| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Moderate -0.5 This appears to be a standard SUVAT kinematics problem involving constant acceleration, likely requiring application of basic equations of motion. The three-part structure suggests routine calculations (finding velocity, using equations of motion, finding time/distance) which are typical textbook exercises for M1 level, making it slightly easier than average but not trivial. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v(8) = 0.25 \times 8 = 2\) | B1 | |
| \(2 = -6.4 + 19.2 - k \Rightarrow k = 10.8\) | B1\(\checkmark\) | [2 marks] ft \((12.8 - v)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([dv/dt = -0.2t + 2.4 (= 0 \text{ when } t = 12)\); \(v_{\max} = -0.1 \times 144 + 2.4 \times 12 - 10.8]\) | M1 | For finding \(t\) when \(dv/dt = 0\) and substituting into \(v(t)\) |
| Maximum speed is \(3.6 \text{ ms}^{-1}\) | A1\(\checkmark\) | [2 marks] ft \((14.4 -\) incorrect \(k)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Displacement \(s_1 = \frac{1}{2} \times 0.25 \times 8^2\ (= 8)\) | B1 | |
| \(s_2 = \int_8^{18}(-0.1t^2 + 2.4t - 10.8)\,dt\) \(= [-0.1t^3/3 + 1.2t^2 - 10.8t]_8^{18}\ (= 26.7)\) | M1 | For using displacement \(s_2 = \int_8^{18}(-0.1t^2 + 2.4t - 10.8)\,dt\) |
| Displacement is \(34.7\) m | A1 | [3 marks] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v(8) = 0.25 \times 8 = 2$ | B1 | |
| $2 = -6.4 + 19.2 - k \Rightarrow k = 10.8$ | B1$\checkmark$ | **[2 marks]** ft $(12.8 - v)$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[dv/dt = -0.2t + 2.4 (= 0 \text{ when } t = 12)$; $v_{\max} = -0.1 \times 144 + 2.4 \times 12 - 10.8]$ | M1 | For finding $t$ when $dv/dt = 0$ and substituting into $v(t)$ |
| Maximum speed is $3.6 \text{ ms}^{-1}$ | A1$\checkmark$ | **[2 marks]** ft $(14.4 -$ incorrect $k)$ |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Displacement $s_1 = \frac{1}{2} \times 0.25 \times 8^2\ (= 8)$ | B1 | |
| $s_2 = \int_8^{18}(-0.1t^2 + 2.4t - 10.8)\,dt$ $= [-0.1t^3/3 + 1.2t^2 - 10.8t]_8^{18}\ (= 26.7)$ | M1 | For using displacement $s_2 = \int_8^{18}(-0.1t^2 + 2.4t - 10.8)\,dt$ |
| Displacement is $34.7$ m | A1 | **[3 marks]** |
---4 A particle $P$ starts from rest and moves in a straight line for 18 seconds. For the first 8 seconds of the motion $P$ has constant acceleration $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Subsequently $P$ 's velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t$ seconds after the motion started, is given by
$$v = - 0.1 t ^ { 2 } + 2.4 t - k ,$$
where $8 \leqslant t \leqslant 18$ and $k$ is a constant.\\
(i) Find the value of $v$ when $t = 8$ and hence find the value of $k$.\\
(ii) Find the maximum velocity of $P$.\\
(iii) Find the displacement of $P$ from its initial position when $t = 18$.
\hfill \mbox{\textit{CAIE M1 2014 Q4}}