| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.3 This is a standard SUVAT/kinematics question requiring reading a velocity-time graph, calculating accelerations from gradients, using area under the graph for distance, and solving linear equations. All techniques are routine for M1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration \(= \dfrac{(-25)}{2.5} = -10\) m s\(^{-2}\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = -15 + 7.5 \times 4\) | M1 | Using \(v\)–\(t\) graph OE |
| \(V = 15\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(v = 0\) at \(t = 4.5\) and \(t = 8\) | B1 | |
| M1 | Attempting to use area to find total distance travelled | |
| \(\frac{1}{2} \times (4.5 + 2) \times 10 + \frac{1}{2} \times (8 - 4.5) \times 15 + \frac{1}{2} \times (T - 8) \times 15 = 100\) | M1 | For setting up an equation for total distance travelled and solving for \(T\) |
| \(T = 13.5\) | A1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration $= \dfrac{(-25)}{2.5} = -10$ m s$^{-2}$ | B1 | AG |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = -15 + 7.5 \times 4$ | M1 | Using $v$–$t$ graph OE |
| $V = 15$ m s$^{-1}$ | A1 | |
---
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $v = 0$ at $t = 4.5$ and $t = 8$ | B1 | |
| | M1 | Attempting to use area to find total distance travelled |
| $\frac{1}{2} \times (4.5 + 2) \times 10 + \frac{1}{2} \times (8 - 4.5) \times 15 + \frac{1}{2} \times (T - 8) \times 15 = 100$ | M1 | For setting up an equation for total distance travelled and solving for $T$ |
| $T = 13.5$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{db1b5f31-1a41-44dd-ae9a-0c67336997eb-05_600_1155_262_497}
The diagram shows the velocity-time graph of a particle which moves in a straight line. The graph consists of 5 straight line segments. The particle starts from rest at a point $A$ at time $t = 0$, and initially travels towards point $B$ on the line.\\
(i) Show that the acceleration of the particle between $t = 3.5$ and $t = 6$ is $- 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) The acceleration of the particle between $t = 6$ and $t = 10$ is $7.5 \mathrm {~ms} ^ { - 2 }$. When $t = 10$ the velocity of the particle is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the value of $V$.\\
(iii) The particle comes to rest at $B$ at time $T$ s. Given that the total distance travelled by the particle between $t = 0$ and $t = T$ is 100 m , find the value of $T$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q4 [7]}}