| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.3 This is a straightforward SUVAT question requiring students to find area under a trapezium-shaped speed-time graph and sketch a corresponding distance-time graph. Both parts involve standard, routine procedures with no problem-solving or novel insight required—simpler than typical A-level questions. |
| 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/Working | Mark | Guidance |
| Complete method to find \(AB\) | M1 | Complete method to find the total area under the graph, with correct structure: triangle + rectangle + triangle, or trapezium, or triangle + trapezium, or trapezium + triangle, or rectangle − triangle − triangle. May use *suvat* on one or more sections. |
| \(= \left(\frac{1}{2} \times 5 \times 5\right) + (5 \times 15) + \left(\frac{1}{2} \times 5 \times 10\right)\) | A1 | Correct unsimplified expression for the distance \(AB\) |
| or \(= \frac{1}{2}(30+15) \times 5\) | ||
| or \(= \left(\frac{1}{2} \times 5 \times 5\right) + \frac{1}{2}(25+15) \times 5\) | ||
| or \(= \frac{1}{2}(20+15) \times 5 + \left(\frac{1}{2} \times 5 \times 10\right)\) | ||
| or \(= (30 \times 5) - \left(\frac{1}{2} \times 5 \times 5\right) - \left(\frac{1}{2} \times 5 \times 10\right)\) | ||
| \(= 112.5 \text{ (m)}\) | A1 | |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Graph with correct shape (curve reaching maximum of 112.5) | B1 | |
| Correct key features on axes (values 5, 20, 30 marked) | B1 | |
| Correct curvature/shape of graph | B1 | |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Accept 110 or better | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Single straight line with positive gradient | SC | Allow B1B0B0 unless clearly first section only; two positive gradient straight lines also allow B1B0B0 |
| One section correct; straight section with positive gradient; curves with first increasing gradient, second decreasing | B1 | — |
| Two sections correct; continuous graph, condone incorrect transition between sections | B1 | — |
| All three sections correct with \(t = 5, 20, 30\) and \(v = 112.5\) shown; continuous graph, condone incorrect transition | B1 | Subtract 1 mark if continuous vertical line at end; ignore figures for first two B marks except SC above |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find $AB$ | M1 | Complete method to find the total area under the graph, with correct structure: triangle + rectangle + triangle, or trapezium, or triangle + trapezium, or trapezium + triangle, or rectangle − triangle − triangle. May use *suvat* on one or more sections. |
| $= \left(\frac{1}{2} \times 5 \times 5\right) + (5 \times 15) + \left(\frac{1}{2} \times 5 \times 10\right)$ | A1 | Correct unsimplified expression for the distance $AB$ |
| or $= \frac{1}{2}(30+15) \times 5$ | | |
| or $= \left(\frac{1}{2} \times 5 \times 5\right) + \frac{1}{2}(25+15) \times 5$ | | |
| or $= \frac{1}{2}(20+15) \times 5 + \left(\frac{1}{2} \times 5 \times 10\right)$ | | |
| or $= (30 \times 5) - \left(\frac{1}{2} \times 5 \times 5\right) - \left(\frac{1}{2} \times 5 \times 10\right)$ | | |
| $= 112.5 \text{ (m)}$ | A1 | |
| **Total: (3)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph with correct shape (curve reaching maximum of 112.5) | B1 | |
| Correct key features on axes (values 5, 20, 30 marked) | B1 | |
| Correct curvature/shape of graph | B1 | |
| **Total: (3)** | | |
**Question 1 Total: (6 marks)**
# Question 1 (velocity-time graph question):
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Accept 110 or better | A1 | — |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Single straight line with positive gradient | SC | Allow B1B0B0 unless clearly first section only; two positive gradient straight lines also allow B1B0B0 |
| One section correct; straight section with positive gradient; curves with first increasing gradient, second decreasing | B1 | — |
| Two sections correct; continuous graph, condone incorrect transition between sections | B1 | — |
| All three sections correct with $t = 5, 20, 30$ and $v = 112.5$ shown; continuous graph, condone incorrect transition | B1 | Subtract 1 mark if continuous vertical line at end; ignore figures for first two B marks except SC above |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{34fc8023-cf31-420a-bb92-a31735fe5bdb-02_630_1537_296_264}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the speed-time graph for the journey of a car moving in a long queue of traffic on a straight horizontal road.
At time $\mathrm { t } = 0$, the car is at rest at the point A .\\
The car then accelerates uniformly for 5 seconds until it reaches a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
For the next 15 seconds the car travels at a constant speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
The car then decelerates uniformly until it comes to rest at the point B.\\
The total journey time is 30 seconds.
\begin{enumerate}[label=(\alph*)]
\item Find the distance AB .
\item Sketch a distance-time graph for the journey of the car from A to B .
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2024 Q1 [6]}}