Challenging +1.2 This is a standard variable acceleration kinematics problem requiring two integrations with given initial conditions. While it involves a moderately complex algebraic expression (5t+4)^{-3}, the method is straightforward: integrate acceleration to get velocity, then integrate velocity to get displacement. The integration itself is routine (chain rule in reverse) and the algebra is manageable for Further Maths students. It's above average difficulty due to the algebraic manipulation required, but follows a well-practiced procedure without requiring novel insight.
A particle \(P\) is moving in a horizontal straight line. Initially \(P\) is at the point \(O\) on the line and is moving with velocity \(25 \text{ m s}^{-1}\). At time \(t\) s after passing through \(O\), the acceleration of \(P\) is \(-\frac{4000}{(5t + 4)^3} \text{ m s}^{-2}\) in the direction \(PO\). The displacement of \(P\) from \(O\) at time \(t\) is \(x\) m.
Find an expression for \(x\) in terms of \(t\). [5]
A particle $P$ is moving in a horizontal straight line. Initially $P$ is at the point $O$ on the line and is moving with velocity $25 \text{ m s}^{-1}$. At time $t$ s after passing through $O$, the acceleration of $P$ is $-\frac{4000}{(5t + 4)^3} \text{ m s}^{-2}$ in the direction $PO$. The displacement of $P$ from $O$ at time $t$ is $x$ m.
Find an expression for $x$ in terms of $t$. [5]
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q3 [5]}}