| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (algebraic) |
| Difficulty | Moderate -0.3 This is a standard two-particle SUVAT problem requiring velocity-time graph sketching, distance calculation via area under graph, and finding acceleration from given constraints. While multi-part with several stages of motion, it follows routine mechanics procedures without requiring novel insight or complex problem-solving—slightly easier than average due to straightforward application of basic kinematics principles. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B1 | Three correct straight lines | |
| \(v = 6 \text{ ms}^{-1}\), \(t = 5\) s and \(t = 17\) s | B1 | Correct trapezium with key values |
| \([\frac{1}{2} \times 6 \times (12+20)]\) or \([\frac{1}{2} \times 5 \times 6 + 12 \times 6 + \frac{1}{2} \times 3 \times 6]\) | M1 | Use of trapezium area or use of suvat formulae |
| Total distance \(= 96\) m | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\frac{1}{2} \times 20 \times v = 96]\) | M1 | Uses area of triangle \(= 96\) or uses \(s = ut + \frac{1}{2}at^2\) to form equation in \(a\) |
| \(v = 9.6 \text{ ms}^{-1}\) or \(48 = \frac{1}{2}a(10)^2\) | A1 | |
| Acceleration \(= 9.6/10 = 0.96 \text{ ms}^{-2}\) | A1 | |
| 3 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1 | Three correct straight lines |
| $v = 6 \text{ ms}^{-1}$, $t = 5$ s and $t = 17$ s | B1 | Correct trapezium with key values |
| $[\frac{1}{2} \times 6 \times (12+20)]$ or $[\frac{1}{2} \times 5 \times 6 + 12 \times 6 + \frac{1}{2} \times 3 \times 6]$ | M1 | Use of trapezium area or use of suvat formulae |
| Total distance $= 96$ m | A1 | AG |
| | 4 | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\frac{1}{2} \times 20 \times v = 96]$ | M1 | Uses area of triangle $= 96$ or uses $s = ut + \frac{1}{2}at^2$ to form equation in $a$ |
| $v = 9.6 \text{ ms}^{-1}$ or $48 = \frac{1}{2}a(10)^2$ | A1 | |
| Acceleration $= 9.6/10 = 0.96 \text{ ms}^{-2}$ | A1 | |
| | 3 | |4 A runner sets off from a point $P$ at time $t = 0$, where $t$ is in seconds. The runner starts from rest and accelerates at $1.2 \mathrm {~ms} ^ { - 2 }$ for 5 s . For the next 12 s the runner moves at constant speed before decelerating uniformly over a period of 3 s , coming to rest at $Q$. A cyclist sets off from $P$ at time $t = 10$ and accelerates uniformly for 10 s , before immediately decelerating uniformly to rest at $Q$ at time $t = 30$.\\
(i) Sketch the velocity-time graph for the runner and show that the distance $P Q$ is 96 m .\\
\includegraphics[max width=\textwidth, alt={}, center]{007ccd92-79ba-409a-97e8-a4cf1f0a6cc5-06_821_1451_708_388}\\
(ii) Find the magnitude of the acceleration of the cyclist.\\
\hfill \mbox{\textit{CAIE M1 2018 Q4 [7]}}