~mm} / \mathrm{s} (constant) at x=60 \mathrm{~mm}, \quad v=?, \quad a= ?\begin{aligned}\because \vec{v} & =v_{x} i+v_{y} j \\v_{x} & =20 \mathrm{~mm} / \mathrm{sec} \\v_{y} & =\frac{d y}{d t}=\frac{d}{d t}\left(\frac{x^{2}}{160}\right)=\frac{(2 x)}{160} \frac{d x}{d t} \\\because v_{x} & =\frac{d x}{d t}=20 \mathrm{~mm} / \mathrm{s}, \quad x=60 \mathrm{~mm} \\\therefore v_{y} & =\frac{(2 \times 60)}{160} \times(20) \\v_{y} & =15 \mathrm{~mm} / \mathrm{sec}\end{aligned}Therefore\begin{array}{l}\vec{v}=v_{x} i+v_{y} j \\\vec{v}=20 i+15 j\end{array}\rightarrow Mggnitude of velocity of the pin\begin{array}{l}v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{20^{2}+15^{2}} \\v=25 \mathrm{~mm} / \mathrm{s} \text { Ans. }\end{array}\longrightarrow Acceleration:-\begin{array}{l}\vec{a}=a_{x} i+a_{y} j \\a_{x}=\frac{d v_{x}}{d t}, \because v_{x}=20 \mathrm{~mm} / \mathrm{s} \text { (constant) } \\\therefore a_{x}=0 \\a_{y}=\frac{d v_{y}}{d t} \\\because v_{y}=\frac{d y}{d t}=\frac{2 x}{160} \times\left(\frac{d x}{d t}\right)=\frac{2 \times x \times 20}{160} \\v_{y}=\frac{d y}{d t}=\frac{x}{4} \\\text { OR } \\a_{y}=\frac{d v_{y}}{d t}=\frac{d^{2} y}{d t}=\frac{1}{4} \times \frac{d x}{d t} \\a_{y}=\frac{1}{4} \times 20 \mathrm{~mm} / \mathrm{sec} \\a_{y}=5 \mathrm{~mm} / \mathrm{s}^{2} \\\therefore \quad a^{2}=0 i+5 j\end{array}Therefore nagnitude of accelerationa=5 \mathrm{~mm} / \mathrm{s}^{2} \text { Ans. } ...