| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Displacement-time graph interpretation or sketching |
| Difficulty | Moderate -0.8 This is a straightforward displacement-time graph interpretation question requiring basic gradient calculations for velocity, sketching a piecewise constant velocity graph, and recognizing that instantaneous velocity changes are physically unrealistic. All parts involve routine application of the relationship between displacement and velocity with no problem-solving insight needed. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=03.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient from \((0,4)\) to \((2,7)\) is \(1.5\), so \(s = 1.5t + 4\) | M1, A1 | Attempt to find the gradient – award if \(\frac{3}{2}\) seen. Must be \(s\) in terms of \(t\). |
| Answer | Marks | Guidance |
|---|---|---|
| Three horizontal line segments at \(1.5\), on axis, and \(-3.5\) | B1* | Three horizontal lines above, on and below the \(t\)-axis. Also allow if third line segment is above first because speed given instead of velocity. |
| Values \(1.5\) and \(-3.5\) seen and \(t = 2\) and \(t = 5\) clear | B1(dep) |
| Answer | Marks | Guidance |
|---|---|---|
| The changes in velocity are instantaneous. In reality, the velocity changes over a period of time. | E1 | Accept "suddenly stops" or similar. Also accept that velocity at \(t = 2\) or \(t = 5\) is ambiguous. |
## Question 5:
**(a)**
Gradient from $(0,4)$ to $(2,7)$ is $1.5$, so $s = 1.5t + 4$ | M1, A1 | Attempt to find the gradient – award if $\frac{3}{2}$ seen. Must be $s$ in terms of $t$.
**(b)**
Three horizontal line segments at $1.5$, on axis, and $-3.5$ | B1* | Three horizontal lines above, on and below the $t$-axis. Also allow if third line segment is above first because speed given instead of velocity.
Values $1.5$ and $-3.5$ seen and $t = 2$ and $t = 5$ clear | B1(dep) |
**(c)**
The changes in velocity are instantaneous. In reality, the velocity changes over a period of time. | E1 | Accept "suddenly stops" or similar. Also accept that velocity at $t = 2$ or $t = 5$ is ambiguous.
---5 The graph shows displacement $s m$ against time $t \mathrm {~s}$ for a model of the motion of a bead moving along a straight wire. The points $( 0,4 ) , ( 2,7 ) , ( 5,7 )$ and $( 9 , - 7 )$ are the endpoints of the line segments.\\
\includegraphics[max width=\textwidth, alt={}, center]{1d1e41f3-a834-4230-b6e1-4b0be9450d30-4_741_1301_404_239}
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the displacement of the bead for $0 \leqslant t \leqslant 2$.
\item Sketch the velocity-time graph for this model.
\item Explain why the model may not be suitable at $t = 2$ and $t = 5$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2023 Q5 [5]}}