| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT/kinematics question requiring standard techniques: calculating area under a trapezoid for distance, using v=u+at to find deceleration, and computing average speed from total distance and time. All steps are routine applications of basic mechanics formulas with no problem-solving insight required, making it easier than average but not trivial due to the two-stage deceleration requiring careful bookkeeping. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(= 50.4\) m | B1 | Allow \(\frac{252}{5}\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v_1 = 12.6 - (62-48)a\) | M1 | Use of suvat for first section of deceleration. \(12.6 \pm (62-48)a\) only. |
| \(0 = v_1 - 2a \times (70-62)\) | M1 | Use of suvat for second section of deceleration. An expression for the velocity at 62 seconds must be \(\pm 2a \times (70-62)\). |
| \(a = 0.42\) | A1 | \(-0.42\) scores A0. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Speed at time \(t = 62\) is \(6.72\ \text{ms}^{-1}\) | B1 | This may be seen in part (b) but must be used in part (c) to get this mark. |
| \(s_2 = (48-8) \times 12.6\ [=504]\) | B2FT | B2 FT for any 2 correct, B1 FT for any 1 correct – follow through *their* value of \(v_1\) where \(0 < v_1 < 12.6\) but must have come from the correct equations seen in part (b). Allow correct value of \(v_1\) from \(a = -0.42\) where \(v_1 = 12.6 + (62-48)a\) and \(v_1 = -2a \times (70-62)\). |
| \(s_3 = 0.5 \times (12.6 + their\ 6.72) \times (62-48)\ \left[= 135.24\ \text{or}\ \frac{3381}{25}\ \text{oe}\right]\) | ||
| or \(their\ 6.72 \times (62-48) + 0.5 \times (62-48) \times (12.6 - their\ 6.72)\) | ||
| \(s_4 = 0.5 \times their\ 6.72 \times (70-62)\ \left[= 26.88\ \text{or}\ \frac{672}{25}\ \text{oe}\right]\) | ||
| Average speed \(= 10.236\ \text{ms}^{-1}\) | B1 | Allow 10.2 or better oe e.g. \(\frac{2559}{250}\), \(10\frac{59}{250}\) |
| 4 |
## Question 3:
### Part 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $= 50.4$ m | **B1** | Allow $\frac{252}{5}$ |
| | **1** | |
### Part 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v_1 = 12.6 - (62-48)a$ | **M1** | Use of suvat for first section of deceleration. $12.6 \pm (62-48)a$ only. |
| $0 = v_1 - 2a \times (70-62)$ | **M1** | Use of suvat for second section of deceleration. An expression for the velocity at 62 seconds must be $\pm 2a \times (70-62)$. |
| $a = 0.42$ | **A1** | $-0.42$ scores A0. |
| | **3** | |
### Part 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Speed at time $t = 62$ is $6.72\ \text{ms}^{-1}$ | **B1** | This may be seen in part (b) but must be used in part (c) to get this mark. |
| $s_2 = (48-8) \times 12.6\ [=504]$ | **B2FT** | B2 FT for any 2 correct, B1 FT for any 1 correct – follow through *their* value of $v_1$ where $0 < v_1 < 12.6$ but must have come from the correct equations seen in part (b). Allow correct value of $v_1$ from $a = -0.42$ where $v_1 = 12.6 + (62-48)a$ and $v_1 = -2a \times (70-62)$. |
| $s_3 = 0.5 \times (12.6 + their\ 6.72) \times (62-48)\ \left[= 135.24\ \text{or}\ \frac{3381}{25}\ \text{oe}\right]$ | | |
| or $their\ 6.72 \times (62-48) + 0.5 \times (62-48) \times (12.6 - their\ 6.72)$ | | |
| $s_4 = 0.5 \times their\ 6.72 \times (70-62)\ \left[= 26.88\ \text{or}\ \frac{672}{25}\ \text{oe}\right]$ | | |
| Average speed $= 10.236\ \text{ms}^{-1}$ | **B1** | Allow 10.2 or better oe e.g. $\frac{2559}{250}$, $10\frac{59}{250}$ |
| | **4** | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{f1f33ef0-0d4d-4a4a-aadb-28de8dc0ea8d-04_666_1278_280_424}
The diagram shows the velocity-time graph for the motion of a bus. The bus starts from rest and accelerates uniformly for 8 seconds until it reaches a speed of $12.6 \mathrm {~ms} ^ { - 1 }$. The bus maintains this speed for 40 seconds. It then decelerates uniformly in two stages. Between 48 and 62 seconds the bus decelerates at $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and between 62 and 70 seconds it decelerates at $2 a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ until coming to rest.
\begin{enumerate}[label=(\alph*)]
\item Find the distance covered by the bus in the first 8 seconds.
\item Find the value of $a$.
\item Find the average speed of the bus for the whole journey.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q3 [8]}}