| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| 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 velocity-time graph question requiring basic area calculations (trapezium rule) and SUVAT application. Part (i) involves simple area under graph, part (ii) uses total distance constraint with clear setup, and part (iii) applies standard deceleration formula. All steps are routine M1 techniques with no problem-solving insight required, making it easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2} \times 6 \times 8.2 + 36 \times 8.2\) or \(\frac{1}{2} \times 8.2 \times (36 + 42)\) | M1 | For using distance = total area under graph |
| Distance \(= 319.8\) m | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = 80.2\) | B1\(\checkmark\) | Distance from \(t = 42\) to \(t = 52\) |
| \(80.2 = \frac{8.2 + V}{2} \times 10\) | M1 | For equating remaining distance to total area under graph between \(t = 42\) and \(t = 52\) |
| \(V = 7.84\) | A1 | AG. Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| M1 | Use gradient property for deceleration | |
| \(d = \frac{8.2 - 7.84}{10} = 0.036\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(80.2 = 8.2 \times 10 + \frac{1}{2}a \times 10^2\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) between \(t = 42\) and \(t = 52\) |
| \(a = -0.036\ \text{ms}^{-2}\) or \(d = 0.036\ \text{ms}^{-2}\) | A1 | Total: 2 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2} \times 6 \times 8.2 + 36 \times 8.2$ or $\frac{1}{2} \times 8.2 \times (36 + 42)$ | M1 | For using distance = total area under graph |
| Distance $= 319.8$ m | A1 | **Total: 2** |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = 80.2$ | B1$\checkmark$ | Distance from $t = 42$ to $t = 52$ |
| $80.2 = \frac{8.2 + V}{2} \times 10$ | M1 | For equating remaining distance to total area under graph between $t = 42$ and $t = 52$ |
| $V = 7.84$ | A1 | AG. **Total: 3** |
## Question 4(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| | M1 | Use gradient property for deceleration |
| $d = \frac{8.2 - 7.84}{10} = 0.036$ | A1 | **Total: 2** |
**Alternative for 4(iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $80.2 = 8.2 \times 10 + \frac{1}{2}a \times 10^2$ | M1 | For using $s = ut + \frac{1}{2}at^2$ between $t = 42$ and $t = 52$ |
| $a = -0.036\ \text{ms}^{-2}$ or $d = 0.036\ \text{ms}^{-2}$ | A1 | **Total: 2** |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{fd2fbf13-912c-46c5-a470-306b2269aa0b-2_522_959_1692_593}
A sprinter runs a race of 400 m . His total time for running the race is 52 s . The diagram shows the velocity-time graph for the motion of the sprinter. He starts from rest and accelerates uniformly to a speed of $8.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 6 s . The sprinter maintains a speed of $8.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for 36 s , and he then decelerates uniformly to a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the end of the race.\\
(i) Calculate the distance covered by the sprinter in the first 42 s of the race.\\
(ii) Show that $V = 7.84$.\\
(iii) Calculate the deceleration of the sprinter in the last 10 s of the race.
\hfill \mbox{\textit{CAIE M1 2016 Q4 [7]}}