| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward multi-stage SUVAT problem with constant accelerations in each phase. Students simply integrate given accelerations to find velocities, then integrate velocities (or use areas under the v-t graph) to find distance. All values are provided, requiring only systematic application of standard mechanics formulas with no problem-solving insight or geometric reasoning needed. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(v(4) = 0.75x4\) \(v(54) = v(4)\) and \(v(60) = v(54) - 0.5(60 - 54)\) Velocity is \(3 \text{ ms}^{-1}\) when \(t = 4\) and \(0\) when \(t = 60\) | B1, B1, B1, M1 | Graph consists of 3 straight line segments with 1st and 3rd having +ve and −ve slopes respectively; \(v\) is single valued and continuous throughout, and \(v(0) = 0\), ft incorrect value(s) for \(v(4)\) and \(v(60)\) |
| (ii) \([XY = \frac{1}{2}(60 + 50)x3\) or \(XY = \frac{1}{2}x0.75x4^2 + 3x50 - \frac{1}{2}x0.5x6^2]\) Distance is \(165 \text{ m}\) | M1, A1 | For using area property for distance or for \(s_1 = \frac{1}{2}a_1t_1^2\), \(s_2 = u_2t_2\), \(s_3 = \frac{1}{2}a_3t_3^2\) and \(XY = s_1 + s_2 - s_3\) |
(i) $v(4) = 0.75x4$ $v(54) = v(4)$ and $v(60) = v(54) - 0.5(60 - 54)$ Velocity is $3 \text{ ms}^{-1}$ when $t = 4$ and $0$ when $t = 60$ | B1, B1, B1, M1 | Graph consists of 3 straight line segments with 1st and 3rd having +ve and −ve slopes respectively; $v$ is single valued and continuous throughout, and $v(0) = 0$, ft incorrect value(s) for $v(4)$ and $v(60)$ | A1 ft | [5]
(ii) $[XY = \frac{1}{2}(60 + 50)x3$ or $XY = \frac{1}{2}x0.75x4^2 + 3x50 - \frac{1}{2}x0.5x6^2]$ Distance is $165 \text{ m}$ | M1, A1 | For using area property for distance or for $s_1 = \frac{1}{2}a_1t_1^2$, $s_2 = u_2t_2$, $s_3 = \frac{1}{2}a_3t_3^2$ and $XY = s_1 + s_2 - s_3$ | [2]
---4 A particle starts from rest at a point $X$ and moves in a straight line until, 60 seconds later, it reaches a point $Y$. At time $t \mathrm {~s}$ after leaving $X$, the acceleration of the particle is
$$\begin{array} { r c c }
0.75 \mathrm {~m} \mathrm {~s} ^ { - 2 } & \text { for } & 0 < t < 4 \\
0 \mathrm {~m} \mathrm {~s} ^ { - 2 } & \text { for } & 4 < t < 54 \\
- 0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 } & \text { for } & 54 < t < 60
\end{array}$$
(i) Find the velocity of the particle when $t = 4$ and when $t = 60$, and sketch the velocity-time graph.\\
(ii) Find the distance $X Y$.
\hfill \mbox{\textit{CAIE M1 2010 Q4 [7]}}