| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Moderate -0.3 This is a straightforward SUVAT application requiring use of v² = u² + 2as to find distance ratios. Part (i) involves simple algebraic manipulation to eliminate acceleration, while part (ii) requires solving for the deceleration first then calculating a specific distance. The question is slightly easier than average as it follows a standard template with clear given values and requires only routine application of a single kinematic equation. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(12^2 = 20^2 - 2a \times AB\) and \(6^2 = 12^2 - 2a \times BC\) | M1 | Use \(v^2 = u^2 + 2(-a)s\) for \(AB\) or \(BC\), where \(a\) is the deceleration |
| \(AB = 128/a\) | A1 | |
| \(BC = 54/a\) | A1 | |
| \(AB : BC = 64:27\) | A1 | Allow equivalent unsimplified ratio |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = 20^2 - 2a \times 80 \rightarrow a = 2.5\) | M1 | Use \(v^2 = u^2 + 2(-a)AD\) to find \(a\) |
| \(BC = 54/2.5\) | M1 | Use \(a\) to find \(BC\) |
| \(BC = 21.6\) m | A1 | |
| Total: | 3 |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $12^2 = 20^2 - 2a \times AB$ and $6^2 = 12^2 - 2a \times BC$ | M1 | Use $v^2 = u^2 + 2(-a)s$ for $AB$ or $BC$, where $a$ is the deceleration |
| $AB = 128/a$ | A1 | |
| $BC = 54/a$ | A1 | |
| $AB : BC = 64:27$ | A1 | Allow equivalent unsimplified ratio |
| **Total: 4** | | |
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = 20^2 - 2a \times 80 \rightarrow a = 2.5$ | M1 | Use $v^2 = u^2 + 2(-a)AD$ to find $a$ |
| $BC = 54/2.5$ | M1 | Use $a$ to find $BC$ |
| $BC = 21.6$ m | A1 | |
| **Total:** | **3** | |
---5 A particle $P$ moves in a straight line $A B C D$ with constant deceleration. The velocities of $P$ at $A , B$ and $C$ are $20 \mathrm {~m} \mathrm {~s} ^ { - 1 } , 12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively.\\
(i) Find the ratio of distances $A B : B C$.\\
(ii) The particle comes to rest at $D$. Given that the distance $A D$ is 80 m , find the distance $B C$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q5 [7]}}