| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring differentiation of a polynomial to find velocity, setting v=0, and substituting back. The algebra is simple (factoring out common terms), and the method is standard textbook material. Slightly above trivial due to the polynomial degree and two-part structure, but easier than average A-level questions. |
| Spec | 1.07a Derivative as gradient: of tangent to curve3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitution of both \(t=0\) and \(t=10\) | M1 | AO2.1 |
| \(s=0\) for both \(t=0\) and \(t=10\) | A1 | AO1.1b |
| Explanation: \(s > 0\) for \(0 < t < 10\) since \(s = \frac{1}{10}t^2(t-10)^2\) | A1 | AO2.4 — since \(s\) is a perfect square, \(s > 0\) for all other \(t\)-values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate displacement \(s\) w.r.t. \(t\) to give velocity \(v\) | M1 | AO1.1a — powers of \(t\) reducing by 1 |
| \(v = \frac{1}{10}(4t^3 - 60t^2 + 200t)\) | A1 | AO1.1b |
| Interpretation of 'rest' to give \(v = \frac{1}{10}(4t^3 - 60t^2 + 200t) = \frac{2}{5}t(t-5)(t-10) = 0\) | M1 | AO1.1b — equating \(v\) to 0 and factorising |
| \(t = 0, 5, 10\) | A1 | AO1.1b |
| Select \(t=5\) and substitute into \(s\) | M1 | AO1.1a |
| Distance \(= 62.5 \text{ m}\) | A1ft | AO1.1b — ft following an incorrect \(t\)-value |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitution of both $t=0$ and $t=10$ | M1 | AO2.1 |
| $s=0$ for both $t=0$ and $t=10$ | A1 | AO1.1b |
| Explanation: $s > 0$ for $0 < t < 10$ since $s = \frac{1}{10}t^2(t-10)^2$ | A1 | AO2.4 — since $s$ is a perfect square, $s > 0$ for all other $t$-values |
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate displacement $s$ w.r.t. $t$ to give velocity $v$ | M1 | AO1.1a — powers of $t$ reducing by 1 |
| $v = \frac{1}{10}(4t^3 - 60t^2 + 200t)$ | A1 | AO1.1b |
| Interpretation of 'rest' to give $v = \frac{1}{10}(4t^3 - 60t^2 + 200t) = \frac{2}{5}t(t-5)(t-10) = 0$ | M1 | AO1.1b — equating $v$ to 0 and factorising |
| $t = 0, 5, 10$ | A1 | AO1.1b |
| Select $t=5$ and substitute into $s$ | M1 | AO1.1a |
| Distance $= 62.5 \text{ m}$ | A1ft | AO1.1b — ft following an incorrect $t$-value |
---\begin{enumerate}
\item A bird leaves its nest at time $t = 0$ for a short flight along a straight line.
\end{enumerate}
The bird then returns to its nest.\\
The bird is modelled as a particle moving in a straight horizontal line.\\
The distance, $s$ metres, of the bird from its nest at time $t$ seconds is given by
$$s = \frac { 1 } { 10 } \left( t ^ { 4 } - 20 t ^ { 3 } + 100 t ^ { 2 } \right) , \quad \text { where } 0 \leqslant t \leqslant 10$$
(a) Explain the restriction, $0 \leqslant t \leqslant 10$\\
(b) Find the distance of the bird from the nest when the bird first comes to instantaneous rest.
\hfill \mbox{\textit{Edexcel AS Paper 2 Q8 [9]}}