| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Easy -1.2 This is a straightforward velocity-time graph interpretation question requiring only basic kinematics: reading gradient for acceleration, calculating area under graph for distance, and identifying maximum velocity. All three parts involve direct application of standard formulas with no problem-solving or conceptual challenges beyond routine graph reading skills. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Acceleration is \(0.09 \text{ ms}^{-2}\) | B1 | |
| [1] | ||
| (ii) \([D = \frac{1}{2}(8 + 4)0.18\) or \(D = (0 + \frac{1}{2}0.09 \times 2^2) + (0.18 \times 4 + \frac{1}{2} \times 0 \times 4^2) + (0.18 \times 2 - \frac{1}{2}0.09 \times 2^2)]\) | M1 | For using the idea that area represents distance or for repeated use of \(s = ut + \frac{1}{2}at^2\) |
| Distance is \(1.08 \text{ m}\) | A1 | |
| [2] | ||
| (iii) \([\frac{1}{2} \times 3V = 1.08]\) | M1 | For using area of triangle = area of trapezium |
| Greatest speed is \(0.72 \text{ ms}^{-1}\) | A1 | |
| [2] | ||
| SR (max 1 out of 2) for candidates who assume (implicitly) that speed is greatest at a specific time (\(t = 11\) or \(t = 9.5\)): \(0.72 \text{ ms}^{-1}\) B1 from \(\frac{1}{2}(0 + V) \times 3 = 1.08\) or from \(\frac{1}{2}(0 + V) \times 1.5 = \frac{1}{2}1.08\) |
**(i)** Acceleration is $0.09 \text{ ms}^{-2}$ | B1 |
| | | **[1]**
**(ii)** $[D = \frac{1}{2}(8 + 4)0.18$ or $D = (0 + \frac{1}{2}0.09 \times 2^2) + (0.18 \times 4 + \frac{1}{2} \times 0 \times 4^2) + (0.18 \times 2 - \frac{1}{2}0.09 \times 2^2)]$ | M1 | For using the idea that area represents distance or for repeated use of $s = ut + \frac{1}{2}at^2$
Distance is $1.08 \text{ m}$ | A1 |
| | | **[2]**
**(iii)** $[\frac{1}{2} \times 3V = 1.08]$ | M1 | For using area of triangle = area of trapezium
Greatest speed is $0.72 \text{ ms}^{-1}$ | A1 |
| | | **[2]**
| | | SR (max 1 out of 2) for candidates who assume (implicitly) that speed is greatest at a specific time ($t = 11$ or $t = 9.5$): $0.72 \text{ ms}^{-1}$ B1 from $\frac{1}{2}(0 + V) \times 3 = 1.08$ or from $\frac{1}{2}(0 + V) \times 1.5 = \frac{1}{2}1.08$ |
---\includegraphics{figure_2}
The diagram shows the velocity-time graph for the motion of a machine's cutting tool. The graph consists of five straight line segments. The tool moves forward for 8 s while cutting and then takes 3 s to return to its starting position. Find
\begin{enumerate}[label=(\roman*)]
\item the acceleration of the tool during the first 2 s of the motion, [1]
\item the distance the tool moves forward while cutting, [2]
\item the greatest speed of the tool during the return to its starting position. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q2 [5]}}