| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find total distance |
| Difficulty | Easy -1.3 This is a straightforward SUVAT question requiring students to sketch a standard velocity-time graph and calculate area under it. All parameters are explicitly given, requiring only direct application of v=u+at and area calculation with no problem-solving or insight needed. Significantly easier than average A-level questions. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Trapezium on the \(t\)-axis starting at \((0, 0)\) | B1 | |
| Fully correct with correct labels | B1 | With height 20, intersecting the \(t\)-axis at 60, horizontal line segment between 10 and 40. Axis need not be labelled with \(t\) and \(v\) but if they are they must be correct. If not labelled, then must assume \(t\) is on the horizontal axis. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Area} = \frac{1}{2}(30+60)\times 20\) or \(\text{Area} = \frac{1}{2}\times10\times20+30\times20+\frac{1}{2}\times20\times20\) | M1 | Must be considering the area of a trapezium. Allow a single slip in one term only. Must be adding all terms together if considering two triangles and a rectangle. Using *their* 20 from (a). If no value for the height shown on the graph in (a), then it must be correct. |
| \(= 900\text{ m}\) | A1FT | FT *their* \(20\ (= 45 \times \textit{their}\ 20)\) but A0 FT if using 2. |
| [2] |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Trapezium on the $t$-axis starting at $(0, 0)$ | **B1** | |
| Fully correct with correct labels | **B1** | With height 20, intersecting the $t$-axis at 60, horizontal line segment between 10 and 40. Axis need not be labelled with $t$ and $v$ but if they are they must be correct. If not labelled, then must assume $t$ is on the horizontal axis. |
| | **[2]** | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Area} = \frac{1}{2}(30+60)\times 20$ **or** $\text{Area} = \frac{1}{2}\times10\times20+30\times20+\frac{1}{2}\times20\times20$ | **M1** | Must be considering the area of a trapezium. Allow a single slip in one term only. Must be adding all terms together if considering two triangles and a rectangle. Using *their* 20 from **(a)**. If no value for the height shown on the graph in **(a)**, then it must be correct. |
| $= 900\text{ m}$ | **A1FT** | FT *their* $20\ (= 45 \times \textit{their}\ 20)$ but A0 FT if using 2. |
| | **[2]** | |
---1 A car starts from rest and accelerates at $2 \mathrm {~ms} ^ { - 2 }$ for 10 s . It then travels at a constant speed for 30 s . The car then uniformly decelerates to rest over a period of 20 s .
\begin{enumerate}[label=(\alph*)]
\item Sketch a velocity-time graph for the motion of the car.\\
\includegraphics[max width=\textwidth, alt={}, center]{2af7fd9a-aa78-4d77-aa4e-c01604c8b0ae-03_762_1081_447_493}
\item Find the total distance travelled by the car.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q1 [4]}}