| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.3 This question requires understanding that distance is the area under a velocity-time graph and applying trapezium rule estimation. While it involves multiple steps and reading from a graph, the concepts are standard M1 material with straightforward arithmetic—slightly easier than average due to the guided nature and basic geometric calculations involved. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(20 \times 7\) or \(140\) and \(\frac{1}{2} \times 4 \times 7\) or \(14\) | B1 | |
| Valid argument that \(s_1 + s_2 > 154\) (AG) | B1 | Total: 2 |
| *Alternatively:* Approx distance is \(20 \times 7 + 4 \times 7k\) (where \(\frac{1}{2} < k < 1\)) | M1 | |
| Whose value (shown) is (clearly) \(> 154\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using area property with correct signs \([140 + 20 - \frac{1}{2} \times 10 \times 8]\) | M1 | Note: \(140 + 20 + 20 - 20\) scores M0 |
| Distance is \(120\text{ m}\) | A1 | Total: 2 |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $20 \times 7$ or $140$ and $\frac{1}{2} \times 4 \times 7$ or $14$ | B1 | |
| Valid argument that $s_1 + s_2 > 154$ (AG) | B1 | **Total: 2** |
| *Alternatively:* Approx distance is $20 \times 7 + 4 \times 7k$ (where $\frac{1}{2} < k < 1$) | M1 | |
| Whose value (shown) is (clearly) $> 154$ | A1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using area property with correct signs $[140 + 20 - \frac{1}{2} \times 10 \times 8]$ | M1 | Note: $140 + 20 + 20 - 20$ scores M0 |
| Distance is $120\text{ m}$ | A1 | **Total: 2** |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-2_649_1244_482_452}
A man runs in a straight line. He passes through a fixed point $A$ with constant velocity $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t = 0$. At time $t \mathrm {~s}$ his velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The diagram shows the graph of $v$ against $t$ for the period $0 \leqslant t \leqslant 40$.\\
(i) Show that the man runs more than 154 m in the first 24 s .\\
(ii) Given that the man runs 20 m in the interval $20 \leqslant t \leqslant 24$, find how far he is from $A$ when $t = 40$.
\hfill \mbox{\textit{CAIE M1 2002 Q2 [4]}}