| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (algebraic) |
| Difficulty | Moderate -0.3 This is a multi-part SUVAT question with straightforward applications: recognizing sign of acceleration, writing v=u+at, sketching a linear velocity graph, finding distance using area under graphs, and commenting on model validity. All parts are routine AS-level mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| The student has found the magnitude of the acceleration and not the actual acceleration which is negative. The argument is not valid. | B1 | Must comment on the validity of the argument. Allow for a comment that student is wrong which includes statement that the acceleration should be negative or similar. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_B = 10 + 0.15t\) | B1 | Award if \(v = 10 + 0.15t\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Line through coordinates \((0, 10)\) and \((20, 13)\) FT their linear (b) | B1 | Line through coordinates \((0,10)\) and \((20,13)\). FT their linear (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Same speed when \(25 - 0.6t = 10 + 0.15t\), giving \(t = 20\) | B1 | Soi FT their (b) |
| Displacement of train A is 380 m; Displacement of train B is 230 m; So distance between them is 150 m | M1, A1 | Attempt to use area or \(suvat\) equation(s) to find the displacement of at least one train at their value for \(t\). cao |
| Answer | Marks | Guidance |
|---|---|---|
| Area of the triangle between graphs is \(\frac{1}{2} \times (25-10) \times 20\) | M1 | Soi FT their (b) |
| So distance between them is 150 m | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| [Velocity is zero at 41.7 s.] For large values of \(t\), the model predicts that the velocity will be negative but the train will not go backwards as the trains are travelling in the same direction | B1 | Must compare what happens in the model with what happens to the train. (See appendix for exemplars) |
## Question 9:
### Part (a):
The student has found the magnitude of the acceleration and not the actual acceleration which is negative. The argument is not valid. | B1 | Must comment on the validity of the argument. Allow for a comment that student is wrong which includes statement that the acceleration should be negative or similar.
**[1]**
### Part (b):
$v_B = 10 + 0.15t$ | B1 | Award if $v = 10 + 0.15t$ seen
**[1]**
### Part (c):
Line through coordinates $(0, 10)$ and $(20, 13)$ FT their linear (b) | B1 | Line through coordinates $(0,10)$ and $(20,13)$. FT their linear (b)
**[1]**
### Part (d):
Same speed when $25 - 0.6t = 10 + 0.15t$, giving $t = 20$ | B1 | Soi FT their (b)
Displacement of train A is 380 m; Displacement of train B is 230 m; So distance between them is 150 m | M1, A1 | Attempt to use area or $suvat$ equation(s) to find the displacement of at least one train at their value for $t$. cao
**Alternative for last 2 marks:**
Area of the triangle between graphs is $\frac{1}{2} \times (25-10) \times 20$ | M1 | Soi FT their (b)
So distance between them is 150 m | A1 | cao
**[3]**
### Part (e):
[Velocity is zero at 41.7 s.] For large values of $t$, the model predicts that the velocity will be negative but the train will not go backwards as the trains are travelling in the same direction | B1 | Must compare what happens in the model with what happens to the train. (See appendix for exemplars)
**[1]**
---9 Two trains are travelling in the same direction on parallel straight tracks and train A overtakes train B . At time $t$ seconds after the front of train A overtakes the front of train B the velocities of trains A and B are $v _ { \mathrm { A } } \mathrm { m } \mathrm { s } ^ { - 1 }$ and $v _ { \mathrm { B } } \mathrm { ms } ^ { - 1 }$ respectively.
The velocity of train A is modelled by $\mathrm { v } _ { \mathrm { A } } = 25 - 0.6 \mathrm { t }$. The velocity-time graph of train A is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b5c47a93-ce43-4aa1-ba7f-fbb650523373-5_664_1399_550_242}
\begin{enumerate}[label=(\alph*)]
\item A student argues that the speed of train A changes by $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 30 seconds so its acceleration is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
Comment on the validity of the student's argument.
\item When the front of train A overtakes the front of train B , train B has a velocity of $10 \mathrm {~ms} ^ { - 1 }$. The acceleration of train $B$ is constant and is modelled as $0.15 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
Write down the equation for $v _ { \mathrm { B } }$ in terms of $t$ that models the velocity of train B .
\item Draw the velocity-time graph of train B on the copy of the diagram in the Printed Answer Booklet.
\item Determine the distance between the fronts of the trains at the time when the trains are travelling at the same velocity.
\item Explain why the model for train A would not be valid for large values of $t$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q9 [7]}}