| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question requiring basic calculus (differentiation to find acceleration, finding maximum by setting dv/dt=0) and interpretation of velocity-time graphs (using areas for displacement). All techniques are standard M1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(v = -4.8\) | B1 | |
| \([\pm 4.8 = 3a]\) | M1 | For using \(v = 0 + at\) |
| Magnitude of acceleration is \(1.6 \text{ ms}^{-2}\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([-0.4t + 4 = 0\) when \(t = 10]\) | M1 | For finding value of \(t\) when \(\text{d}v/\text{d}t = 0\) |
| M1 | For evaluating \(v(10)\) as \(v_{\max}\) (graph excludes possibility of \(v(10)\) as \(v_{\min}\)) | |
| \(v_{\max} = -0.2 \times 100 + 4 \times 10 - 15 \rightarrow\) Maximum velocity is \(5 \text{ ms}^{-1}\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Distance 0 to \(3\text{ s} = \frac{1}{2} \times 3 \times 4.8\ (= 7.2)\) | B1 | |
| Distance 3 to \(5\text{ s} = -\int_3^5(-0.2t^2 + 4t - 15)\,\text{d}t\) | M1 | Attempt to integrate and use limits |
| Distance \(= \pm 4.5333\ldots\text{ m}\) | A1 | |
| Average speed \(= (7.2 + 4.533) \div 5 = 2.35 \text{ ms}^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distance \(BC = \left[-\dfrac{0.2t^3}{3} + 2t^2 - 15t\right]_5^{15}\) | M1 | ft for errors in coefficients in cubic expression |
| and Av speed \(= (AB + BC) \div 15\) | ||
| Av speed \(= (45.066 \div 15) = 3.00 \text{ ms}^{-1}\) | A1 | [6] |
## Question 7:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $v = -4.8$ | B1 | |
| $[\pm 4.8 = 3a]$ | M1 | For using $v = 0 + at$ |
| Magnitude of acceleration is $1.6 \text{ ms}^{-2}$ | A1 | 3 marks total |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $[-0.4t + 4 = 0$ when $t = 10]$ | M1 | For finding value of $t$ when $\text{d}v/\text{d}t = 0$ |
| | M1 | For evaluating $v(10)$ as $v_{\max}$ (graph excludes possibility of $v(10)$ as $v_{\min}$) |
| $v_{\max} = -0.2 \times 100 + 4 \times 10 - 15 \rightarrow$ Maximum velocity is $5 \text{ ms}^{-1}$ | A1 | 3 marks total |
### Part (iii)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| Distance 0 to $3\text{ s} = \frac{1}{2} \times 3 \times 4.8\ (= 7.2)$ | B1 | |
| Distance 3 to $5\text{ s} = -\int_3^5(-0.2t^2 + 4t - 15)\,\text{d}t$ | M1 | Attempt to integrate and use limits |
| Distance $= \pm 4.5333\ldots\text{ m}$ | A1 | |
| Average speed $= (7.2 + 4.533) \div 5 = 2.35 \text{ ms}^{-1}$ | B1 | |
## Question (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance $BC = \left[-\dfrac{0.2t^3}{3} + 2t^2 - 15t\right]_5^{15}$ | M1 | ft for errors in coefficients in cubic expression |
| and Av speed $= (AB + BC) \div 15$ | | |
| Av speed $= (45.066 \div 15) = 3.00 \text{ ms}^{-1}$ | A1 | **[6]** |7\\
\includegraphics[max width=\textwidth, alt={}, center]{c7133fc4-9a14-43fd-b5ed-788da72291cd-4_512_1351_998_397}
The diagram shows the velocity-time graph for the motion of a particle $P$ which moves on a straight line $B A C$. It starts at $A$ and travels to $B$ taking 5 s. It then reverses direction and travels from $B$ to $C$ taking 10 s . For the first 3 s of $P$ 's motion its acceleration is constant. For the remaining 12 s the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$ after leaving $A$, where
$$v = - 0.2 t ^ { 2 } + 4 t - 15 \text { for } 3 \leqslant t \leqslant 15$$
(i) Find the value of $v$ when $t = 3$ and the magnitude of the acceleration of $P$ for the first 3 s of its motion.\\
(ii) Find the maximum velocity of $P$ while it is moving from $B$ to $C$.\\
(iii) Find the average speed of $P$,
\begin{enumerate}[label=(\alph*)]
\item while moving from $A$ to $B$,
\item for the whole journey.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2014 Q7 [12]}}