| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Moderate -0.8 This is a straightforward AS-level mechanics question requiring basic interpretation of velocity-time graphs and application of SUVAT equations. Part (a) is conceptual recall (area = distance), parts (b) and (c) involve simple calculations with given values, and part (d) requires equating areas of two trapezoids—all standard textbook exercises with no novel problem-solving required. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Because the distances travelled or displacements are equal (oe). If they mention the times are the same as well, ignore it. | B1 | Must mention distances being equal specifically. |
| (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.8\) or \(\frac{4}{5}\) (m s\(^{-2}\)) | B1 | cao |
| (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}\times5\times4+(4\times22.5)\) OR \(\frac{1}{2}(27.5+22.5)\times4\) OR \(27.5\times4-\frac{1}{2}\times5\times4\) | M1 | Clear attempt to find the total area under the \(P\) graph, correct structure (triangle + rectangle) OR trapezium OR (rectangle − triangle); must see use of \(\frac{1}{2}\) where appropriate. OR may use *suvat* for distance in one or more sections. NB: M0 for single *suvat* formula for whole motion. |
| \(100\) (m) | A1 | cao |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total area under graph = their answer for part (c) | M1 | Clear attempt to equate total area under \(S\) graph with correct structure (triangle + rectangle) OR trapezium OR (rectangle − triangle), must use \(\frac{1}{2}\) where appropriate, to give a quadratic equation in \(X\) only. NB: M0 for single *suvat* formula for whole motion. |
| \(\frac{1}{2}X\times X+X(27.5-X)=100\) | A1ft | Correct unsimplified quadratic in \(X\) only with at most one error, follow their answer for (c). |
| OR \(\frac{1}{2}(27.5+27.5-X)\times X=100\) | A1ft | Correct unsimplified quadratic in \(X\) only, follow their answer for (c). |
| OR \(27.5X-\frac{1}{2}X^2=100\) | ||
| \(X=3.92\) to 3 sf | A1 | cao |
| (4 marks) |
# Question 1:
## Part 1(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Because the distances travelled or displacements are equal (oe). If they mention the times are the same as well, ignore it. | B1 | Must mention distances being equal specifically. |
| **(1 mark)** | | |
## Part 1(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.8$ or $\frac{4}{5}$ (m s$^{-2}$) | B1 | cao |
| **(1 mark)** | | |
## Part 1(c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}\times5\times4+(4\times22.5)$ **OR** $\frac{1}{2}(27.5+22.5)\times4$ **OR** $27.5\times4-\frac{1}{2}\times5\times4$ | M1 | Clear attempt to find the **total** area under the $P$ graph, correct structure (triangle + rectangle) **OR** trapezium **OR** (rectangle − triangle); must see use of $\frac{1}{2}$ where appropriate. **OR** may use *suvat* for distance in one or more sections. **NB**: M0 for single *suvat* formula for whole motion. |
| $100$ (m) | A1 | cao |
| **(2 marks)** | | |
## Part 1(d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Total area under graph = their answer for part (c) | M1 | Clear attempt to equate total area under $S$ graph with correct structure (triangle + rectangle) **OR** trapezium **OR** (rectangle − triangle), must use $\frac{1}{2}$ where appropriate, to give a quadratic equation in $X$ only. **NB**: M0 for single *suvat* formula for whole motion. |
| $\frac{1}{2}X\times X+X(27.5-X)=100$ | A1ft | Correct unsimplified quadratic in $X$ only with at most one error, follow their answer for (c). |
| **OR** $\frac{1}{2}(27.5+27.5-X)\times X=100$ | A1ft | Correct unsimplified quadratic in $X$ only, follow their answer for (c). |
| **OR** $27.5X-\frac{1}{2}X^2=100$ | | |
| $X=3.92$ to 3 sf | A1 | cao |
| **(4 marks)** | | |
**(8 marks total)**1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d9615c4f-d8fa-4e44-978a-cf34b2b1c0b5-02_720_1490_283_299}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Two children, Pat $( P )$ and Sam $( S )$, run a race along a straight horizontal track.\\
Both children start from rest at the same time and cross the finish line at the same time.\\
In a model of the motion:\\
Pat accelerates at a constant rate from rest for 5 s until reaching a speed of $4 \mathrm {~ms} ^ { - 1 }$ and then maintains a constant speed of $4 \mathrm {~ms} ^ { - 1 }$ until crossing the finish line.
Sam accelerates at a constant rate of $1 \mathrm {~ms} ^ { - 2 }$ from rest until reaching a speed of $X \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and then maintains a constant speed of $X \mathrm {~m} \mathrm {~s} ^ { - 1 }$ until crossing the finish line.
Both children take 27.5 s to complete the race.\\
The velocity-time graphs shown in Figure 1 describe the model of the motion of each child from the instant they start to the instant they cross the finish line together.
Using the model,
\begin{enumerate}[label=(\alph*)]
\item explain why the areas under the two graphs are equal,
\item find the acceleration of Pat during the first 5 seconds,
\item find, in metres, the length of the race,
\item find the value of $X$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2023 Q1 [8]}}