Jim places an $8$ m rod $XY$ against a vertical wall, as shown in the diagram below. The points

Question

Jim places an $8$ m rod $XY$ against a vertical wall, as shown in the diagram below. The points $O$, $X$ and $Y$ are coplanar, with $O$ being the point where the vertical wall and the flat ground meet. At time $t$ seconds, the ends $X$ and $Y$ are $x$ and $y$ meters from $O$ respectively. Jim makes a conjecture that as the rod slips, $y$ decreases at a rate proportional to $y$. Based on Jim’s conjecture, show that $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{k\left( 64-{{x}^{2}} \right)}{x}$, where $k$ is a positive constant.[4] It is given that $X$ is $4$ meters aways from $O$ initially, and that $k=3$. Find an expression for $x$ in terms of $t$. Hence, find the time taken for $Y$ to be $3$ meters from $O$ from the instance the rod starts sliding from the initial position, giving your answer to $1$ decimal place.[5] Sketch the graph of $x$ against $t$.[2] Comment, with justification, on whether Jim’s conjecture is appropriate for the behaviour of the rod.[1]

Accepted Answer

(a) $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{k(64-x^2)}x$ $(\text{shown})$ (b) $x=\sqrt{64-48e^{-6t}}$, and the time taken is $0.3\text{ s}$ $(1\text{ d.p.})$ (c) Increasing from $(0,4)$ towards the horizontal asymptote $x=8$ (d) Not appropriate, since the model predicts $y>0$ for every finite $t$

2023 DHS P1 Q10

2023 DHS P1 Q10

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