| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Displacement expressions and comparison |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring basic calculus (differentiation to find acceleration) and area under velocity-time graphs. Part (i) involves equating areas under two graphs (one piecewise linear, one quadratic), which is routine. Part (ii) requires differentiating v = 3t - 0.3t² to get acceleration and equating to the constant acceleration from the graph. Both parts use standard M1 techniques 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 area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | For using \(s_Q = \int v_Q\, dt\) | |
| \(s_Q = 1.5t^2 - 0.1t^3\ (+C)\) | A1 | |
| M1 | For using limits 0 to 10 or equivalent (or 0 to 5 if candidate states or implies \(v_Q\) is symmetric about \(t = 5\)) | |
| \(s_Q(10) = 50\) (or \(s_Q(5) = 25\)) | A1ft | May be implied in subsequent working |
| M1 | For using \(\frac{1}{2}\cdot 10 v_{\max} = s_Q(10)\) (or \(\frac{1}{2}\cdot 5 v_{\max} = s_Q(5)\)) | |
| Greatest velocity is \(10\ \text{ms}^{-1}\) | A1 | AG; Special case (max 1 mark): \(5v = 50 \Rightarrow v = 10\) → B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(a_P = 10/5\) | B1 | |
| \([3 - 0.6t = 2]\) | M1 | For differentiating to find \(a_Q(t)\) and equating to \(a_P\) |
| \(t = 1.67\) (or \(1\frac{2}{3}\)) | A1 |
## Question 6:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For using $s_Q = \int v_Q\, dt$ |
| $s_Q = 1.5t^2 - 0.1t^3\ (+C)$ | A1 | |
| | M1 | For using limits 0 to 10 or equivalent (or 0 to 5 if candidate states or implies $v_Q$ is symmetric about $t = 5$) |
| $s_Q(10) = 50$ (or $s_Q(5) = 25$) | A1ft | May be implied in subsequent working |
| | M1 | For using $\frac{1}{2}\cdot 10 v_{\max} = s_Q(10)$ (or $\frac{1}{2}\cdot 5 v_{\max} = s_Q(5)$) |
| Greatest velocity is $10\ \text{ms}^{-1}$ | A1 | AG; Special case (max 1 mark): $5v = 50 \Rightarrow v = 10$ → B1 |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $a_P = 10/5$ | B1 | |
| $[3 - 0.6t = 2]$ | M1 | For differentiating to find $a_Q(t)$ and equating to $a_P$ |
| $t = 1.67$ (or $1\frac{2}{3}$) | A1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-4_593_746_269_701}
A particle $P$ starts from rest at the point $A$ and travels in a straight line, coming to rest again after 10 s . The velocity-time graph for $P$ consists of two straight line segments (see diagram). A particle $Q$ starts from rest at $A$ at the same instant as $P$ and travels along the same straight line as $P$. The velocity of $Q$ is given by $v = 3 t - 0.3 t ^ { 2 }$ for $0 \leqslant t \leqslant 10$. The displacements from $A$ of $P$ and $Q$ are the same when $t = 10$.\\
(i) Show that the greatest velocity of $P$ during its motion is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the value of $t$, in the interval $0 < t < 5$, for which the acceleration of $Q$ is the same as the acceleration of $P$.
\hfill \mbox{\textit{CAIE M1 2007 Q6 [9]}}