| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Standard +0.3 This is a standard SUVAT problem requiring students to work with a trapezoid velocity-time graph, set up algebraic expressions involving T, and solve for distances. While it has multiple parts and requires careful algebraic manipulation, it follows a very familiar pattern for M1 students with no novel insights needed—slightly easier than average. |
| 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, right-hand steeper than left-hand slope | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Deceleration \(0.5T\) | B1 | May be implied |
| Constant speed \(180 - 1.5T\) | B1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5[180 + (180 - 1.5T)] \times 25 = 3300\) | M1 | Uses area property |
| \(T = 64\) | A1 | |
| Distance decelerating \(= [0.5 \times 32 \times 25 =]\ 400\) m | B1 | |
| Total: 3 |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Trapezium, right-hand steeper than left-hand slope | B1 | |
| **Total: 1** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Deceleration $0.5T$ | B1 | May be implied |
| Constant speed $180 - 1.5T$ | B1 | |
| **Total: 2** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5[180 + (180 - 1.5T)] \times 25 = 3300$ | M1 | Uses area property |
| $T = 64$ | A1 | |
| Distance decelerating $= [0.5 \times 32 \times 25 =]\ 400$ m | B1 | |
| **Total: 3** | | |
---3 A train travels between two stations, $A$ and $B$. The train starts from rest at $A$ and accelerates at a constant rate for $T$ s until it reaches a speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It then travels at this constant speed before decelerating at a constant rate, coming to rest at $B$. The magnitude of the train's deceleration is twice the magnitude of its acceleration. The total time taken for the journey is 180 s .\\
(i) Sketch the velocity-time graph for the train's journey from $A$ to $B$.\\
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(ii) Find an expression, in terms of $T$, for the length of time for which the train is travelling with constant speed.\\
(iii) The distance from $A$ to $B$ is 3300 m . Find how far the train travels while it is decelerating.\\
\hfill \mbox{\textit{CAIE M1 2017 Q3 [6]}}