IB AA HL May 2026 P3 TZC Q2

A particle moves back and forth in a horizontal line between two end points. A fixed origin \(O\) is the midpoint between these end points. Its displacement \(x\) metres from \(O\) at time \(t\) seconds is \(x=4\sin(2t+\frac\pi3)\), where \(0\leq t\leq\pi\).

1. Determine
   1. the amplitude;[1]
   2. the initial displacement;[2]
   3. the value of \(t\) when the particle first passes through \(O\).[2]
2. The acceleration \(a\) at displacement \(s\) satisfies \(a=-n^2s\), where \(n>0\).
   The displacement is \(s=A\sin(nt+b)\), where \(t\geq0\), \(A,n>0\) and \(-\pi\leq b\leq\pi\).
   By finding expressions for \(\frac{\mathrm{d}s}{\mathrm{d}t}\) and \(\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\), verify that \(s=A\sin(nt+b)\) satisfies \(a=-n^2s\).[2]
3. 1. Use the chain rule to show that \(a=v\frac{\mathrm{d}v}{\mathrm{d}s}\), where \(v\) is velocity.[1]
   2. By solving \(v\frac{\mathrm{d}v}{\mathrm{d}s}=-n^2s\), show that \(v^2=n^2(A^2-s^2)\).[5]
   3. Hence, or otherwise, find the particle’s maximum speed.[2]
4. The continuous random variable \(S\) denotes the particle’s displacement \(s\) from \(O\) at time \(t\).
   \[f(s)=\begin{cases}\frac1{\pi\sqrt{A^2-s^2}},&-A<s<A,\\0,&\text{otherwise}.\end{cases}\]
   Show that \(P(-\frac A2\leq S\leq\frac A2)=\frac13\).[4]
5. For \(-A<s<A\), \(f(s)\) can be expressed as \(\frac{k}{|v(s)|}\), where \(k>0\) and \(v(s)\) is the particle’s velocity at displacement \(s\) from \(O\).
   Find the value of \(k\).[3]
6. 1. Determine \(E(S)\), justifying your answer.[2]
   2. Interpret the result in the context of the particle’s motion.[1]