| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT/kinematics question requiring students to calculate distance from a speed-time graph (trapezium area), then find time differences. It involves only basic arithmetic and standard graph interpretation with no conceptual challenges—easier than average but not trivial due to unit conversion and multi-step calculation. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find how much less distance travelled in the \(50\) s | M1 | Sensible attempt at method including finding distance as an area |
| Distance is area of trapezium: \(\frac{(25-10)\times(50+20)}{2} = 525\) m | A1 | cao. Need not be evaluated. Many correct routes. |
| This distance is made up at \(25\text{ ms}^{-1}\) to give extra time | M1 | |
| Extra time is \(\frac{525}{25} = 21\) | A1 | FT their area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find the distance travelled in the \(50\) s | M1 | Sensible attempt including finding distance as an area |
| Find the time for rest of journey \(+ 50\) and subtract \(\frac{5000}{25} = 200\) | M1 | May be scored later |
| Distance travelled in the \(50\) s is \(725\) m | A1 | cao. Many correct routes to find area |
| Extra time is \(\frac{(5000-725)}{25} + 50 - 200 = 21\) | A1 | FT their area. Award full marks for 21 seen www |
# Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find how much less distance travelled in the $50$ s | M1 | Sensible attempt at method including finding distance as an area |
| Distance is area of trapezium: $\frac{(25-10)\times(50+20)}{2} = 525$ m | A1 | cao. Need not be evaluated. Many correct routes. |
| This distance is made up at $25\text{ ms}^{-1}$ to give extra time | M1 | |
| Extra time is $\frac{525}{25} = 21$ | A1 | FT their area |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find the distance travelled in the $50$ s | M1 | Sensible attempt including finding distance as an area |
| Find the time for rest of journey $+ 50$ and subtract $\frac{5000}{25} = 200$ | M1 | May be scored later |
| Distance travelled in the $50$ s is $725$ m | A1 | cao. Many correct routes to find area |
| Extra time is $\frac{(5000-725)}{25} + 50 - 200 = 21$ | A1 | FT their area. Award full marks for 21 seen www |
---7 A car is usually driven along the whole of a 5 km stretch of road at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. On one occasion, during a period of 50 seconds, the speed of the car is as shown by the speed-time graph in Fig. 7.\\
The rest of the 5 km is travelled at $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{970d2349-7705-4b66-9931-83613e5d852f-5_510_1016_589_296}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
How much more time than usual did the journey take on this occasion?\\
Show your working clearly.
\hfill \mbox{\textit{OCR MEI AS Paper 1 Q7 [4]}}