| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 12 |
| 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 kinematics question involving velocity-time graphs with constant speeds and simple linear motion. Students need to calculate distances using area under graphs and apply the basic principle that distance = speed × time. All parts involve routine calculations with no conceptual challenges or problem-solving insight required—easier than average A-level. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Total distance is \(3 \times 30 + 3 \times 30 + 3 \times 30 = 270\) m | M1, M1, A1 | For any calculation of a rectangular area; For addition of three positive areas; For correct value 270 |
| Subtotal: 3 marks | ||
| (ii) Distance at \(t = 50\) is \(90 - 30 = 30\) m | M1, A1 | For correct use of signed areas; For correct value 30 |
| Distance at \(t = 80\) is 60 m | A1 | For correct value 60 |
| Subtotal: 3 marks | ||
| (iii) Child's speed is \(\frac{30}{50} = 0.6\) m s\(^{-1}\) | B1, M1, A1 | For distance 30 m; For dividing by 50; For correct value 0.6 |
| Subtotal: 3 marks | ||
| (iv) Child walks \(60 - 30 = 30\) m in next 30 s | B1 | For child's distance gone from \(t = 50\) to 80 |
| Hence \(30 = \frac{1}{2}(0.6 + v) \times 30\) | M1 | For suitable use of \(s = \frac{1}{2}(u + v)t\) or equiv |
| i.e. child's speed is 1.4 m s\(^{-1}\) | A1 | For correct value 1.4 |
| Subtotal: 3 marks | ||
| Total: 12 marks |
**(i)** Total distance is $3 \times 30 + 3 \times 30 + 3 \times 30 = 270$ m | M1, M1, A1 | For any calculation of a rectangular area; For addition of three positive areas; For correct value 270
| | **Subtotal: 3 marks**
**(ii)** Distance at $t = 50$ is $90 - 30 = 30$ m | M1, A1 | For correct use of signed areas; For correct value 30
Distance at $t = 80$ is 60 m | A1 | For correct value 60
| | **Subtotal: 3 marks**
**(iii)** Child's speed is $\frac{30}{50} = 0.6$ m s$^{-1}$ | B1, M1, A1 | For distance 30 m; For dividing by 50; For correct value 0.6
| | **Subtotal: 3 marks**
**(iv)** Child walks $60 - 30 = 30$ m in next 30 s | B1 | For child's distance gone from $t = 50$ to 80
Hence $30 = \frac{1}{2}(0.6 + v) \times 30$ | M1 | For suitable use of $s = \frac{1}{2}(u + v)t$ or equiv
i.e. child's speed is 1.4 m s$^{-1}$ | A1 | For correct value 1.4
| | **Subtotal: 3 marks**
| | **Total: 12 marks**\includegraphics{figure_3}
A woman runs from $A$ to $B$, then from $B$ to $A$ and then from $A$ to $B$ again, on a straight track, taking 90 s. The woman runs at a constant speed throughout. Fig. 1 shows the $(t, v)$ graph for the woman.
\begin{enumerate}[label=(\roman*)]
\item Find the total distance run by the woman. [3]
\item Find the distance of the woman from $A$ when $t = 50$ and when $t = 80$, [3]
\end{enumerate}
\includegraphics{figure_4}
At time $t = 0$, a child also starts to move, from $A$, along $AB$. The child walks at a constant speed for the first 50 s and then at an increasing speed for the next 40 s. Fig. 2 shows the $(t, v)$ graph for the child; it consists of two straight line segments.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item At time $t = 50$, the woman and the child pass each other, moving in opposite directions. Find the speed of the child during the first 50 s. [3]
\item At time $t = 80$, the woman overtakes the child. Find the speed of the child at this instant. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 Q4 [12]}}