| Exam Board | SPS |
|---|---|
| Module | SPS SM Mechanics (SPS SM Mechanics) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward mechanics question involving differentiation to find when acceleration is zero, identifying maximum displacement by setting velocity to zero, and integration to find displacement. All steps are standard A-level techniques with clear signposting, though it requires careful handling of piecewise functions and multiple integrations across different time intervals. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
\begin{enumerate}
\item At time $t$ seconds, where $0 \leq t \leq T$, a particle, $P$, moves so that its velocity $v m s ^ { - 1 }$ is given by
\end{enumerate}
$$v = 7.2 t - 0.45 t ^ { 2 }$$
When $t = 0$ the particle is sitting stationary at a displacement of $\mathrm { d } m$ from a point O .\\
The particle's acceleration is zero when $t = T$.\\
(i) Find the value of $T$.
For $t \geq T$, the particle moves with a velocity $v = 48 - 2.4 t m s ^ { - 1 }$.\\
(ii) Find the time when $P$ is at its maximum displacement from O . The particle passes through the point O when $t = 38$.\\
(iii) Find $d$.
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\hfill \mbox{\textit{SPS SPS SM Mechanics 2023 Q5 [8]}}